Question

我有一个名为list_to_book_for的有效json对象：

{"fold1-key1":{"name":"Van Antwerpen"},"fold2-key2":{"name":"Delporte"},"fold3-key3":{"name":"Lallemand"}}

我想在此对象中添加变量的内容：

"fold1":
    {
        "name": "Too book for ",
        "items":
        {

        }
    },

我想将变量插入到项目中吗？

Answer 1

当然可以

var x = 'your items';

var obj = { fold1: { name: "Too book for ", items: {} } };

obj.fold1.items = x;

Answer 2

当然：

var jsonObj = {"fold1": {
    "name": "Too book for ", 
    "items": {} 
}}; // or do JSON.parse on your original json
var items = {"fold1-key1":{"name":"Van Antwerpen"},"fold2-key2":{"name":"Delporte"},"fold3-key3":{"name":"Lallemand"}};

jsonObj.fold1.items = items;
var result = JSON.stringify(jsonObj);

将返回：

{"fold1":
    {"name":"Too book for ",
     "items":
         {"fold1-key1":{"name":"Van Antwerpen"},
          "fold2-key2":{"name":"Delporte"},
          "fold3-key3":{"name":"Lallemand"}
          }
    }
}

Answer 3

也许这就是你想要做的事情：

for( var key in list_to_book_for ) {
  if( list_to_book_for.hasOwnProperty(key) ){
     obj.fold1.items[key] = list_to_book_for[key];
  }
}

Answer 4

如果您的JSON对象仍然采用原始JSON格式，请执行以下操作：

list_to_book_for = '{"fold1-key1":{"name":"Van Antwerpen"},"fold2-key2":{"name":"Delporte"},"fold3-key3":{"name":"Lallemand"}}';

然后，您可以使用JSON.parse()

将JSON解析为javascript

"fold1": {
    "name": "To book for ",
    "items": JSON.parse(list_to_book_for)
},

Answer 5

谢谢大家的帮助，这是解决方案：

我在代码的顶部声明了一个全局变量：

    chaine_book_for = {};

    chaine_book_for = function()
    {
        var tmp;
        $.ajax(
        {
            method:'POST',
            dataType: 'json',
            async: false,
            url:'ajax/liste_users.php',
                                                                                    success:function(retour_php)
    {
                                                                                                                chaine_book_for = retour_php;
                                                                                                                tmp = retour_php;
    }
});
return tmp;
}();

重要的是将 async：false

我在子菜单（json对象）中插入内容：

"fold1":
{
    "name": "Too book for ",
    "items": chaine_book_for,
    callback: function(key, options)

Answer 6

尝试使用eval，对于此代码工作正常，请查看

 <script>
var obj={"fold1-key1":{"name":"Van Antwerpen"},"fold2-key2":{"name":"Delporte"},"fold3-key3":{"name":"Lallemand"}};
var s='{"fold1": {"name": "Too book for ","items": { } }}';
eval('var u='+s);
for(x in u)
{
    obj[x]=u[x];
}
</script>

将js变量插入json对象

6 个答案: