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Php Json数据循环问题_Php_Jquery - Fatal编程技术网

Php Json数据循环问题

Php Json数据循环问题,php,jquery,Php,Jquery,我使用for循环来处理数据,但它没有正确显示。我真的被卡住了。数据来自另一个脚本,但问题在for循环中。我错过了什么 $.ajax({ type:"POST", url: "more.php", data:"&max_id="+maxid, //Forms name datatype: "json", success: function(data, textStatus, xhr) { data = JSON.parse(xhr.r

我使用for循环来处理数据,但它没有正确显示。我真的被卡住了。数据来自另一个脚本,但问题在for循环中。我错过了什么

$.ajax({
    type:"POST",
    url: "more.php",
    data:"&max_id="+maxid, //Forms name
    datatype: "json",
    success: function(data, textStatus, xhr) {
        data = JSON.parse(xhr.responseText);
        //var image = data.images;
        for (var i = 0, len = data.length; i < len; i++) {
            var image = data.images;
            var likes = data[i].likes[i];
            var comments = data[i].comments[i];
            var nextid=data[i].next_id;
            $(".row").append("<div class='col-md-2 col-sm-3'><div class='thumbnail'><img src='"+image+"' style='width: 100%;' alt='Gallery Image'><p><br/> Likes:"+likes+"  <br/> Comments: "+comments+" </p></div></div>");
        }
    }
});

由于
data
不是数组而是对象,因此可能没有定义
data[i]

我想你应该去掉那些括号:
data[I].likes[I]
=>
data.likes[I]

你的for循环应该是这样的..在你的代码中,你用数据的长度检查了I。。相反,您必须使用
图像
注释
长度(如果您的数据对象不是数组..) 请参见下面编辑的代码

for (var i = 0; i < data.images.length; i++) {
                var image = data.images[i];
                var likes = data.likes[i];
                var comments = data.comments[i];
                var nextid=data.next_id;
                $(".row").append("<div class='col-md-2 col-sm-3'><div class='thumbnail'><img src='"+image+"' style='width: 100%;' alt='Gallery Image'><p><br/> Likes:"+likes+"  <br/> Comments: "+comments+" </p></div></div>");
    }
for(var i=0;i喜欢:“+Likes+”
评论:“+Comments+”

”; }
您是否以数组的形式获取数据。。?或者你得到的数据只发布在这里..?我只是复制你的for循环部分,并替换为我的for循环部分,也就是说,我要求你共享JSFIDLE,以便我可以删除错误
for (var i = 0; i < data.images.length; i++) {
                var image = data.images[i];
                var likes = data.likes[i];
                var comments = data.comments[i];
                var nextid=data.next_id;
                $(".row").append("<div class='col-md-2 col-sm-3'><div class='thumbnail'><img src='"+image+"' style='width: 100%;' alt='Gallery Image'><p><br/> Likes:"+likes+"  <br/> Comments: "+comments+" </p></div></div>");
    }