Home >Backend Development >PHP Tutorial >Select delete with ajax?

Select delete with ajax?

WBOY
WBOYOriginal
2016-09-12 17:44:441209browse

Attach my code first

<code><button class="box" data-removetype="select_del">刪除</button>
<form id="selectid">
<? while ($row = mysql_fetch_array($s))
{
       <input type="checkbox" name="id[]" value="<? echo $row['mail_id'];?>">
 }
 ?>
 </form>
 
 $(".box").click(function(e){
    var removetype = e.currentTarget.dataset.removetype;
    /*mailbox*/
    if (removetype=='select_del')
    {
        $.ajax({
                type: "POST",
                url:"del?to="+removetype,
                data:$("#selectid").serialize(),
                cache: false,
                success: function(){
                
                }
            });
    }
});
</code>

del

<code>$user = $_GET['to'];
if ($user=='select_del')
{
    $getid=$_POST['id'];
    foreach($getid as $value)
    {
        $DEL = mysql_query("
        DELETE FROM `資料表名稱`
        WHERE `id` = '".$value."'
        ");
     }
}
</code>

The above is my code...
But it doesn't work
So I want to ask if this is the right way to do it?
Make sure the ID and data table name are correct

Reply content:

Attach my code first

<code><button class="box" data-removetype="select_del">刪除</button>
<form id="selectid">
<? while ($row = mysql_fetch_array($s))
{
       <input type="checkbox" name="id[]" value="<? echo $row['mail_id'];?>">
 }
 ?>
 </form>
 
 $(".box").click(function(e){
    var removetype = e.currentTarget.dataset.removetype;
    /*mailbox*/
    if (removetype=='select_del')
    {
        $.ajax({
                type: "POST",
                url:"del?to="+removetype,
                data:$("#selectid").serialize(),
                cache: false,
                success: function(){
                
                }
            });
    }
});
</code>

del

<code>$user = $_GET['to'];
if ($user=='select_del')
{
    $getid=$_POST['id'];
    foreach($getid as $value)
    {
        $DEL = mysql_query("
        DELETE FROM `資料表名稱`
        WHERE `id` = '".$value."'
        ");
     }
}
</code>

The above is my code...
But it doesn't work
So I want to ask if this is the right way to do it?
Make sure the ID and data table name are correct

<code>data:$("#selectid").serialize()</code>

You can try to replace with

<code>data:$("#selectid").find('input').serialize()</code>

If it doesn’t work, I think you need to check how the script is written to see if it complies with the specifications. For example:
<a class=" <?php if($status==1): ?>selected<?php endif ; ?>" >RETURN</a>

You wrote like this

<code><? while ($row = mysql_fetch_array($s))
{
       <input type="checkbox" name="id[]" value="<? echo $row['mail_id'];?>">
 }
 ?></code>

I don’t think it will work properly....

Statement:
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn