单击时禁用表格或提交按钮，然后提交表单

Question

I am creating a dynamic table with a query on my BD, the problem is that I want to disable a button on a table every time I make a submit, I would like to get the form submitted without page load so that the disabled button is visible.

<?php
            $theCounter=0;
            while($row = mysql_fetch_object($result))
            {

        ?>

            <form id="id" action="table.php" method="post">
            <input type="hidden" value="<?php  echo $row->id_table;  ?>" name="idUser">
                <tr>
                    <td><?php  echo $row->id_userS;  ?></td>
                    <td><?php  echo $row->username_s;  ?></td>
                    <td><?php  echo $row->num_people;     ?></td>
                    <td><?php  echo $row->start_time;  ?></td>
                    <td> 
                        <input type="submit" name="delete_check" value="Eliminar"> 
                    </td>
                    <td> 
                        <input type="submit" name="push_check" value="Enviar Push">
                    </td>
              </form>
        <?php
            $theCounter++;
            }

        ?>

The button I want to disable or hide its the one named "push_check". It doesn't matter if I disable the button or hide the button

Thanks in advance!

Answer 1

PHP:-

<?php
        $theCounter=0;
        while($row = mysql_fetch_object($result))
        {

Echo "

        <form id='id' action='table.php' method='post'>
        <input type='hidden' value='$row->id_table;' name='idUser'>
            <tr>
                <td>$row->id_userS</td>
                <td>$row->username_s</td>
                <td>$row->num_people</td>
                <td>$row->start_time</td>
                <td> 
                    <input type='submit' id='del$thecounter' name='delete_check' value='Eliminar' onclick='hide(del$thecounter)'> 
                </td>
                <td> 
                    <input type='submit' id='push$thecounter' name='push_check' value='Enviar Push' onclick='hide(push$thecounter)'>
                </td>
          </form>
";
        $theCounter++;
        }
?>

Now use Javascript:-

function hide(a)
{
     document.getElementById(a).style.visibility="hidden";
}

Answer 2

If you want the button to be disabled on all the page loads after the first submit then you need to store some flag in the session which will indicate that the button has already been clicked. And you need to check for that flag on every page load.

Answer 3

I'm not sure if I understood everything you want to achive, but here is what I made out of your question:

<script>

    document.forms[0].onsubmit = function(e){

        e.preventDefault();  // at the beginning stop page from being reloaded by the script
        var form = e.currentTarget; // get form
        form.push_check.disabled = true; // set submit-button disabled

        if(form.idUser.value){ // send data if there is data to send
           document.forms[0].onsubmit = null; // set event listener null otherwise onsubmit is a forever-loop
           document.forms[0].submit();  // send finally data to the php-script
        }
        else{
           form.mysubmit.disabled = false; // there was no data to send
        }
    }
</script>

You can put this little script after or within the body-tag. It should work in each browser.

Answer 4

One more method to prevent loading the page is using

<form action='' method='POST' target='upload_target' onsubmit='hide(dynamicid)'>

fields between form.....
<input type='submit' id='dynamicid'/>
<iframe id='upload_target' name='upload_target' src='#' style='width:0;height:0;border:0px solid #fff;'></iframe>
</form>

Answer 5

Hi i manage to do it by this:

<?php
                        $push_button = $row->time_push;
                        if($push_button==0)
                        {
                      ?>
                    <td> 
                        <input type="submit" id='push$thecounter' name="push_check" value="Enviar Push">
                    </td>
                    <?php
                        }
                        ?>

Since i've already making a query to my DB, and the button is making a change, im checking if that change its already made, so if not, im displaying the button! thanks a lot for all your comments!