Question

我在html中有这个按钮，调用jquery来确认删除

<div class='delete' name='<?php echo $story_id; ?>'>delete book</div>

这是我的jquery文件

$(document).ready(function(){   
    $('.delete').click(function(){
       var del_id = $(this).attr('name');

        $.confirm({
            'type':'POST',                  
            'data':'delete_id='+del_id,
            'title'     : 'Delete Book Confirmation',
            'message'   : 'You are about to delete this book. <br />It cannot be restored at a later time! Do you still want to continue?',
            'buttons'   : {
                'Yes'   : {
                    'class' : 'blue',
                    'action': function(){
                              document.location.href="sample2.php";
                    }

                },
                'No'    : {
                    'class' : 'gray',
                }
            }
        });

    });

});

该按钮应将值传递给jquery，如果单击警报中的yes按钮，则该值应随后传递给sample2.php。我该怎么办？对不起，我是jquery的新手。 var del_id = $(this).attr('name');无效。

Answer 1

$（this）.attr（'name'）工作正常。尝试提醒以打印值。

我认为你应该从$ .confirm（）中删除data属性。并写这样的。

 $(document).ready(function(){  
       $('.delete').click(function(){
       var del_id = $(this).attr('name');

        $.confirm({
            'type':'POST',
            'title'     : 'Delete Book Confirmation',
            'message'   : 'You are about to delete this book. <br />It cannot be restored at a later time! Do you still want to continue?',
            'buttons'   : {
                'Yes'   : {
                    'class' : 'blue',
                    'action': function(){
                              document.location.href="sample2.php?delete_id="+del_id;
                    }

                },
                'No'    : {
                    'class' : 'gray',
                }
            }
        });

    });

});'

我认为它会奏效。

Answer 2

JQuery.post的data参数应该是普通的Javascript对象：

'data': { delete_id: del_id },

Answer 3

尝试change div element至button喜欢，

<button class='delete' name='<?php echo $story_id; ?>'>delete book</button>

或使用data- attribute代替name attribute喜欢，

<div class='delete' data-name='<?php echo $story_id; ?>'>delete book</div>

使用data()喜欢

获取它

var del_id=$(this).data('name');

将值传递给jquery和php

3 个答案: