When I try to send id value of row via POST to another page, this page will always GET id value 1

Question

I started learning webdeveloping and i tried to send "id" of one of the rows generated from database to another page. The rows are clickable thanks to the javascript code, so i can choose whichever row i want. The problem is, that even though the POST method seems right:

<form id="send" method="POST" action=<?php echo "secondpage.php?id=". $row['id']; ?> ></form>

// In inspect of the main page it gets the value.

However second page always receive id value of 1. Doesn't matter if i click on the row with id=18 or any other. It will always recieve value of 1...

I heard that this could be a problem with javascript code which i put under PHP code.

Here is a code with PHP:

<div id="content">
<table id="usersTable" class="table table-bordered table-hover table-sm ">
<form action=http://localhost/dodawanie.php>

    <input class="btn btn-primary" type="submit" value="Dodawanie">

</form>
<?php if ($result->num_rows > 0) {
     while($row = $result->fetch_assoc()) {?>

    <tr>         

        <td><?php echo "id: ". $row['id']; ?> </td> 
    <td><?php echo "Name: ". $row["first_name"]; ?> </td> 
        <td><?php echo "Last: ". $row["last_name"];?>  </td>
    <form id="send" method="POST" action=<?php echo "secondpage.php?id=". $row['id']; ?> >
</form> 
    </tr>

   <?php }
} else {
     echo "0 results";
}

$conn->close();
   ?>
</table>
</div>

Here is javascript:

<script type = "text/javascript">
$(document).ready(function(){
    $('#usersTable').find('tr').click( function(){
    //  alert('You clicked row ' + ($(this).index()+1) );
            $('#send').submit();
        });
    });
</script>

I would gladly accept any help to find an error here.

I don't recommend sending parameters in the query string when doing a POST, try to adhere to the REST principles. But anyway, what's the rendered `action` attribute? — MaxArt, Nov 23 '16 at 22:14
Any time you send parameters via the URL, they will be sent via GET. You also should not nest a `
` inside a `` either: http://stackoverflow.com/questions/5967564/form-inside-a-table — WillardSolutions, Nov 23 '16 at 22:18
As for the secondpage.php there is not much of a code right now, however i try to check what i sent by `$id= $_GET['id'];` — Peter, Nov 23 '16 at 22:24

score 1 · Accepted Answer · edited Nov 23 '16 at 22:32

1

Change the <form id="send" id value as unique

or use a class some thing like below:
<form class="form_send" then in your javascript search for the form_class inside the clicked tr:

$(document).ready(function(){
    $('#usersTable').find('tr').click( function(){
          $(this).find('form.form_send').submit();
        });
    });

edited Nov 23 '16 at 22:32

Funk Forty Niner

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answered Nov 23 '16 at 22:21

tradebel123

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score 0 · Answer 2 · answered Nov 23 '16 at 22:18

0

Ids have to be unique. $('#send').submit() only finds and submits the first form with that id.

You could add your row id to the form's id attribute to make them unique for example.

answered Nov 23 '16 at 22:18

Andy

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When I try to send id value of row via POST to another page, this page will always GET id value 1

2 Answers2