在一个表单中有多个按钮，每个按钮应该能够更新数据库

Hi Friends please tell me where I am wrong cause the database is still unchanged even though update query is executed successfully Thank you

here is the form code

echo'<form action="processor.php" method="post" id ="post_form">';
echo '<input type="hidden" name= "status" id="status" value="">';
$upload_arr=array("1");
foreach($upload_arr as $upload_id)
{
    echo '<input type="button" name="accept-<?=$upload_id?>"   value="accept"             onclick="submit_this(this.name)"/><br>';
    echo    '<input type="button" name="reject-<?=$upload_id?>"  value="reject"  onclick="submit_this(this.name)"/><br>';
    echo    '<input type="button" name="saccept-<?=$upload_id?>"   value="saccept" onclick="submit_this(this.name)"/><br>';
    echo    '<input type="button" name="sreject-<?=$upload_id?>"   value="sreject"       onclick="submit_this(this.name)"/><br>';
 }
echo '</form>';

Note :here i have used $upload_id a php variable which has row[upload id] i.e. it is like accept 1,reject1,saccept1 ans sreject1 for first image and for second image it is accept2 reject2 and so on

Now the code for processor.php

$status_pass = isset($_POST['status'])?$_POST['status']:NULL;
if(!empty($status_pass)){
$status_arr = explode('-', $status_pass);
$action  = $status_arr[0];
$upload_id = $status_arr[1];

if($action == 'accept'){
    $status = 1;
}
if($action == 'reject'){
    $status = 2;
}
if($action == 'saccept'){
    $status = 3;
}
if($action == 'sreject'){
    $status = 4;
}
echo $status;
$sql="UPDATE upload SET status='$status' where upload_id = '$upload_id' ";
echo "update success";
 $result = mysql_query("$sql") or die("Invalid query: " . mysql_error());

}

in your form

<form action="processor.php" method="post" id ='post_form'>
  <input type="hidden" name= "status" id="status" value="">
  <?php
    foreach($upload_arr as $upload_id){
  ?>
      <input type="button" name="accept-<?=$upload_id?>"   value="accept" onclick='submit_this(this.name);'/><br>
      <input type="button" name="reject-<?=$upload_id?>"  value="reject" onclick='submit_this(this.name);'/><br>
      <input type="button" name="saccept-<?=$upload_id?>"   value="saccept" onclick='submit_this(this.name);'/><br>
      <input type="button" name="sreject-<?=$upload_id?>"   value="sreject" onclick='submit_this(this.name);'/><br>
  <?php
    }
  ?>
</form>
<script>
function submit_this(name){
    document.getElementById('status').value = name;
    document.getElementById('post_form').submit();
}
</script>

in you php file

<?php
$status_pass = isset($_POST['status'])?$_POST['status']:NULL;

if(!empty($status_pass)){
    $status_arr = explode('-', $status_pass);
    $action  = $status_arr[0];
    $upload_id = $status_arr[1];

    if($action == 'accept'){
        $status = 1;
    }
    if($action == 'reject'){
        $status = 2;
    }
    if($action == 'saccept'){
        $status = 3;
    }
    if($action == 'sreject'){
        $status = 4;
    }

    $sql="UPDATE upload SET status='$status' where upload_id = '$upload_id' ";
    //execute sql here
}


?>

can you please replace $_POST["accept'.$upload_id.'"] this with $_POST["accept".$upload_id]

Don't use '' quotes for $upload_id input tag, It will work then properly

there's some problems of your code

First

if you want to multiple submit form, the type shuold be button, you can use javascript to contorl submit form, and before submit, you can also pass some value to a hidden input field, as your example, you can set hidden input value = "accept'.$upload_id.'"

Second

in your php file,

$sql="UPDATE upload SET status='1'";

has no where condition, of course, this will update all your records.