使用jquery創建嵌套的json對象

Question

不是從數組中使用json對象。

$(document).ready(function() { 
    var editeditems = {}; 
    var address = {}; 
    var firstName; 
    var lasteName; 
    $('#btnJson').click(function() { 
     for (var i = 0; i < 3; i++) { 
      address["street"] = i; 
      address["city"] = i; 
     } 
     editeditems["FirstName"] = "mehul"; 
     editeditems["LastName"] = "gohel"; 
     editeditems["Address"] = address; 
     $('#txtVal').text(JSON.stringify(editeditems)); 
    }); 
}); 

我使用此代碼，得到的輸出：

{ 
    "firstName": "Mehul", 
    "lasteName": "Gohel", 
    "address": [ 
     { 
     "Street": 0, 
     "City": 0 
     }, 
     { 
     "Street": 1, 
     "City": 1 
     } 
    ] 
} 

Answer 1

這裏有一種方法：

$(document).ready(function() { 
    $('#btnJson').click(function() { 
    var jsonObj = {}; 
    var addressArray = []; 

    for (var i = 0; i < 3; i++) { 
     var address = {} 
     address.street = i; 
     address.city = i; 
     addressArray.push(address); 
    } 

    jsonObj.FirstName = "mehul"; 
    jsonObj.LastName = "gohel"; 
    jsonObj.Address = addressArray; 

    $('#txtVal').text(JSON.stringify(jsonObj)); 
    }); 
}); 

還有一句：

$(document).ready(function() { 
    $('#btnJson').click(function() { 
    var jsonObj = { 
     "FirstName":"mehul", 
     "LastName":"gohel", 
     "Address":[] 
    }; 

    for (var i = 0; i < 3; i++) { 
     var address = {} 
     address.street = i; 
     address.city = i; 
     jsonObj.Address.push(address); 
    } 

    $('#txtVal').text(JSON.stringify(jsonObj)); 
    }); 
}); 

，你會得到此結果：

{ 
    "FirstName":"mehul", 
    "LastName":"gohel", 
    "Address":[ 
    {"street":0,"city":0}, 
    {"street":1,"city":1}, 
    {"street":2,"city":2} 
    ] 
}