我有以下jquery函数来解析json数组“projectDetail”,但是当我尝试从显示为undefined的数组中打印ApprovalDesc的值时,我能否知道什么是正确的语法?
function show(projectDetail) {
var project = projectDetail;
$.each(project, function (i, tweet) {
$.each(tweet.ProjectUpdates, function (i, tweet1) {
$.each(tweet1.ProjectApproval, function (i, tweet2) {
alert(i + ': ' + tweet2.ApprovalDesc);
var PStatus = tweet.ApprovalDesc;
});
});
});
}
{ "d" : {
"ProjectUpdates" : { "ExceptionId" : 0,
"ProjectApproval" : [ { "ApprovalDesc" : "ABCApproved",
"CreatedBy" : null,
"CreatedDate" : "/Date(-62135596800000)/",
"ModifiedDate" : "/Date(-62135596800000)/",
"ApprovalDesc" : "welcome",
"SortOrder" : 1
} ],
"ProjectStatusGallery" : [ { "CategoryId" : 0,
"CreatedBy" : null,
"CreatedDate" : "/Date(-62135596800000)/",
"ImageName" : "flower",
"ImageUrl" : "D://Images",
"ModifiedBy" : null,
"ModifiedDate" : "/Date(-62135596800000)/",
},
{ "CategoryId" : 0,
"CreatedBy" : null,
"CreatedDate" : "/Date(-62135596800000)/",
"ImageName" : "flower2",
"ImageUrl" : "D://Images",
"ModifiedBy" : null,
"ModifiedDate" : "/Date(-62135596800000)/"
}
]
},
"__type" : "sample"
} }
答案 0 :(得分:1)
welcome 应引用为字符串。
答案 1 :(得分:0)
你有没有在json字符串上调用parseJSON把它放在一个对象中?然后它易于访问,无需循环。
http://api.jquery.com/jQuery.parseJSON/
var json = $.parseJSON(your_json_string);
console.log(json.__type.sample);
console.log(json.d.ProjectUpdates.ProjectStatusGallery.ImageName);