在一个表单中有多个按钮，每个按钮都应该能够更新数据库。

Question

嗨，朋友们，请告诉我哪里错了，因为数据库仍然没有变化，即使成功地执行了更新查询

这是表单代码

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>';

注意:这里我使用了$upload_id (一个php变量)，它具有行上传id，例如，对于第一个图像，reject1，saccept1和sreject1，对于第二个图像，它是accept2 reject2等等。

现在是processor.php的代码

$status_pass = isset($_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());

}

Answer 1

以你的形式

<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>

在你的php文件

<?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
}


?>

Answer 2

请用$_POST"accept'.$upload_id.'"替换$_POST"accept".$upload_id好吗？

Answer 3

不要为$upload_id输入标签使用'‘引号，它会正常工作的