我怎样才能在php中对ajax响应应用条件

Question

function getProject(){
    var postData = {
        'project' : proj_id
    };

    //console.log(postData);
    var site_url = "<?php echo site_url('/'); ?>";
    $.ajax({
        type: "POST",
        url: site_url + "maincontroller/project",
        data: postData, //assign the var here 
        success: function(msg) {
            $("#project").html(msg);
            getSubProject();                              
        }
    });

假设这是一个函数现在我想对AJAX响应应用一个条件，响应id为<div id="project"></div>。如果响应成功则生成此div，否则不生成。有人可以帮帮我吗？

Answer 1

<强>替换

$("#project")=.html(msg);

。通过

$("#project").html(msg);

Answer 2

试试这个

success: function(msg, status){
       if (status == 200) {// ok 

       $("#project").html(msg);

            getSubProject();                             
      }
}