php表单到javascript

Question

I need to pass the form id to javascript, but always shows the same data.

<script>
$(document).ready(function() {
    $('input[name="submit"]').on('click', function(){
        var vauid = $( "input[name='uid']"     ).val(); alert("
 ID:  "+vauid);
    return false;
    }); 
});
</script>                                               
<? include "../includes/config.php";
$sql = "select * from table";
$reg = mysql_query($sql);
while($row = mysql_fetch_array($reg)){ ?>
<form action="" method="POST">
        <input type="text"   name="uid"     value="<?  echo $row['uid']?>" />
        <input type="submit" name="submit" value="Show"/>
</form> 
<? }?>

Answer 1

Part of the problem is that you're inserting numerous forms with elements having the same names - that is a bad idea as names are intended to be unique.

Second, when you click you have to reference the currently clicked info:

$(document).ready(function() {
    $('input[name="submit"]').on('click', function(e){
        e.preventDefault();
        var vauid = $(this).prev("input[name='uid']").val(); 
        console.log("ID:  "+vauid);
    }); 
});

Here is a working demo.

$(this) will be the clicked button from which we look for the previous input's value. I'm also using preventDefault() to stop the button's default action, rather than returning false.

In addition you should quit using alert() for troubleshooting., use console.log() instead.