Problem getting results with php and mysqli
I've a problem by running this php script:
<?php
$link = mysqli_connect("localhost", "root", "*******", "adsb");
/* check connection */
if ($conn->connect_error)
die("Connection failed: " . $link->connect_error);
$query = "SELECT aircrafts.id, heli.reg, heli.hex, heli.typ, heli.opp, aircrafts.flight, aircrafts.altitude, aircrafts.lat, aircrafts.lon, aircrafts.squawk, aircrafts.message_date FROM `aircrafts` JOIN `heli` ON aircrafts.hex=heli.hex WHERE aircrafts.id='2414';";
$query .= "SELECT aircrafts.id, plane.reg, plane.hex, plane.typ, plane.opp, aircrafts.flight, aircrafts.altitude, aircrafts.lat, aircrafts.lon, aircrafts.squawk, aircrafts.message_date FROM `aircrafts` JOIN `plane` ON aircrafts.hex=plane.hex WHERE aircrafts.id='2414'";
$result = mysqli_multi_query($query);
/* execute multi query */
if ($result->num_rows > 0)
echo "<h1>INFO ABOUT FLIGHT RECORD " . $id . "</h1>";
echo "<table><th>Registratie</th><th>ICAO24</th><th>Type</th><th>Operator</th><th>Callsign</th><th>Squawk</th><th>Time</th></tr>";
while ($row = $result->fetch_assoc())
echo "<tr><td><a href='aircraft.php?hex=" . $row["hex"] . "'>" . $row["reg"] . "</a></td><td>" . $row["hex"] . "</td><td><img src='/database/SilhouttesLogos/" . $row["typ"] . ".bmp' /></td><td><img src='/database/OperatorFlags/" . $row["opp"] . ".bmp' /></td><td>" . $row["flight"] . "</td><td>" . $row["squawk"] . "</td><td>" . $row["message_date"] . "</td></tr>";
echo "</table>";
else
echo "0 results";
/* close connection */
mysqli_close($link);
When I run the 2 query's in phpmyadmin I get 1 result.
(When I run 1 query on my site I get a result too, see code at the bottom of this post)
But when I run it on my site it's shows "0 results".
So... There must be a fould in the $result section.
Who can help? :-)
<?php
$servername = "localhost";
$username = "root";
$password = "*****";
$dbname = "adsb";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error)
die("Connection failed: " . $conn->connect_error);
$sql = "SELECT aircrafts.id, heli.reg, heli.hex, heli.typ, heli.opp, aircrafts.flight, aircrafts.altitude, aircrafts.lat, aircrafts.lon, aircrafts.squawk, aircrafts.message_date FROM `aircrafts` JOIN `heli` ON aircrafts.hex=heli.hex WHERE aircrafts.id=2414";
$result = $conn->query($sql);
if ($result->num_rows > 0)
echo "<table><th>Registratie</th><th>ICAO24</th><th>Type</th><th>Operator</th><th>Callsign</th><th>Squawk</th><th>Time</th></tr>";
while ($row = $result->fetch_assoc())
echo "<tr><td><a href='aircraft.php?hex=" . $row["hex"] . "'>" . $row["reg"] . "</a></td><td>" . $row["hex"] . "</td><td><img src='/database/SilhouttesLogos/" . $row["typ"] . ".bmp' /></td><td><img src='/database/OperatorFlags/" . $row["opp"] . ".bmp' /></td><td>" . $row["flight"] . "</td><td>" . $row["squawk"] . "</td><td>" . $row["message_date"] . "</td></tr>";
else
echo "0 results";
$conn->close();
php mysql mysqli
|
show 1 more comment
I've a problem by running this php script:
<?php
$link = mysqli_connect("localhost", "root", "*******", "adsb");
/* check connection */
if ($conn->connect_error)
die("Connection failed: " . $link->connect_error);
$query = "SELECT aircrafts.id, heli.reg, heli.hex, heli.typ, heli.opp, aircrafts.flight, aircrafts.altitude, aircrafts.lat, aircrafts.lon, aircrafts.squawk, aircrafts.message_date FROM `aircrafts` JOIN `heli` ON aircrafts.hex=heli.hex WHERE aircrafts.id='2414';";
$query .= "SELECT aircrafts.id, plane.reg, plane.hex, plane.typ, plane.opp, aircrafts.flight, aircrafts.altitude, aircrafts.lat, aircrafts.lon, aircrafts.squawk, aircrafts.message_date FROM `aircrafts` JOIN `plane` ON aircrafts.hex=plane.hex WHERE aircrafts.id='2414'";
$result = mysqli_multi_query($query);
/* execute multi query */
if ($result->num_rows > 0)
echo "<h1>INFO ABOUT FLIGHT RECORD " . $id . "</h1>";
echo "<table><th>Registratie</th><th>ICAO24</th><th>Type</th><th>Operator</th><th>Callsign</th><th>Squawk</th><th>Time</th></tr>";
while ($row = $result->fetch_assoc())
echo "<tr><td><a href='aircraft.php?hex=" . $row["hex"] . "'>" . $row["reg"] . "</a></td><td>" . $row["hex"] . "</td><td><img src='/database/SilhouttesLogos/" . $row["typ"] . ".bmp' /></td><td><img src='/database/OperatorFlags/" . $row["opp"] . ".bmp' /></td><td>" . $row["flight"] . "</td><td>" . $row["squawk"] . "</td><td>" . $row["message_date"] . "</td></tr>";
echo "</table>";
else
echo "0 results";
/* close connection */
mysqli_close($link);
When I run the 2 query's in phpmyadmin I get 1 result.
(When I run 1 query on my site I get a result too, see code at the bottom of this post)
But when I run it on my site it's shows "0 results".
So... There must be a fould in the $result section.
Who can help? :-)
<?php
$servername = "localhost";
$username = "root";
$password = "*****";
$dbname = "adsb";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error)
die("Connection failed: " . $conn->connect_error);
$sql = "SELECT aircrafts.id, heli.reg, heli.hex, heli.typ, heli.opp, aircrafts.flight, aircrafts.altitude, aircrafts.lat, aircrafts.lon, aircrafts.squawk, aircrafts.message_date FROM `aircrafts` JOIN `heli` ON aircrafts.hex=heli.hex WHERE aircrafts.id=2414";
$result = $conn->query($sql);
if ($result->num_rows > 0)
echo "<table><th>Registratie</th><th>ICAO24</th><th>Type</th><th>Operator</th><th>Callsign</th><th>Squawk</th><th>Time</th></tr>";
while ($row = $result->fetch_assoc())
echo "<tr><td><a href='aircraft.php?hex=" . $row["hex"] . "'>" . $row["reg"] . "</a></td><td>" . $row["hex"] . "</td><td><img src='/database/SilhouttesLogos/" . $row["typ"] . ".bmp' /></td><td><img src='/database/OperatorFlags/" . $row["opp"] . ".bmp' /></td><td>" . $row["flight"] . "</td><td>" . $row["squawk"] . "</td><td>" . $row["message_date"] . "</td></tr>";
else
echo "0 results";
$conn->close();
php mysql mysqli
WARNING: It's important that you DO NOT usemysqli_multi_query. That function does not support placeholder values, something critical to securing your application. It's also extremely important to verify that each statement succeeds before proceeding to the next, slamming in a bunch of queries and ignoring errors is how you create huge problems.
– tadman
Nov 14 '18 at 20:21
1
You also can't chain twoSELECTqueries together without fetching data from each in turn. What you probably mean to do isJOINtogether the two tables into a single query you then fetch.
– tadman
Nov 14 '18 at 20:22
Note: The object-oriented interface tomysqliis significantly less verbose, making code easier to read and audit, and is not easily confused with the obsoletemysql_queryinterface. Before you get too invested in the procedural style it’s worth switching over. Example:$db = new mysqli(…)and$db->prepare("…")The procedural interface is an artifact from the PHP 4 era whenmysqliAPI was introduced and should not be used in new code. You're mixing and matching styles here for no apparent reason.
– tadman
Nov 14 '18 at 20:22
2
Plus,$result = mysqli_multi_query($query);the connection was never made.
– Funk Forty Niner
Nov 14 '18 at 20:23
Check your error log as this code should be producing multiple errors due to incorrect arguments, as Funky Forty Niner points out.
– tadman
Nov 14 '18 at 20:23
|
show 1 more comment
I've a problem by running this php script:
<?php
$link = mysqli_connect("localhost", "root", "*******", "adsb");
/* check connection */
if ($conn->connect_error)
die("Connection failed: " . $link->connect_error);
$query = "SELECT aircrafts.id, heli.reg, heli.hex, heli.typ, heli.opp, aircrafts.flight, aircrafts.altitude, aircrafts.lat, aircrafts.lon, aircrafts.squawk, aircrafts.message_date FROM `aircrafts` JOIN `heli` ON aircrafts.hex=heli.hex WHERE aircrafts.id='2414';";
$query .= "SELECT aircrafts.id, plane.reg, plane.hex, plane.typ, plane.opp, aircrafts.flight, aircrafts.altitude, aircrafts.lat, aircrafts.lon, aircrafts.squawk, aircrafts.message_date FROM `aircrafts` JOIN `plane` ON aircrafts.hex=plane.hex WHERE aircrafts.id='2414'";
$result = mysqli_multi_query($query);
/* execute multi query */
if ($result->num_rows > 0)
echo "<h1>INFO ABOUT FLIGHT RECORD " . $id . "</h1>";
echo "<table><th>Registratie</th><th>ICAO24</th><th>Type</th><th>Operator</th><th>Callsign</th><th>Squawk</th><th>Time</th></tr>";
while ($row = $result->fetch_assoc())
echo "<tr><td><a href='aircraft.php?hex=" . $row["hex"] . "'>" . $row["reg"] . "</a></td><td>" . $row["hex"] . "</td><td><img src='/database/SilhouttesLogos/" . $row["typ"] . ".bmp' /></td><td><img src='/database/OperatorFlags/" . $row["opp"] . ".bmp' /></td><td>" . $row["flight"] . "</td><td>" . $row["squawk"] . "</td><td>" . $row["message_date"] . "</td></tr>";
echo "</table>";
else
echo "0 results";
/* close connection */
mysqli_close($link);
When I run the 2 query's in phpmyadmin I get 1 result.
(When I run 1 query on my site I get a result too, see code at the bottom of this post)
But when I run it on my site it's shows "0 results".
So... There must be a fould in the $result section.
Who can help? :-)
<?php
$servername = "localhost";
$username = "root";
$password = "*****";
$dbname = "adsb";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error)
die("Connection failed: " . $conn->connect_error);
$sql = "SELECT aircrafts.id, heli.reg, heli.hex, heli.typ, heli.opp, aircrafts.flight, aircrafts.altitude, aircrafts.lat, aircrafts.lon, aircrafts.squawk, aircrafts.message_date FROM `aircrafts` JOIN `heli` ON aircrafts.hex=heli.hex WHERE aircrafts.id=2414";
$result = $conn->query($sql);
if ($result->num_rows > 0)
echo "<table><th>Registratie</th><th>ICAO24</th><th>Type</th><th>Operator</th><th>Callsign</th><th>Squawk</th><th>Time</th></tr>";
while ($row = $result->fetch_assoc())
echo "<tr><td><a href='aircraft.php?hex=" . $row["hex"] . "'>" . $row["reg"] . "</a></td><td>" . $row["hex"] . "</td><td><img src='/database/SilhouttesLogos/" . $row["typ"] . ".bmp' /></td><td><img src='/database/OperatorFlags/" . $row["opp"] . ".bmp' /></td><td>" . $row["flight"] . "</td><td>" . $row["squawk"] . "</td><td>" . $row["message_date"] . "</td></tr>";
else
echo "0 results";
$conn->close();
php mysql mysqli
I've a problem by running this php script:
<?php
$link = mysqli_connect("localhost", "root", "*******", "adsb");
/* check connection */
if ($conn->connect_error)
die("Connection failed: " . $link->connect_error);
$query = "SELECT aircrafts.id, heli.reg, heli.hex, heli.typ, heli.opp, aircrafts.flight, aircrafts.altitude, aircrafts.lat, aircrafts.lon, aircrafts.squawk, aircrafts.message_date FROM `aircrafts` JOIN `heli` ON aircrafts.hex=heli.hex WHERE aircrafts.id='2414';";
$query .= "SELECT aircrafts.id, plane.reg, plane.hex, plane.typ, plane.opp, aircrafts.flight, aircrafts.altitude, aircrafts.lat, aircrafts.lon, aircrafts.squawk, aircrafts.message_date FROM `aircrafts` JOIN `plane` ON aircrafts.hex=plane.hex WHERE aircrafts.id='2414'";
$result = mysqli_multi_query($query);
/* execute multi query */
if ($result->num_rows > 0)
echo "<h1>INFO ABOUT FLIGHT RECORD " . $id . "</h1>";
echo "<table><th>Registratie</th><th>ICAO24</th><th>Type</th><th>Operator</th><th>Callsign</th><th>Squawk</th><th>Time</th></tr>";
while ($row = $result->fetch_assoc())
echo "<tr><td><a href='aircraft.php?hex=" . $row["hex"] . "'>" . $row["reg"] . "</a></td><td>" . $row["hex"] . "</td><td><img src='/database/SilhouttesLogos/" . $row["typ"] . ".bmp' /></td><td><img src='/database/OperatorFlags/" . $row["opp"] . ".bmp' /></td><td>" . $row["flight"] . "</td><td>" . $row["squawk"] . "</td><td>" . $row["message_date"] . "</td></tr>";
echo "</table>";
else
echo "0 results";
/* close connection */
mysqli_close($link);
When I run the 2 query's in phpmyadmin I get 1 result.
(When I run 1 query on my site I get a result too, see code at the bottom of this post)
But when I run it on my site it's shows "0 results".
So... There must be a fould in the $result section.
Who can help? :-)
<?php
$servername = "localhost";
$username = "root";
$password = "*****";
$dbname = "adsb";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error)
die("Connection failed: " . $conn->connect_error);
$sql = "SELECT aircrafts.id, heli.reg, heli.hex, heli.typ, heli.opp, aircrafts.flight, aircrafts.altitude, aircrafts.lat, aircrafts.lon, aircrafts.squawk, aircrafts.message_date FROM `aircrafts` JOIN `heli` ON aircrafts.hex=heli.hex WHERE aircrafts.id=2414";
$result = $conn->query($sql);
if ($result->num_rows > 0)
echo "<table><th>Registratie</th><th>ICAO24</th><th>Type</th><th>Operator</th><th>Callsign</th><th>Squawk</th><th>Time</th></tr>";
while ($row = $result->fetch_assoc())
echo "<tr><td><a href='aircraft.php?hex=" . $row["hex"] . "'>" . $row["reg"] . "</a></td><td>" . $row["hex"] . "</td><td><img src='/database/SilhouttesLogos/" . $row["typ"] . ".bmp' /></td><td><img src='/database/OperatorFlags/" . $row["opp"] . ".bmp' /></td><td>" . $row["flight"] . "</td><td>" . $row["squawk"] . "</td><td>" . $row["message_date"] . "</td></tr>";
else
echo "0 results";
$conn->close();
php mysql mysqli
php mysql mysqli
edited Nov 15 '18 at 1:36
userlond
2,42622232
2,42622232
asked Nov 14 '18 at 20:17
Darius123Darius123
1
1
WARNING: It's important that you DO NOT usemysqli_multi_query. That function does not support placeholder values, something critical to securing your application. It's also extremely important to verify that each statement succeeds before proceeding to the next, slamming in a bunch of queries and ignoring errors is how you create huge problems.
– tadman
Nov 14 '18 at 20:21
1
You also can't chain twoSELECTqueries together without fetching data from each in turn. What you probably mean to do isJOINtogether the two tables into a single query you then fetch.
– tadman
Nov 14 '18 at 20:22
Note: The object-oriented interface tomysqliis significantly less verbose, making code easier to read and audit, and is not easily confused with the obsoletemysql_queryinterface. Before you get too invested in the procedural style it’s worth switching over. Example:$db = new mysqli(…)and$db->prepare("…")The procedural interface is an artifact from the PHP 4 era whenmysqliAPI was introduced and should not be used in new code. You're mixing and matching styles here for no apparent reason.
– tadman
Nov 14 '18 at 20:22
2
Plus,$result = mysqli_multi_query($query);the connection was never made.
– Funk Forty Niner
Nov 14 '18 at 20:23
Check your error log as this code should be producing multiple errors due to incorrect arguments, as Funky Forty Niner points out.
– tadman
Nov 14 '18 at 20:23
|
show 1 more comment
WARNING: It's important that you DO NOT usemysqli_multi_query. That function does not support placeholder values, something critical to securing your application. It's also extremely important to verify that each statement succeeds before proceeding to the next, slamming in a bunch of queries and ignoring errors is how you create huge problems.
– tadman
Nov 14 '18 at 20:21
1
You also can't chain twoSELECTqueries together without fetching data from each in turn. What you probably mean to do isJOINtogether the two tables into a single query you then fetch.
– tadman
Nov 14 '18 at 20:22
Note: The object-oriented interface tomysqliis significantly less verbose, making code easier to read and audit, and is not easily confused with the obsoletemysql_queryinterface. Before you get too invested in the procedural style it’s worth switching over. Example:$db = new mysqli(…)and$db->prepare("…")The procedural interface is an artifact from the PHP 4 era whenmysqliAPI was introduced and should not be used in new code. You're mixing and matching styles here for no apparent reason.
– tadman
Nov 14 '18 at 20:22
2
Plus,$result = mysqli_multi_query($query);the connection was never made.
– Funk Forty Niner
Nov 14 '18 at 20:23
Check your error log as this code should be producing multiple errors due to incorrect arguments, as Funky Forty Niner points out.
– tadman
Nov 14 '18 at 20:23
WARNING: It's important that you DO NOT use
mysqli_multi_query. That function does not support placeholder values, something critical to securing your application. It's also extremely important to verify that each statement succeeds before proceeding to the next, slamming in a bunch of queries and ignoring errors is how you create huge problems.– tadman
Nov 14 '18 at 20:21
WARNING: It's important that you DO NOT use
mysqli_multi_query. That function does not support placeholder values, something critical to securing your application. It's also extremely important to verify that each statement succeeds before proceeding to the next, slamming in a bunch of queries and ignoring errors is how you create huge problems.– tadman
Nov 14 '18 at 20:21
1
1
You also can't chain two
SELECT queries together without fetching data from each in turn. What you probably mean to do is JOIN together the two tables into a single query you then fetch.– tadman
Nov 14 '18 at 20:22
You also can't chain two
SELECT queries together without fetching data from each in turn. What you probably mean to do is JOIN together the two tables into a single query you then fetch.– tadman
Nov 14 '18 at 20:22
Note: The object-oriented interface to
mysqli is significantly less verbose, making code easier to read and audit, and is not easily confused with the obsolete mysql_query interface. Before you get too invested in the procedural style it’s worth switching over. Example: $db = new mysqli(…) and $db->prepare("…") The procedural interface is an artifact from the PHP 4 era when mysqli API was introduced and should not be used in new code. You're mixing and matching styles here for no apparent reason.– tadman
Nov 14 '18 at 20:22
Note: The object-oriented interface to
mysqli is significantly less verbose, making code easier to read and audit, and is not easily confused with the obsolete mysql_query interface. Before you get too invested in the procedural style it’s worth switching over. Example: $db = new mysqli(…) and $db->prepare("…") The procedural interface is an artifact from the PHP 4 era when mysqli API was introduced and should not be used in new code. You're mixing and matching styles here for no apparent reason.– tadman
Nov 14 '18 at 20:22
2
2
Plus,
$result = mysqli_multi_query($query); the connection was never made.– Funk Forty Niner
Nov 14 '18 at 20:23
Plus,
$result = mysqli_multi_query($query); the connection was never made.– Funk Forty Niner
Nov 14 '18 at 20:23
Check your error log as this code should be producing multiple errors due to incorrect arguments, as Funky Forty Niner points out.
– tadman
Nov 14 '18 at 20:23
Check your error log as this code should be producing multiple errors due to incorrect arguments, as Funky Forty Niner points out.
– tadman
Nov 14 '18 at 20:23
|
show 1 more comment
1 Answer
1
active
oldest
votes
This is probably your specific problem.
$result = mysqli_multi_query($query);
Here is the PHP Doc
You need to establish the connection.
$result = mysqli_multi_query($link, $query);
Like the others mentioned in the comments, I would refactor this code.
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "1"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53308127%2fproblem-getting-results-with-php-and-mysqli%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
This is probably your specific problem.
$result = mysqli_multi_query($query);
Here is the PHP Doc
You need to establish the connection.
$result = mysqli_multi_query($link, $query);
Like the others mentioned in the comments, I would refactor this code.
add a comment |
This is probably your specific problem.
$result = mysqli_multi_query($query);
Here is the PHP Doc
You need to establish the connection.
$result = mysqli_multi_query($link, $query);
Like the others mentioned in the comments, I would refactor this code.
add a comment |
This is probably your specific problem.
$result = mysqli_multi_query($query);
Here is the PHP Doc
You need to establish the connection.
$result = mysqli_multi_query($link, $query);
Like the others mentioned in the comments, I would refactor this code.
This is probably your specific problem.
$result = mysqli_multi_query($query);
Here is the PHP Doc
You need to establish the connection.
$result = mysqli_multi_query($link, $query);
Like the others mentioned in the comments, I would refactor this code.
answered Nov 14 '18 at 20:30
Christopher JohnstonChristopher Johnston
967
967
add a comment |
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53308127%2fproblem-getting-results-with-php-and-mysqli%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
WARNING: It's important that you DO NOT use
mysqli_multi_query. That function does not support placeholder values, something critical to securing your application. It's also extremely important to verify that each statement succeeds before proceeding to the next, slamming in a bunch of queries and ignoring errors is how you create huge problems.– tadman
Nov 14 '18 at 20:21
1
You also can't chain two
SELECTqueries together without fetching data from each in turn. What you probably mean to do isJOINtogether the two tables into a single query you then fetch.– tadman
Nov 14 '18 at 20:22
Note: The object-oriented interface to
mysqliis significantly less verbose, making code easier to read and audit, and is not easily confused with the obsoletemysql_queryinterface. Before you get too invested in the procedural style it’s worth switching over. Example:$db = new mysqli(…)and$db->prepare("…")The procedural interface is an artifact from the PHP 4 era whenmysqliAPI was introduced and should not be used in new code. You're mixing and matching styles here for no apparent reason.– tadman
Nov 14 '18 at 20:22
2
Plus,
$result = mysqli_multi_query($query);the connection was never made.– Funk Forty Niner
Nov 14 '18 at 20:23
Check your error log as this code should be producing multiple errors due to incorrect arguments, as Funky Forty Niner points out.
– tadman
Nov 14 '18 at 20:23