Question

如何使用jQuery通过以下JSON转储从 fields 键获取所有值：

{
  "fields": [
    {
      "label": "Name",
      "field_type": "text",
      "required": true,
      "field_options": {
        "size": "medium"
      },
      "cid": "c1"
    }, {
      "label": "Email",
      "field_type": "text",
      "required": true,
      "field_options": {
        "size": "medium"
      },
      "cid": "c2"
    }
  ]
}

我想要的只是获得：

{
  "label": "Name",
  "field_type": "text",
  "required": true,
  "field_options": {
    "size": "medium"
  },
  "cid": "c1"
}, {
  "label": "Email",
  "field_type": "text",
  "required": true,
  "field_options": {
    "size": "medium"
  },
  "cid": "c2"
}

Answer 1

将主json存储到变量中。您可以使用以下方式访问其密钥：

variable.fields

Answer 2

var originalJson = '{"fields":[{"label":"Name","field_type":"text","required":true,"field_options":{"size":"medium"},"cid":"c1"},{"label":"Email","field_type":"text","required":true,"field_options":{"size":"medium"},"cid":"c2"}]}'

var fieldsJson = JSON.parse(originalJson).fields;

如果你想将它作为json字符串添加

var fieldsJsonString = JSON.stringify(fieldsJson)

Answer 3

var o = {"fields":[{"label":"Name","field_type":"text","required":true,"field_options":{"size":"medium"},"cid":"c1"},{"label":"Email","field_type":"text","required":true,"field_options":{"size":"medium"},"cid":"c2"}]};
var labels = o.fields

然后，您可以通过for。

访问标签

for(var i = 0; i < labels.length; i++){
    console.log(labels[i]);
}

Answer 4

当我使用

时

var obj = jQuery.parseJSON(payload);
     console.log(obj.fields);

我明白了：

[Object {label =“Email”，field_type =“text”，required = true，more ...}，Object {label =“Name”，field_type =“text”，required = true，more ... }，Object {label =“Subject”，field_type =“text”，required = true，more ...}，Object {label =“Message”，field_type =“paragraph”，required = true，more ...}]

如何获得：

{"label":"Name","field_type":"text","required":true,"field_options":{"size":"medium"},"cid":"c1"},{"label":"Email","field_type":"text","required":true,"field_options":{"size":"medium"},"cid":"c2"}

从JSON Key获取所有值（jQuery）

4 个答案: