使用jQuery获取链接的属性，并将其作为表单发布到另一个脚本文件 - Using jQuery to grab attributes of link and spit as form post to another script file

Using jQuery to grab attributes of link and spit as form post to another script file

我想获取<a href>链接的属性，并将这些属性作为post数据吐到另一个脚本文件中。以下是我到目前为止所做的事情的细节

Html链接

<a href="#" class="submit-button" id="deleteRecord" attr1="xyz" attr2="abc" >Delete</a>

现在jQuery代码如下

$(document).ready(function(){
  //Other code here
  $(document).on('click','#deleteRecord',function(){
    //I want to submit variables attr1 and attr2 as post data to delete_post.php
  });
});

提前谢谢。

如果您不确定要发布多少属性，请使用data属性和循环：

<a href="#" class="submit-button" id="deleteRecord" data-something="blah" data-attr1="xyz" data-attr2="abc" >Delete</a>

脚本：

$(document).ready(function(){
     $(document).on('click','#deleteRecord',function(e){
        e.preventDefault();
        var postData = {};
        $.each($(this).data(), function(k,v) {
            postData[k] = v;
        });
        $.ajax({
            type    : 'post',
            url     : 'delete_post.php',
            data    : postData,
            success : function (data) {
                // do something with data...
                console.log(data);
            },
            error   : function (obj,status,error) {
                // do something when error...
            }
        });
    });
});

，你可以这么做

$(document).ready(function(){
  //Other code here
  $(document).on('click','#deleteRecord',function(){
    //I want to submit variables attr1 and attr2 as post data to delete_post.php
    
    var link = $("#deleteRecord");
    var formData = new FormData();
    
    // perform validations
    
    // append to formData
    formData.append("attr1", link.attr("attr1");
    formData.append("attr2", link.attr("attr2");
    
    // ajax post
    $.post("") 
    
  });
});

试试这个：

<a href="#" class="submit-button" id="deleteRecord" data-attr1="xyz" data-attr2="abc" >Delete</a>

JQuery：

$(document).on('click','#deleteRecord',function(){
   var attr1= $(this).data('attr1');
   var attr2= $(this).data('attr2');
    $.ajax({
    type: "POST",
    url: 'delete_record.php',
    contentType: "application/json; charset=utf-8",
    dataType: "json",
    data: '{"attr1":' + attr1+ ', "attr2":"' + attr1+ '"}',
    success: function(result){
    var data = result.d;
    },
    error: function (xhr, reason, ex) {
        var err = $.parseJSON(xhr.responseText);
        if (err != null && xhr.status.toString() != "0") {
            showerror('Error Code : ' + xhr.status + "'nError Message :" + xhr.statusText);
        }
    }
});
});

参考：

http://api.jquery.com/data/