while loop set only first value in input field in php

Question

I want to add users present in a given table. I am iterating whole table and sending each value to javascript file.

<?php 
          $sql = "select * from user where user_id not in (select second_user_id from friends where first_user_id = '$user_id'\n"
    . " union\n"
    . " select first_user_id from friends where second_user_id = '$user_id') limit 20";


              $result_not_friends=mysqli_query($dbc,$sql)
              or die("error in fetching");

            // print_r($row_not_friends);

           ?>
           <table class="table table-hover table-bordered">
            <h1>Users</h1>
              <tbody>
                <?php 
                  while ( $row_not_friends = mysqli_fetch_array($result_not_friends))
                  {
                    if ( $row_not_friends['user_id'] != $user_id )
                    {
                 ?>
                  <tr>
                    <td>
                      <?php echo $row_not_friends['user_name']; ?>
                    </td>
                    <!-- here I am sending request and processing it via ajax -->
                    <td><i class="fa fa-user-plus send_request"></i></td>
                    <input type="hidden" class="send_second" value="<?php echo $row_not_friends['user_id']; ?>">
                    <input type="hidden" class="send_first" value="<?php echo $user_id; ?>">
                  </tr>

                  <?php 
                    }
                    }
                   ?>
              </tbody>
          </table>

Now I am accessing each value in a javascript file as follow: // Here a request is send

$('.send_request').on('click',
    function (e) {
        e.preventDefault();

        var first = $('.send_first').val();
        var second = $('.send_second').val();
        alert('firt id is ' + first);
        alert('second id is ' + second);
        $.ajax(
        {

            url:'send_process.php',
            type:'POST',
            dataType:"json",
            data: { first: first, second: second },
            success:function(data)
            {
                if(data.send_success)
                {
                  window.location.href = "friend.php";

                }
                else
                {
                    alert("something went wrong");
                    window.location.href = "friend.php";
                }

            },
            error : function() { console.log(arguments); }
        }
            );
    });

But here var second = $('.send_second').val(); gives only top-most element value of $row_not_friends['user_id'] . When I am echoing the value, it gives correct result. Please help me.

epascarello · Accepted Answer · 2015-04-10 13:29:52Z

1

Because you are selecting ALL the elements in the page and the default behavior of val() is it returns the first item. It has no clue you want the nth item.

First thing is you need to fix your HTML it is invalid. You can not have an input as a sibling of a tr element. You need to move it inside of a TD.

                <!-- here I am sending request and processing it via ajax -->
                <td><i class="fa fa-user-plus send_request"></i> <!-- removed the closing td from here  -->
                <input type="hidden" class="send_second" value="<?php echo $row_not_friends['user_id']; ?>">
                <input type="hidden" class="send_first" value="<?php echo $user_id; ?>"></td>  <!-- moved the closing td to here  -->
              </tr>

You need to find the elements in the same row as the button you clicked. Since the hidden inputs are npw siblings of the button you can use the siblings() method.

var btn = $(this);
var first = btn.siblings('.send_first').val();
var second = btn.siblings('.send_second').val();

edited Apr 10, 2015 at 13:29

answered Apr 10, 2015 at 13:21

epascarello

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5 Comments

kero Over a year ago

Thanks for introducing .siblings() to me! Now I can finally write that .parent().child() stuff in a nicer way

epascarello Over a year ago

Your HTML is invalid that is why!

khan shah Over a year ago

How can I make it valid?

khan shah Over a year ago

Perfect, It worked.. You deserve an oscar :P . I can't upvote, it needs 15 reputation..

epascarello Over a year ago

But you can mark it as an answer since it is your question and get the points to allow you to vote in the future.

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