如何迭代JSON对象并打印其属性及其值?

时间:2014-09-23 14:44:32

标签: javascript json

我想在JavaScript中导航下面JSON中的每个属性。下面的JSON包含两个记录供参考,但实时会有很多这样的记录。

{" Record_0":[{"状态":" CREATED"" CreatorLoginId":"桑德亚" "名称":" G1"}]," Record_1":[{"状态":" CREATED&#34 ;, " CreatorLoginId":"圣""名称":" G2"}]}

我想得到字段的值" Status"," CreatorLoginId"和"姓名"将它们分配给其他东西。

我该怎么做?

4 个答案:

答案 0 :(得分:1)

            var myJSON = JSON.parse('{"Record_0":[{"Status":"CREATED","CreatorLoginId":"sandhya","Name":"G1"}],"Record_1":[{"Status":"CREATED","CreatorLoginId":"San","Name":"G2"}]}');
            for(var pr in myJSON)
            {
                console.log(myJSON[pr][0].Status);
                console.log(myJSON[pr][0].CreatorLoginId);
                console.log(myJSON[pr][0].Name);
            }

答案 1 :(得分:0)

打印怎么样?如果你的意思是输出到js控制台,它将是

for (index in object) {
    console.log(index + ': ' + object[index]);
}

如果您的意思是将其添加到网页,只需用一个小标记替换console.log:

var parent = document.getElementById('parentID');
for (index in object) {
    parent.innerHTML += index + ': ' + object[index] + '<br>';
}

对于嵌套对象(包括数组)

function print(object, parent) {
    for (index in object) {
        if (typeof object[index] == 'object') {
            print(object[index});
        }
        parent.innerHTML += index + ': ' + object[index] + '<br>';
    }
}

编辑:在迭代之前不要忘记JSON.parse():字符串

答案 2 :(得分:0)

//遍历群组

for(组信息中的var currentRecord)

{

store.data.items.push({serial: {}, groupName: {}, createdBy: {}, status: {} });
store.data.items[iNoOfGroups].serial = iNoOfGroups + 1;
store.data.items[iNoOfGroups].groupName = groupInformation[currentRecord][0].Name;
store.data.items[iNoOfGroups].createdBy = groupInformation[currentRecord][0].CreatorLoginId;
store.data.items[iNoOfGroups].status = groupInformation[currentRecord][0].Status;
iNoOfGroups++;

}

答案 3 :(得分:0)

var myJSON = JSON.parse('{"Record_0":[{"Status":"CREATED","CreatorLoginId":"sandhya","Name":"G1"}],"Record_1":[{"Status":"CREATED","CreatorLoginId":"San","Name":"G2"}]}');

for(var key in myJSON){
   console.log(myJSON[key][0].Status);
   console.log(myJSON[key][0].CreatorLoginId);
   console.log(myJSON[key][0].Name);
}`