jquery ajax调用删除mysql中的行

Question

如何通过仅使用jquery ajax调用传递ID来删除行，这就是delete.php

<?php

require_once('DB.class.php');
$dbconnect = new DB('schoollife', 'root', '');

$id=$_GET['chapter_id'];

$del="DELETE FROM sl_chapter WHERE chapter_id=".$id ;
$result=mysql_query($del);

// if successfully deleted
if($result){
echo "<font color='red'>record deleted successful</font>";
//echo "<BR>";
//echo "<a href='delete.php'>Back to main page</a>";
}
else {
echo "<br/>";
die('Error: ' . mysql_error());
} 

?>

这是Chapter.html

<script>
 $(document).on('click','.delbutton',function(){ 
        var del_id = element.attr("id");
        var info = 'id=' + del_id;
        if(confirm("Are you sure you want to delete!"))
        {
        $.ajax({
        type: "json",
        url: "delete.php",
        //data: info,
        success: function(){
        }
        });
        $(this).parents(".record").animate({ backgroundColor: "#fbc7c7" }, "fast")
        .animate({ opacity: "hide" }, "slow");
        }
        return false;

        });
            </script>

我如何通过仅使用jquery ajax调用传递ID来删除行，我认为删除查询确实出错了。

Answer 1

您已经快到了，您需要将要删除的ID传递给delete.php

Chapter.html：

<script>
 $(document).on('click','.delbutton',function(){ 
        var del_id = element.attr("id");
        if(confirm("Are you sure you want to delete!"))
        {
        $.ajax({
        type: "POST", //changed
        url: "delete.php",
        data: 'id=' + del_id, // changed
        success: function(){
        }
        });
        $(this).parents(".record").animate({ backgroundColor: "#fbc7c7" }, "fast")
        .animate({ opacity: "hide" }, "slow");
        }
        return false;

        });
            </script>

delete.php：

require_once('DB.class.php');
$dbconnect = new DB('schoollife', 'root', '');

$id=$_POST['id']; //changed

$del="DELETE FROM sl_chapter WHERE chapter_id=".$id ;
$result=mysql_query($del);

// if successfully deleted
if($result){
echo "<font color='red'>record deleted successful</font>";
//echo "<BR>";
//echo "<a href='delete.php'>Back to main page</a>";
}
else {
echo "<br/>";
die('Error: ' . mysql_error());
} 

?>

Answer 2

您发送JSON jQuery的类型，但type应该是POST或GET

$.ajax({
    type: "GET", // should be GET or POST not JSON
    url: "delete.php",
    data: info,
    success: function(){
    }
});

还要在delete.php更改它

$id=$_GET['chapter_id'];

至

$id=$_GET['id'];

更多信息

Answer 3

好，几个修复：

1. 您正在发送JSON而不是GET
2. 我假设您只想在删除后执行动画，因此将动画放置在success 。 （如果没有，请忽略此）。
3. 不要注释掉data

4. 将id替换为info变量中的chapter_id

    $(document).on('click','.delbutton',function(){ var del_id = element.attr("id"); var info = 'chapter_id=' + del_id; if(confirm("Are you sure you want to delete!")) { $.ajax({ type: "get", url: "delete.php", data: info, success: function(){ $(this).parents(".record").animate({ backgroundColor: "#fbc7c7" }, "fast").animate({ opacity: "hide" }, "slow"); } }); } return false; });