CODEWITHSUNDEEP
phpjqueryajax
mhamad aras
mhamad aras

Reputation: 39

send id to another form when button clicked and change value

i am trying to send the id to another page then select the data by that id from the other page so it works but for only 1 data like if i have server 1 and server 2 server 1 will work but server 2 wont work can any one tell me what is the problem here is my codes

PHP

<?php
$get = $data->show(" SELECT * FROM servers where film_name='avengers' ");
foreach ($get as $row) {
 $id=$row['server_id'];
 $name=$row['server_name'];
 $link=$row['link'];
 $movie=$row['film_name'];
?>
<button type="button" id="btn"  value="<?php echo $id ?>" class="btn btn-warning"><?php echo "$name"; ?></button> 
<?php } ?>

ajax

    <script type="text/javascript">
    $(document).ready(function (){

        $('#btn').click(function (){

var serverid = $('#btn').val();


$.ajax({
    url:"../../control/operation/view_movie.php",
    method:"POST",
    data:{serverid:serverid},
    success:function(data)
    {
$("#show").html(data);
    }
});

        });
    });
</script>

viewmovie page 

    <?php

    $id=$_POST['serverid'];
    $getuser = $data->getdata("SELECT * FROM servers WHERE server_id='$id'");
    $link=$getuser['link'];
    $name=$getuser['film_name'];

     ?>
  <iframe class="embed-responsive-item" src="<?php echo $link ?>" allowfullscreen></iframe>

it is not like when i load the first one the other wont load even if i didnt load the first one the others wont work thanks for helping img for detail

Upvotes: 0

Views: 1253

Answers (2)

Ibrahim Abou Khalil
Ibrahim Abou Khalil

Reputation: 322

The id is a number and you are identifying it as a string in the query.

So change this: $getuser = $data->getdata("SELECT * FROM servers WHERE server_id='$id'");

To this: $getuser = $data->getdata("SELECT * FROM servers WHERE server_id=$id");

just remove the quotations (' ').

Solution 2

For the button use this:

<button type="button" value="<?php echo $id ?>" class="btn btn-warning" onclick="getmovie(<?php echo $id ?>)"><?php echo "$name"; ?></button>

where onclick="getmovie(<?php echo $id ?>)" is the funtion to view the movie and we are sending the id as parameter. And the ajax is as following:

<script type="text/javascript">

    function getmovie(serverid){
        $.ajax({
        url:"../../control/operation/view_movie.php",
        method:"POST",
        data:{serverid:serverid},
        success:function(data)
            {
                $("#show").html(data);
            }
        });
    }
    </script>

Upvotes: 1

jithu thomas
jithu thomas

Reputation: 289

You have to check that required params are available on the second page.

<?php
    if(!empty($_POST['serverid'])){
    $id=$_POST['serverid'];
    $getuser = $data->getdata("SELECT * FROM servers WHERE server_id='$id'");
    $link=$getuser['link'];
    $name=$getuser['film_name'];
?>
     <iframe class="embed-responsive-item" src="<?php echo $link ?>" allowfullscreen></iframe>

<?php }else{
?>
<iframe class="embed-responsive-item" allowfullscreen> No data</iframe>
<?php } ?>

Upvotes: 0

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