Question

JavaScript的：

function loadDoc(url) {
    $.ajax({
        url: 'mytestjson',
        dataType: 'json',
        cache: false
    }).success(function (result) {
        console.log(result);
        //var txt = result.newBranches[0].newNonBranchLocation;
        $.each(result.newBranches, function(index, txt) 
        {
                // change this to FALSE when ready
                if (newNonBranchLocation.indexOf('TRUE') > -1) {
                        $('#btbumpin').hide();
                    }
            else {
                $('#btbumpin').show();
            }
        });
    })
}

JSON：

{
    "errorNode": {
        "errorCode": 0,
        "errorMsg": "",
        "errorSev": ""
    },
    "newBranches": [{
        "cn": "11111101",
        "newNonBranchLocation": "TRUE",
        "newBranchNumber": "111111"
       }, {
        "cn": "11111102",
        "newNonBranchLocation": "TRUE",
        "newBranchNumber": "111111"
       }, {
        "cn": "11111103",
        "newNonBranchLocation": "TRUE",
        "newBranchNumber": "111111"
       }, {
        "cn": "11111181",
        "newNonBranchLocation": "TRUE",
        "newBranchNumber": "111111"
    }]
}

我有上面的jQuery和JSON。我想返回newBranches中的所有值，但我目前只获取第一个newBranches对象的值

    "cn": "11111101",
    "newNonBranchLocation": "TRUE",
    "newBranchNumber": "111111"

我想返回11111101TRUE11111111111102TRUE111111等等即返回与newBranches中的所有值连接的字符串。

Answer 1

您可以使用双$.Each()来交叉JSON并将值添加到缓冲区。

$(function() {
        
            var json = {
                "errorNode": {
                "errorCode": 0,
                "errorMsg": "",
                "errorSev": ""
                },
                "newBranches": [{
                    "cn": "11111101",
                    "newNonBranchLocation": "TRUE",
                    "newBranchNumber": "111111"
                }, {
                    "cn": "11111102",
                    "newNonBranchLocation": "TRUE",
                    "newBranchNumber": "111111"
                }, {
                    "cn": "11111103",
                    "newNonBranchLocation": "TRUE",
                    "newBranchNumber": "111111"
                }, {
                    "cn": "11111181",
                    "newNonBranchLocation": "TRUE",
                    "newBranchNumber": "111111"
                }]};
        
                $('body').append(dumpContentOfJSON(json.newBranches));
        
        });
        
        
        
        function dumpContentOfJSON(json) {
            var buffer = '';
            $.each(json, function(index, array) 
            {
                buffer += dumpArray(array);
            });
            return buffer;
        }
        
        function dumpArray(array) {
            var buffer = '';
            $.each(array, function(index,value){
                buffer += value;
            });
            return buffer;
        }

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<body>
</body>

Answer 2

在循环外声明一个变量，对于ex

var output = '';

在循环内部，您只需将所需的所有值附加到变量中即可。

output + = txt.cn + txt.newNonBranchLocation + txt.newBranchNumber;

循环完成后，您可以使用此变量，该变量将包含附加的字符串。

jQuery返回JSON中的所有项目

2 个答案: