Question

如何通过jquery循环获取内部循环的每个信息的info1或info2的id。例如，我想从info1获取id 1，然后id 1中的所有信息与info2中的相同。我需要输出显示在浏览器中。

＆＃13;

   var data = {
    "info1": {
        "1": {
            "clientname": "ruhul yahoo",
            "clientemail": "ruhul080@yahoo.com",
            "clientusername": "ruhulya"
        },
        "2": {
            "clientname": "kaosar yahoo",
            "clientemail": "kaosar080@yahoo.com",
            "clientusername": "kaosar"
        }
    },
    "info2": {
        "3": {
            "id": "24",
            "receiver": "5",
            "content": "chat system",
            "time": "2015-08-19 12:09:19"
        },
        "4": {
            "id": "23",
            "receiver": "4",
            "content": "chat system",
            "time": "2015-08-19 12:09:19"
        }
    },

}

＆＃13;

提前致谢。

Answer 1

迭代json数组并像下面的代码一样访问对象。

 var data = {
    "info1": {
        "1": {
            "clientname": "ruhul yahoo",
            "clientemail": "ruhul080@yahoo.com",
            "clientusername": "ruhulya"
        },
        "2": {
            "clientname": "kaosar yahoo",
            "clientemail": "kaosar080@yahoo.com",
            "clientusername": "kaosar"
        }
    },
    "info2": {
        "3": {
            "id": "24",
            "receiver": "5",
            "content": "chat system",
            "time": "2015-08-19 12:09:19"
        },
        "4": {
            "id": "23",
            "receiver": "4",
            "content": "chat system",
            "time": "2015-08-19 12:09:19"
        }
    },

};


for(var j in data){
   for(var k in data[j]){
      console.log(data[j][k]); 
  }
}

如果您运行上面的示例，您的浏览器的控制台将记录以下对象。

    Object {clientname: "ruhul yahoo", clientemail: "ruhul080@yahoo.com", clientusername: "ruhulya"}
    Object {clientname: "kaosar yahoo", clientemail: "kaosar080@yahoo.com", clientusername: "kaosar"}
    Object {id: "24", receiver: "5", content: "chat system", time: "2015-08-19 12:09:19"}
   Object {id: "23", receiver: "4", content: "chat system", time: "2015-08-19 12:09:19"}

然后您可以像普通对象console.log(data[j][k].clientname);

一样访问这些值

Answer 2

你可以像这样迭代这个对象

for(var i in data){
    for(var j in data[i]){
        console.log(data[i][j]);
    }
}

Answer 3

此函数将为您找到对象中变量名的第一个实例。如果您需要在特定路径中查找变量，则可以相当容易地修改此函数。当然，该功能通过了您提供的测试用例。

function findVar(data, varName) {    
    for (var i in data) {
        if (i === varName) return data[i];

        if (typeof (data[i]) === 'object') {
            var findResult = findVar(data[i], varName)
            if (typeof(findResult) !== 'undefined') 
            {
                return findResult;
            }
        }
    }

    return undefined;
}

Answer 4

首先，这不是有效的JSON，在最后,之前移动{

其次，将其解析为JSON并将信息作为

data.info1[1].clientname

＆＃13;

var data = JSON.parse('{"info1":{"1":{"clientname":"ruhul yahoo","clientemail":"ruhul080@yahoo.com","clientusername":"ruhulya"},"2":{"clientname":"kaosar yahoo","clientemail":"kaosar080@yahoo.com","clientusername":"kaosar"}},"info2":{"3":{"id":"24","receiver":"5","content":"chat system","time":"2015-08-19 12:09:19"},"4":{"id":"23","receiver":"4","content":"chat system","time":"2015-08-19 12:09:19"}}}');
alert(data.info1[1].clientname);
alert(data.info1[2].clientname);

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>

＆＃13;

通过jquery检索json数据

4 个答案: