Question

我试图在我的JSON对象中显示嵌套项目。现在它们的值是[object Object]。我使用$ .each循环，但它似乎没有得到嵌套项的值。我究竟做错了什么？

JSON

{
    "AssetGUID":"00000000-0000-0000-0000-000000000000",
    "AwayForRepair":false,
    "BooleanDataItems":[
      {
         "Column":null,
         "DisplayValue":null,
         "TableName":null,
         "Value":false
      }
    ],
    "ConditionID":0,
    "DecimalDataItems":[
      {
         "Column":null,
         "DisplayValue":null,
         "TableName":null,
         "Value":0
      }
    ],
    "DeviceName":null,
    "Faulty":false,
    "ForDisposal":false,
    "ImageDataItems":[
      {
         "Column":null,
         "DisplayValue":null,
         "TableName":null,
         "Value":null
      }
    ],
    "InspectionDate":"\/Date(-62135596800000+0000)\/",
    "InspectionPassed":false,
    "InspectionType":0,
    "IntegerDataItems":[
      {
         "Column":null,
         "DisplayValue":null,
         "TableName":null,
         "Value":0
      }
    ],
    "LocationGUID":"00000000-0000-0000-0000-000000000000",
    "StringDataItems":[
      {
         "Column":null,
         "DisplayValue":null,
         "TableName":null,
         "Value":null
      }
    ],
    "TagTypeID":0,
    "TransactionGUID":null,
    "UserID":0
}

的jQuery

var content = '';
$.each(data, function(i, post) {
    content += '<li>' + i + " : " + post + '</li>';
});

$("#addJSON").html(content);

输出

AssetGUID : 00000000-0000-0000-0000-000000000000
AwayForRepair : false
BooleanDataItems : [object Object]
ConditionID : 0
DecimalDataItems : [object Object]
DeviceName : null
Faulty : false
ForDisposal : false
ImageDataItems : [object Object]
InspectionDate : /Date(-62135596800000+0000)/
InspectionPassed : false
InspectionType : 0
IntegerDataItems : [object Object]
LocationGUID : 00000000-0000-0000-0000-000000000000
StringDataItems : [object Object]
TagTypeID : 0
TransactionGUID : null
UserID : 0

Answer 1

当post是数组或对象时，您需要递归到它中。

function display_json(data) {
    var content = '<ul>';
    $.each(data, function(i, post) {
        if (post !== null && typeof post == 'object') {
            content += display_json(post);
        } else {
            content += '<li>' + i + " : " + post + '</li>';
        }
    });
    content += '</ul>';
    return content;
}

使用jQuery在JSON对象中显示嵌套项

1 个答案: