Convert Fahrenheit to Celsius
I need some help for an web app I make for a school pjoject.
I want to convert the temperature from Fahrenheit to celsius.
Now I wrote this code:
$("#desc").html('Sky: ' + data.weather[0].description);
var fahr = data.main.temp + " °F";
var cels = ("fahr" - 32) * 5 / 9 + " °C";
$("#temp").html(data.main.temp + " °F");
$("#celButt").on("click", function()
$("#temp").html(cels););
$("#fahrButt").on("click", function()
$("#temp").html(fahr););
If I click te button to convert the data I get this as result:
NaN °C
Someone who can help me?
Thanks!!
javascript openweathermap converters
asked Nov 13 '18 at 10:51
Imo SetzImo Setz
31
add a comment |
I need some help for an web app I make for a school pjoject.
I want to convert the temperature from Fahrenheit to celsius.
Now I wrote this code:
$("#desc").html('Sky: ' + data.weather[0].description);
var fahr = data.main.temp + " °F";
var cels = ("fahr" - 32) * 5 / 9 + " °C";
$("#temp").html(data.main.temp + " °F");
$("#celButt").on("click", function()
$("#temp").html(cels););
$("#fahrButt").on("click", function()
$("#temp").html(fahr););
If I click te button to convert the data I get this as result:
NaN °C
Someone who can help me?
Thanks!!
javascript openweathermap converters
asked Nov 13 '18 at 10:51
Imo SetzImo Setz
31
2
Change("fahr" - 32) * 5 / 9 + " °C"to(data.main.temp - 32) * 5 / 9 + " °C". Don't know what you were trying to do by subtracting a number from a string!
– phuzi
Nov 13 '18 at 10:53
1
I don't understand why there is string("fahr" - 32)
– Ayaz Shah
Nov 13 '18 at 10:53
add a comment |
I need some help for an web app I make for a school pjoject.
I want to convert the temperature from Fahrenheit to celsius.
Now I wrote this code:
$("#desc").html('Sky: ' + data.weather[0].description);
var fahr = data.main.temp + " °F";
var cels = ("fahr" - 32) * 5 / 9 + " °C";
$("#temp").html(data.main.temp + " °F");
$("#celButt").on("click", function()
$("#temp").html(cels););
$("#fahrButt").on("click", function()
$("#temp").html(fahr););
If I click te button to convert the data I get this as result:
NaN °C
Someone who can help me?
Thanks!!
javascript openweathermap converters
asked Nov 13 '18 at 10:51
Imo SetzImo Setz
31
I need some help for an web app I make for a school pjoject.
I want to convert the temperature from Fahrenheit to celsius.
Now I wrote this code:
$("#desc").html('Sky: ' + data.weather[0].description);
var fahr = data.main.temp + " °F";
var cels = ("fahr" - 32) * 5 / 9 + " °C";
$("#temp").html(data.main.temp + " °F");
$("#celButt").on("click", function()
$("#temp").html(cels););
$("#fahrButt").on("click", function()
$("#temp").html(fahr););
If I click te button to convert the data I get this as result:
NaN °C
Someone who can help me?
Thanks!!
javascript openweathermap converters
javascript openweathermap converters
asked Nov 13 '18 at 10:51
Imo SetzImo Setz
31
asked Nov 13 '18 at 10:51
Imo SetzImo Setz
31
asked Nov 13 '18 at 10:51
Imo SetzImo Setz
31
asked Nov 13 '18 at 10:51
Imo SetzImo Setz
31
asked Nov 13 '18 at 10:51
Imo SetzImo Setz
31
31
2
Change("fahr" - 32) * 5 / 9 + " °C"to(data.main.temp - 32) * 5 / 9 + " °C". Don't know what you were trying to do by subtracting a number from a string!
– phuzi
Nov 13 '18 at 10:53
1
I don't understand why there is string("fahr" - 32)
– Ayaz Shah
Nov 13 '18 at 10:53
add a comment |
2
Change("fahr" - 32) * 5 / 9 + " °C"to(data.main.temp - 32) * 5 / 9 + " °C". Don't know what you were trying to do by subtracting a number from a string!
– phuzi
Nov 13 '18 at 10:53
1
I don't understand why there is string("fahr" - 32)
– Ayaz Shah
Nov 13 '18 at 10:53
2
2
Change
("fahr" - 32) * 5 / 9 + " °C" to (data.main.temp - 32) * 5 / 9 + " °C". Don't know what you were trying to do by subtracting a number from a string!– phuzi
Nov 13 '18 at 10:53
Change
("fahr" - 32) * 5 / 9 + " °C" to (data.main.temp - 32) * 5 / 9 + " °C". Don't know what you were trying to do by subtracting a number from a string!– phuzi
Nov 13 '18 at 10:53
1
1
I don't understand why there is string
("fahr" - 32)– Ayaz Shah
Nov 13 '18 at 10:53
I don't understand why there is string
("fahr" - 32)– Ayaz Shah
Nov 13 '18 at 10:53
add a comment |
2 Answers
2
active
oldest
votes
You cannot add or subtract or divide or multiply a string, which, by the time you try to calculate Celsius, is exactly what you are doing.
You are trying to subtract the number 32 from the string "80 °F" for example, which doesn't make sense.
Instead, you can do the following :
var fahr = data.main.temp + " °F";
var cels = (data.main.temp - 32) * 5 / 9 + " °C";
Also you may as well change :
$("#temp").html(data.main.temp + " °F");
To
$("#temp").html(fahr);
For increased readability and to keep things consistent.
answered Nov 13 '18 at 10:55
cmprogramcmprogram
1,173619
add a comment |
Instead of "fahr" try var cels = (data.main.temp - 32) * 5 / 9 + " °C";
answered Nov 13 '18 at 10:57
James EleJames Ele
1918
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You cannot add or subtract or divide or multiply a string, which, by the time you try to calculate Celsius, is exactly what you are doing.
You are trying to subtract the number 32 from the string "80 °F" for example, which doesn't make sense.
Instead, you can do the following :
var fahr = data.main.temp + " °F";
var cels = (data.main.temp - 32) * 5 / 9 + " °C";
Also you may as well change :
$("#temp").html(data.main.temp + " °F");
To
$("#temp").html(fahr);
For increased readability and to keep things consistent.
answered Nov 13 '18 at 10:55
cmprogramcmprogram
1,173619
add a comment |
You cannot add or subtract or divide or multiply a string, which, by the time you try to calculate Celsius, is exactly what you are doing.
You are trying to subtract the number 32 from the string "80 °F" for example, which doesn't make sense.
Instead, you can do the following :
var fahr = data.main.temp + " °F";
var cels = (data.main.temp - 32) * 5 / 9 + " °C";
Also you may as well change :
$("#temp").html(data.main.temp + " °F");
To
$("#temp").html(fahr);
For increased readability and to keep things consistent.
answered Nov 13 '18 at 10:55
cmprogramcmprogram
1,173619
add a comment |
You cannot add or subtract or divide or multiply a string, which, by the time you try to calculate Celsius, is exactly what you are doing.
You are trying to subtract the number 32 from the string "80 °F" for example, which doesn't make sense.
Instead, you can do the following :
var fahr = data.main.temp + " °F";
var cels = (data.main.temp - 32) * 5 / 9 + " °C";
Also you may as well change :
$("#temp").html(data.main.temp + " °F");
To
$("#temp").html(fahr);
For increased readability and to keep things consistent.
answered Nov 13 '18 at 10:55
cmprogramcmprogram
1,173619
You cannot add or subtract or divide or multiply a string, which, by the time you try to calculate Celsius, is exactly what you are doing.
You are trying to subtract the number 32 from the string "80 °F" for example, which doesn't make sense.
Instead, you can do the following :
var fahr = data.main.temp + " °F";
var cels = (data.main.temp - 32) * 5 / 9 + " °C";
Also you may as well change :
$("#temp").html(data.main.temp + " °F");
To
$("#temp").html(fahr);
For increased readability and to keep things consistent.
answered Nov 13 '18 at 10:55
cmprogramcmprogram
1,173619
edited Nov 13 '18 at 11:00
answered Nov 13 '18 at 10:55
cmprogramcmprogram
1,173619
answered Nov 13 '18 at 10:55
cmprogramcmprogram
1,173619
answered Nov 13 '18 at 10:55
cmprogramcmprogram
1,173619
1,173619
add a comment |
add a comment |
Instead of "fahr" try var cels = (data.main.temp - 32) * 5 / 9 + " °C";
answered Nov 13 '18 at 10:57
James EleJames Ele
1918
add a comment |
Instead of "fahr" try var cels = (data.main.temp - 32) * 5 / 9 + " °C";
answered Nov 13 '18 at 10:57
James EleJames Ele
1918
add a comment |
Instead of "fahr" try var cels = (data.main.temp - 32) * 5 / 9 + " °C";
answered Nov 13 '18 at 10:57
James EleJames Ele
1918
Instead of "fahr" try var cels = (data.main.temp - 32) * 5 / 9 + " °C";
answered Nov 13 '18 at 10:57
James EleJames Ele
1918
answered Nov 13 '18 at 10:57
James EleJames Ele
1918
answered Nov 13 '18 at 10:57
James EleJames Ele
1918
answered Nov 13 '18 at 10:57
James EleJames Ele
1918
1918
add a comment |
add a comment |
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2
Change
("fahr" - 32) * 5 / 9 + " °C"to(data.main.temp - 32) * 5 / 9 + " °C". Don't know what you were trying to do by subtracting a number from a string!– phuzi
Nov 13 '18 at 10:53
1
I don't understand why there is string
("fahr" - 32)– Ayaz Shah
Nov 13 '18 at 10:53