我有一个JSON对象,其中包含以下内容:
[
{
"_id":"5078c3a803ff4197dc81fbfb",
"email":"user1@gmail.com",
"image":"some_image_url",
"name":"Name 1"
},
{
"_id":"5078c3a803ff4197dc81fbfc",
"email":"user2@gmail.com",
"image":"some_image_url",
"name":"Name 2"
}
]
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我想将"_id"键更改为"id",这样就可以了
[
{
"id":"5078c3a803ff4197dc81fbfb",
"email":"user1@gmail.com",
"image":"some_image_url",
"name":"Name 1"
},
{
"id":"5078c3a803ff4197dc81fbfc",
"email":"user2@gmail.com",
"image":"some_image_url",
"name":"Name 2"
}
]
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我怎么能用Javascript,jQuery或Ruby,Rails做到这一点?
谢谢.
Pau*_*aul 86
var json = '[{"_id":"5078c3a803ff4197dc81fbfb","email":"user1@gmail.com","image":"some_image_url","name":"Name 1"},{"_id":"5078c3a803ff4197dc81fbfc","email":"user2@gmail.com","image":"some_image_url","name":"Name 2"}]';
var obj = JSON.parse(json)[0];
obj.id = obj._id;
delete obj._id;
json = JSON.stringify([obj]);
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这会将json更改为:
[{"email":"user1@gmail.com","image":"some_image_url","name":"Name 1","id":"5078c3a803ff4197dc81fbfb"}]
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由于顺序在JSON中的名称/值对中没有意义,因此这与您想要的结果相同.
eva*_*nal 25
在这种情况下,最简单的方法是使用字符串替换.序列化JSON将无法正常工作,因为_id将成为对象的属性名称,更改属性名称不是一项简单的任务(至少在大多数语言中都没有,在javascript中也不是那么糟糕).而只是做;
jsonString = jsonString.replace("\"_id\":", "\"id\":");
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Str*_*ger 16
正如evanmcdonnal所提到的,最简单的解决方案是将其作为字符串而不是JSON处理,
var json = [{"_id":"5078c3a803ff4197dc81fbfb","email":"user1@gmail.com","image":"some_image_url","name":"Name 1"},{"_id":"5078c3a803ff4197dc81fbfc","email":"user2@gmail.com","image":"some_image_url","name":"Name 2"}];
json = JSON.parse(JSON.stringify(json).split('"_id":').join('"id":'));
document.write(JSON.stringify(json));Run Code Online (Sandbox Code Playgroud)
这会将给定的JSON数据转换为字符串,并将"_id"替换为"id",然后将其转换回所需的JSON格式.但我使用split而join不是replace,因为replace只会替换第一次出现的字符串.
如果你想重命名某个键的所有出现,你可以使用带有 g 选项的正则表达式。例如:
var json = [{"_id":"1","email":"user1@gmail.com","image":"some_image_url","name":"Name 1"},{"_id":"2","email":"user2@gmail.com","image":"some_image_url","name":"Name 2"}];
str = JSON.stringify(json);
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现在我们在str中有字符串格式的json。
使用带有g选项的正则表达式将所有出现的“_id”替换为“id” :
str = str.replace(/\"_id\":/g, "\"id\":");
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并返回json格式:
json = JSON.parse(str);
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现在我们有了我们想要的密钥名称的 json。
JSON.parse有两个参数。第二个参数reviver是一个转换函数,可以格式化输出我们想要的格式。请参阅此处的ECMA 规范。
在复活功能中:
this 是包含作为此函数处理的属性的对象,属性名称作为字符串,属性值作为此函数的参数。const json = '[{"_id":"5078c3a803ff4197dc81fbfb","email":"user1@gmail.com","image":"some_image_url","name":"Name 1"},{"_id":"5078c3a803ff4197dc81fbfc","email":"user2@gmail.com","image":"some_image_url","name":"Name 2"}]';
const obj = JSON.parse(json, function(k, v) {
if (k === "_id") {
this.id = v;
return; # if return undefined, orignal property will be removed
}
return v;
});
const res = JSON.stringify(obj);
console.log(res)
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输出:
[{"email":"user1@gmail.com","image":"some_image_url","name":"Name 1","id":"5078c3a803ff4197dc81fbfb"},{"email":"user2@gmail.com","image":"some_image_url","name":"Name 2","id":"5078c3a803ff4197dc81fbfc"}]
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尝试这个:
let jsonArr = [
{
"_id":"5078c3a803ff4197dc81fbfb",
"email":"user1@gmail.com",
"image":"some_image_url",
"name":"Name 1"
},
{
"_id":"5078c3a803ff4197dc81fbfc",
"email":"user2@gmail.com",
"image":"some_image_url",
"name":"Name 2"
}
]
let idModified = jsonArr.map(
obj => {
return {
"id" : obj._id,
"email":obj.email,
"image":obj.image,
"name":obj.name
}
}
);
console.log(idModified);
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