如何使用angularjs将json数据添加到另一个json中的特定索引

时间:2015-06-26 07:46:50

标签: javascript json angularjs

我的回复json来自API如下:

{
    "columnHeaders": [
        {
            "columnName": "id",
            "columnType": "bigint",
            "columnLength": 0,
            "columnDisplayType": "INTEGER",
            "isColumnNullable": false,
            "isColumnPrimaryKey": true,
            "columnValues": [],
            "visibilityCriteria": []
        },
        {
            "columnName": "client_id",
            "columnType": "bigint",
            "columnLength": 0,
            "columnDisplayType": "INTEGER",
            "isColumnNullable": false,
            "isColumnPrimaryKey": false,
            "columnValues": [],
            "visibilityCriteria": []
        },

        {
            "columnName": "countries",
            "columnType": "int",
            "columnLength": 0,
            "columnDisplayType": "CODELOOKUP",
            "isColumnNullable": false,
            "isColumnPrimaryKey": false,
            "displayName": "null",
            "dependsOn": 0,
            "orderPosition": 0,
            "visible": false,
            "mandatoryIfVisible": false,
            "columnValues": [
                {
                    "id": 19,
                    "value": "India",
                    "score": 0,
                    "parentId": 0
                },
                {
                    "id": 20,
                    "value": "USA",
                    "score": 0,
                    "parentId": 0
                }
            ],

            "columnCode": "countries",
            "visibilityCriteria": []
        }
    }
]

现在我想将以下json添加到我现有的json数据中。

"columnValuesLookup": [
                {
                    "id": 19,
                    "value": "English",
                    "score": 0,
                    "parentId": 0
                },
                {
                    "id": 20,
                    "value": "SA",
                    "score": 0,
                    "parentId": 0
                }
            ],

如何将json数据添加到现有json对象中的特定索引,以便我的最终json看起来像

{
    "columnHeaders": [
        {
            "columnName": "id",
            "columnType": "bigint",
            "columnLength": 0,
            "columnDisplayType": "INTEGER",
            "isColumnNullable": false,
            "isColumnPrimaryKey": true,
            "columnValues": [],
            "visibilityCriteria": []
        },
        {
            "columnName": "client_id",
            "columnType": "bigint",
            "columnLength": 0,
            "columnDisplayType": "INTEGER",
            "isColumnNullable": false,
            "isColumnPrimaryKey": false,
            "columnValues": [],
            "visibilityCriteria": []
        },

        {
            "columnName": "countries",
            "columnType": "int",
            "columnLength": 0,
            "columnDisplayType": "CODELOOKUP",
            "isColumnNullable": false,
            "isColumnPrimaryKey": false,
            "displayName": "null",
            "dependsOn": 0,
            "orderPosition": 0,
            "visible": false,
            "mandatoryIfVisible": false,
            "columnValues": [
                {
                    "id": 19,
                    "value": "India",
                    "score": 0,
                    "parentId": 0
                },
                {
                    "id": 20,
                    "value": "USA",
                    "score": 0,
                    "parentId": 0
                }
            ],
            "columnValuesLookup": [
                {
                    "id": 19,
                    "value": "English",
                    "score": 0,
                    "parentId": 0
                },
                {
                    "id": 20,
                    "value": "SA",
                    "score": 0,
                    "parentId": 0
                }
            ],
            "columnCode": "countries",
            "visibilityCriteria": []
        }
    }
]

1 个答案:

答案 0 :(得分:0)

var o = JSON.parse(json1)
o.columnHeaders[2].columnValuesLookup = JSON.parse(json2).columnValuesLookup 
JSON.stringify(o)