我正在尝试从Servlet传递的JSON对象中检索数据。但是我的钥匙已编号。我需要将这些数字添加到键名中。
我的回应看起来像这样。
{"shareInfo1":[{"uname":"xyz","image":"iVBO..","imname":"ryty","senderPicture":"iVBOR"}],"shareInfo2":[{"uname":"sds","image":"iVBO","imname":"ryty","senderPicture":"iVBOR}],....}
我尝试使用以下代码:
$.ajax({
type: "GET",
url: "RetrieveShares",
data:'action='+encodeURIComponent(action),
dataType: "json",
success: function( data, textStatus, jqXHR)
{
if(data.success)
{
for(var z=1;z<=data.loops;z++)
{
len=data["shareInfo"+z].length;
for(var k = len-1;k>=0;k -= 1){
var newcommhtml = '<div id="S0'+thecid+'" class="snew">';
newcommhtml = newcommhtml + '<div class="author-image"><img src="data:image/jpeg;base64,'+data["shareInfo"+z][k].senderPicture+'" alt="'+data["shareInfo"+z][k].uname+'" width="100%" height="100%" class="ava"></div><span>'+data["shareInfo"+z][k].uname+' shared the image '+data["shareInfo"+z][k].imname+'</span>';
newcommhtml = newcommhtml + '<div class="s-content"><div class="s-message"><span>'+nl2br(data["shareInfo"+z][k].message+'</span></div><div class="shpicture">');
newcommhtml = newcommhtml + '<img src="data:image/jpeg;base64,'+data["shareInfo"+z][k].image+'" alt="'+data["shareInfo"+z][k].imname+'" width="100%" height="100%" class="SharedImage" class="SharedImage" data-id="'+data["shareInfo"+z][k].id+'" data-alid="'+data["shareInfo"+z][k].alid+'" data-shareid="'+data["shareInfo"+z][k].shareId+'">';
newcommhtml = newcommhtml + '</div></div>';
var thelm = "#S0"+thecid;
$('#spscrl').append(newcommhtml);
$(thelm).hide().fadeIn('slow');
newcommhtml ='<div class="SPcommentbox">';
var i=0;
if(Object.keys(data["shareInfo"+z][k]).length>8)
{
var x;
for(x=data["shareInfo"+z][k].loopreq-1;x>=0;x-=1)
{
newcommhtml = newcommhtml + '<div class="comment"><div class="commenter-image"><img src="data:image/jpeg;base64,'+data["shareInfo"+z][k]["commenterPicture"+i]+'" alt="'+data["shareInfo"+z][k]["whocommented"+i]+'" width="100%" height="100%" class="ava"></div><div class="commentername">'+data["shareInfo"+z][k]["whocommented"+i]+'</div><span>'+data["shareInfo"+z][k]["commented"+i]+'</span></div>';
i+=1;
}
}
newcommhtml = newcommhtml + '<div class="comment"><div class="commenter-image">';
newcommhtml = newcommhtml +'</div><div class="addcomment"><input type ="text" placeholder="Write a comment..." class="commentbox"></input></div></div>';
$('#spscrl').append(newcommhtml);
thecid++;
}
$(".primg > img").first().clone().appendTo(".commenter-image");
}
}
},
error: function(jqXHR, textStatus, errorThrown)
{
alert("error"+errorThrown);
console.log("Something really bad happened " + textStatus);
},
});
但这会产生错误
TypeError: data[("shareInfo" + z)] is undefined
请任何人帮助我解决这个问题...谢谢...
最佳答案
在开始修改DOM元素之前,请使结构更小,更轻松。
这是使用jQuerys example函数的$.each();
var data = {
loops: {
"shareInfo1": [
{
"uname": "xyz",
"image": "iVBO..",
"imname": "ryty",
"senderPicture": "iVBOR"
},
{
"uname": "abc",
"image": "iVBO..",
"imname": "smile",
"senderPicture": "iVBOR"
}
],
"shareInfo2": [
{
"uname": "sds",
"image": "iVBO",
"imname": "happy",
"senderPicture": "iVBOR"
}
]
}
};
console.log('data', data); // DEBUG
// Iterating the data
$.each(data.loops, function(key, shareInfo) {
console.log('shareInfo', key, shareInfo); // DEBUG
$.each(shareInfo, function(key, user) {
console.log('user', key, user); // DEBUG
// Username: user.uname
});
});