我有以下JSON字符串:
{
"responseHeader":{
"status":0,
"QTime":2,
"params":{
"facet":"false",
"fl":"id,title,friendlyurl,avatar,locpath,objectid,objecttype",
"indent":"off",
"q":"title_search:*castle*",
"wt":"json",
"fq":"userid:\"C325D42C-A777-4275-BDD2-D7810A8AB9AB\"",
"defType":"lucene"
}
},
"response":{
"numFound":2,
"start":0,
"docs":[
{
"title":"castle a",
"objecttype":1,
"friendlyurl":"castle-a",
"avatar":"6_887_castle-a.JPG",
"objectid":6
},
{
"title":"castle b",
"objecttype":1,
"friendlyurl":"castle-b",
"avatar":"794_360_13j-castle-by-night.jpg",
"objectid":794
}
]
}
}
这里有2个搜索结果: 城堡A和城堡B
我想循环遍历所有结果并获取属性title,objectype和friendlyurl
的值出于测试目的,我只是将JSON字符串分配给变量'data' 然后我尝试了:
for (var i = 0, l = data.items.length; i < l; i++) {
console.log(data.items[i].title);
console.log(data.items[i].objecttype);
console.log(data.items[i].friendlyurl);
}
但后来我得到:Uncaught TypeError:无法读取未定义的属性'length'
如果我这样做:
data = $.parseJSON(data);
for (var i = 0, l = data.items.length; i < l; i++) {
console.log(data.items[i].title);
console.log(data.items[i].objecttype);
console.log(data.items[i].friendlyurl);
}
我得到:未捕获的SyntaxError:意外的令牌C jquery-1.8.3.min.js:2
js文件中的行:e.JSON.parse(t);
我也尝试过:
var data = '{"responseHeader":{"status":0,"QTime":1,"params":{"facet":"false","fl":"id,title,friendlyurl,avatar,locpath,objectid,objecttype","indent":"off","start":"0","q":"title_search:*castle*","wt":"json","fq":"userid:\"C325D42C-A777-4275-BDD2-D7810A8AB9AB\"","rows":"10","defType":"lucene"}},"response":{"numFound":2,"start":0,"docs":[{"title":"castle a","objecttype":1,"friendlyurl":"castle-a","avatar":"6_887_castle-a.JPG","objectid":6},{"title":"castle b","objecttype":1,"friendlyurl":"castle-b","avatar":"794_360_13j-castle-by-night.jpg","objectid":794}]}}'
var result,
size = data.result.docs.length,
index;
for (index = 0; index < size; index++) {
result = data.result.docs[index];
console.log(result.title);
}
但是会导致错误:未捕获TypeError:无法读取未定义的属性'docs'。
答案 0 :(得分:3)
这是非常基本的JSON处理。这样的事情应该给你一个开始:
data = $.parseJSON(json);
$.each(data.response.docs, function(i, item) {
console.log(item.title);
console.log(item.objecttype);
console.log(item.friendlyurl);
});
答案 1 :(得分:1)
你不需要jQuery,你可以像这样使用vanilla JavaScript(假设这个json被分配给一个名为json的var)
var result,
size = json.result.docs.length,
index;
for(index=0; index<size;index++) {
result = json.result.docs[index];
console.log(result.title);
}
答案 2 :(得分:0)
您必须循环遍历所有json对象属性并查找所需的键值。基本操作将包含一个ajax调用,该调用具有调用返回的对象作为参数,然后对于每个循环,您可以检查所需的对象属性。
这是一个应该如何看待基本操作:
$.ajax({
url: 'url_to_json.json',
dataType: 'json', // Choosing a JSON datatype
success: function(data) // Variable data contains the data we get from JSON
{
$.each(data, function(filterGroupKey, filterGroupValue){
if (filterGroupValue && filterGroupValue.hasOwnProperty('title')){
var filterTagModel = filterGroupValue;
filterTagModel.title = filterGroupValue['title'];
filterTagModel.objectType = filterGroupValue['objecttype'] ;
filterTagModel.friendlyUrl = filterGroupValue['friendlyurl'];
}
});
}
});
然后你可以使用filterTagModel及其所有属性来操纵所获得的数据。