typeahead bootstrap selected value post









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I am using bootstrap typeahead to suggest values from database for a form field. Here is the type ahead code.



 $(document).ready(function()

$('#client_id').typeahead(
source: function(query, result)

$.ajax(
url:'client_search.php?extraParams=<?php echo $merchant_id;?>&',
method:"GET",
data:query:query,
dataType:"json",
success:function(data)

console.log(data);
result($.map(data, function(item)
return item;

));


)

)


);


The client_search.php file returns two values in json format -



$clientdata = array();
if($query->num_rows > 0)
while($row = $query->fetch_assoc())
$json['id'] = $row['id'];
$json['name'] = $row['name'];
array_push($clientdata, $data);



// Return results as json encoded array
echo json_encode($json);


Everything is working fine except that on form submit i want to post the ID field of the client instead of name.
Any help appreciated :)










share|improve this question

























    up vote
    0
    down vote

    favorite












    I am using bootstrap typeahead to suggest values from database for a form field. Here is the type ahead code.



     $(document).ready(function()

    $('#client_id').typeahead(
    source: function(query, result)

    $.ajax(
    url:'client_search.php?extraParams=<?php echo $merchant_id;?>&',
    method:"GET",
    data:query:query,
    dataType:"json",
    success:function(data)

    console.log(data);
    result($.map(data, function(item)
    return item;

    ));


    )

    )


    );


    The client_search.php file returns two values in json format -



    $clientdata = array();
    if($query->num_rows > 0)
    while($row = $query->fetch_assoc())
    $json['id'] = $row['id'];
    $json['name'] = $row['name'];
    array_push($clientdata, $data);



    // Return results as json encoded array
    echo json_encode($json);


    Everything is working fine except that on form submit i want to post the ID field of the client instead of name.
    Any help appreciated :)










    share|improve this question























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I am using bootstrap typeahead to suggest values from database for a form field. Here is the type ahead code.



       $(document).ready(function()

      $('#client_id').typeahead(
      source: function(query, result)

      $.ajax(
      url:'client_search.php?extraParams=<?php echo $merchant_id;?>&',
      method:"GET",
      data:query:query,
      dataType:"json",
      success:function(data)

      console.log(data);
      result($.map(data, function(item)
      return item;

      ));


      )

      )


      );


      The client_search.php file returns two values in json format -



      $clientdata = array();
      if($query->num_rows > 0)
      while($row = $query->fetch_assoc())
      $json['id'] = $row['id'];
      $json['name'] = $row['name'];
      array_push($clientdata, $data);



      // Return results as json encoded array
      echo json_encode($json);


      Everything is working fine except that on form submit i want to post the ID field of the client instead of name.
      Any help appreciated :)










      share|improve this question













      I am using bootstrap typeahead to suggest values from database for a form field. Here is the type ahead code.



       $(document).ready(function()

      $('#client_id').typeahead(
      source: function(query, result)

      $.ajax(
      url:'client_search.php?extraParams=<?php echo $merchant_id;?>&',
      method:"GET",
      data:query:query,
      dataType:"json",
      success:function(data)

      console.log(data);
      result($.map(data, function(item)
      return item;

      ));


      )

      )


      );


      The client_search.php file returns two values in json format -



      $clientdata = array();
      if($query->num_rows > 0)
      while($row = $query->fetch_assoc())
      $json['id'] = $row['id'];
      $json['name'] = $row['name'];
      array_push($clientdata, $data);



      // Return results as json encoded array
      echo json_encode($json);


      Everything is working fine except that on form submit i want to post the ID field of the client instead of name.
      Any help appreciated :)







      twitter-bootstrap autocomplete typeahead






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      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 9 at 12:52









      Sohail Ahmed

      61




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