PHP variables not returning to success AJAX/jQuery POST correctly using json_encode()

Question

I've been trying for hours to get this to work. I have a <div id=""data_friends> tag and a hidden input field that I want to update using AJAX. The divtag is as follows:

<div class="friends-tab-list" id="data_friends">

    <?php 
    //Default query limits results to 8

    $sql = ("SELECT * FROM users WHERE FIND_IN_SET(?, friend_array) > 0 LIMIT 0,8");
    $query = mysqli_prepare($con, $sql);
    $query->bind_param('s', $username);
    $query->execute();

    $result = $query->get_result();

    $num_rows = $result->num_rows;

        while ($row = $result->fetch_assoc()) {
        $row['profile_pic']; 
        $row['username'];

        echo "<a class='profile-img-item' href='" . $row['username'] . "'>
              <img src='" . $row['profile_pic'] . "' title='" . $row['username'] . "'>
              </a>";
                }
        $query->close();

          ?>
</div>

The hidden input is as follows: <input id='max' type='hidden' value='<?php echo $num_rows; ?>'>

I'm clicking on a View More Friends button and sending data to includes/handlers/ajax_load_profile_friends.php using the following:

$.ajax({

    url:'includes/handlers/ajax_load_profile_friends.php',
    type:'POST',
    dataType: 'json',
    data:{'username':username, 'num_friends':num_friends, 'counter':counter},

        success: function(data) {
            $('#data_friends').html(data.html);
            $('#max').val(data.num_rows);
                }
        });

The data coming from ajax_load_profile_friends.php looks like this:

$query = mysqli_prepare($con,"SELECT * FROM users WHERE FIND_IN_SET(?, friend_array) LIMIT $counter"); 
$query->bind_param('s', $username);
$query->execute();

$result = $query->get_result();

$num_rows = $result->num_rows;
}       

while ($row = $result->fetch_assoc()) {
$row['profile_pic']; 
$row['username'];

$html = "<a class='profile-img-item' href='" . $row['username'] . "'>
          <img src='" . $row['profile_pic'] . "' title='" . $row['username'] . "'>
         </a>";

}   

echo json_encode(array('num_rows' => $num_rows, 'html' => $html));

When I run this, I get a single return in my when I'm suppose to get a return of 16 records with each click I thought by doing this in my success function $('#data_friends').html(data.html);

The value in my hidden input field <input id='max' type='hidden' value='<?php echo $num_rows; ?>'> is not updating using this $('#max').val(data.num_rows);

Is there something I'm missing in ajax_load_profile_friends.php that is causing these behaviors?

**Keep in mind that I can get this to work when I don't use json_encode & write success function like this $('#data_friends').html(data.html); and remove the dataType: 'json', from AJAX. The problem here is that in both ways, I'm not able to update my hidden input value. I figured I would try and correct this method since most examples specify json_encode() as the way to return data.

When you use json_encode you get the num_rows null but the html is correct? — Batsheva Hansav
– Batsheva Hansav, Commented Feb 27, 2019 at 9:56

far2005 · Accepted Answer · 2019-02-26 07:23:06Z

2

header( "Content-Type: application/json", TRUE );
$query = mysqli_prepare($con,"SELECT * FROM users WHERE FIND_IN_SET(?, friend_array) LIMIT $counter"); 
$query->bind_param('s', $username);
$query->execute();

$result = $query->get_result();

$num_rows = $result->num_rows;

$html='';

while ($row = $result->fetch_assoc()) {


$html .= "<a class='profile-img-item' href='" . $row['username'] . "'>
          <img src='" . $row['profile_pic'] . "' title='" . $row['username'] . "'>
         </a>";

}   

echo json_encode(array('num_rows' => $num_rows, 'html' => $html));

edited Feb 26, 2019 at 7:23

answered Feb 26, 2019 at 7:02

far2005

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15 Comments

Raylene Over a year ago

When I concatenate $html as suggested the button does nothing. No errors thrown or errors in console, just a button that does nothing. When I remove .= and replace with = I get a single record returned.

far2005 Over a year ago

what are getting when you run this console.log(data) ?

Raylene Over a year ago

Without .= I'm getting on first click num_rows: 24, html: "<a class='profile-img-item' href='la_fway'>…elion2.jpg' title='la_fway'> on second click I'm getting same thing, only num_rows=40. So it is counting up, but only returning a single record. When I use .= I get nothing.

far2005 Over a year ago

Are you getting your result from ajax like you wanted?

Barmar Over a year ago

I'm really running out of ideas about what could be going wrong? Can I get access to your web page?

|

JohnnyB1988 · Accepted Answer · 2019-02-26 07:20:53Z

1

You are not declaring $html variable before while loop. Try this one

<?php

$query = mysqli_prepare($con,"SELECT * FROM users WHERE FIND_IN_SET(?, friend_array) LIMIT $counter"); 
$query->bind_param('s', $username);
$query->execute();

$result = $query->get_result();

$num_rows = $result->num_rows;
}       
$html = '';
while ($row = $result->fetch_assoc()) {
// $row['profile_pic']; 
// $row['username'];

$html .= "<a class='profile-img-item' href='" . $row['username'] . "'>
          <img src='" . $row['profile_pic'] . "' title='" . $row['username'] . "'>
         </a>";

}   

echo json_encode(array('num_rows' => $num_rows, 'html' => $html));

answered Feb 26, 2019 at 7:20

JohnnyB1988

1789 bronze badges

3 Comments

Raylene Over a year ago

this way does something. I'm getting to 40 records returning, but it won't display all 43. The value in my hidden input field is still not updating. hmm...weird.

Barmar Over a year ago

@Raylene What is the value of counter?

Raylene Over a year ago

The counter begins at 8 - the initial query num_rows for hidden input - value='<?php echo $num_rows; ?>' I have it increasing with this: $("#viewAllFriends").click(function(){ counter = counter+16; I echoed this on the page and it works. I have this in the ajax page to evaluate for the final query: if($counter > $num_friends || $counter == $num_friends){ $counter = "No More Friends to Show"; When I use the json_encode method however, it stops at 40 (there are 43 records). My bootstrap alert doesn't show, b/c my if statement nvr evaluates to true.

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