我有一个 json 对象,如下所示。我想使用下面的代码删除“otherIndustry”条目及其值,但该代码不起作用。
var updatedjsonobj = delete myjsonobj['otherIndustry'];
如何删除 Json 对象特定的键及其值。 下面是我的示例 json 对象,我想在其中删除“otherIndustry”键及其值。
var myjsonobj = {
"employeeid": "160915848",
"firstName": "tet",
"lastName": "test",
"email": "[email protected]",
"country": "Brasil",
"currentIndustry": "aaaaaaaaaaaaa",
"otherIndustry": "aaaaaaaaaaaaa",
"currentOrganization": "test",
"salary": "1234567"
};
delete myjsonobj ['otherIndustry'];
console.log(myjsonobj);
日志仍然打印相同的对象,而不从对象中删除“otherIndustry”条目。
delete
运算符用于 remove
对象 property
。
delete
运算符不返回新对象,仅返回boolean
:true或false。
另一方面,解释器执行
var updatedjsonobj = delete myjsonobj['otherIndustry'];
后,updatedjsonobj
变量将存储一个 boolean
值。
如何删除Json对象特定的键及其值?
您只需要知道属性名称即可将其从对象的属性中删除。
delete myjsonobj['otherIndustry'];
let myjsonobj = {
"employeeid": "160915848",
"firstName": "tet",
"lastName": "test",
"email": "[email protected]",
"country": "Brasil",
"currentIndustry": "aaaaaaaaaaaaa",
"otherIndustry": "aaaaaaaaaaaaa",
"currentOrganization": "test",
"salary": "1234567"
}
delete myjsonobj['otherIndustry'];
console.log(myjsonobj);
如果您想在知道值时删除
key
,可以使用 Object.keys
函数,该函数返回给定对象自己的可枚举属性的数组。
let value="test";
let myjsonobj = {
"employeeid": "160915848",
"firstName": "tet",
"lastName": "test",
"email": "[email protected]",
"country": "Brasil",
"currentIndustry": "aaaaaaaaaaaaa",
"otherIndustry": "aaaaaaaaaaaaa",
"currentOrganization": "test",
"salary": "1234567"
}
Object.keys(myjsonobj).forEach(function(key){
if (myjsonobj[key] === value) {
delete myjsonobj[key];
}
});
console.log(myjsonobj);
有几种方法可以做到这一点,让我们一一看看:
const myObject = {
"employeeid": "160915848",
"firstName": "tet",
"lastName": "test",
"email": "[email protected]",
"country": "Brasil",
"currentIndustry": "aaaaaaaaaaaaa",
"otherIndustry": "aaaaaaaaaaaaa",
"currentOrganization": "test",
"salary": "1234567"
};
delete myObject['currentIndustry'];
// OR delete myObject.currentIndustry;
console.log(myObject);
let myObject = {
"employeeid": "160915848",
"firstName": "tet",
"lastName": "test",
"email": "[email protected]",
"country": "Brasil",
"currentIndustry": "aaaaaaaaaaaaa",
"otherIndustry": "aaaaaaaaaaaaa",
"currentOrganization": "test",
"salary": "1234567"
};
myObject.currentIndustry = undefined;
myObject = JSON.parse(JSON.stringify(myObject));
console.log(myObject);
const myObject = {
"employeeid": "160915848",
"firstName": "tet",
"lastName": "test",
"email": "[email protected]",
"country": "Brasil",
"currentIndustry": "aaaaaaaaaaaaa",
"otherIndustry": "aaaaaaaaaaaaa",
"currentOrganization": "test",
"salary": "1234567"
};
const {currentIndustry, ...filteredObject} = myObject;
console.log(filteredObject);
或者如果你可以使用 underscore js 库的 omit():
const filteredObject = _.omit(currentIndustry, 'myObject');
console.log(filteredObject);
什么时候用什么??
如果您不想创建新的过滤对象,只需选择选项 1 或 2。确保使用 let 定义对象,同时使用第二个选项,因为我们要覆盖这些值。或者您可以使用其中任何一个。
希望这有帮助:)
按照这个,它可以像你所看到的那样:
var obj = {
Objone: 'one',
Objtwo: 'two'
};
var key = "Objone";
delete obj[key];
console.log(obj); // prints { "objtwo": two}
这里还有一个例子。 (查看参考)
const myObject = {
"employeeid": "160915848",
"firstName": "tet",
"lastName": "test",
"email": "[email protected]",
"country": "Brasil",
"currentIndustry": "aaaaaaaaaaaaa",
"otherIndustry": "aaaaaaaaaaaaa",
"currentOrganization": "test",
"salary": "1234567"
};
const {otherIndustry, ...otherIndustry2} = myObject;
console.log(otherIndustry2);
.as-console-wrapper {
max-height: 100% !important;
top: 0;
}
function omit(obj, key) {
const {[key]:ignore, ...rest} = obj;
return rest;
}
您可以像这样使用 ES6 扩展运算符。要删除您的钥匙,只需致电
const newJson = omit(myjsonobj, "otherIndustry");
在 javascript 中处理
type=object
时,如果保持纯函数总是更好。
我在尝试删除返回的 JSON 对象时遇到问题,发现它实际上是一个字符串。如果您在删除之前使用 JSON.parse() ,则可以确定您的密钥将被删除。
let obj;
console.log(this.getBody()); // {"AED":3.6729,"AZN":1.69805,"BRL":4.0851}
obj = this.getBody();
delete obj["BRL"];
console.log(obj) // {"AED":3.6729,"AZN":1.69805,"BRL":4.0851}
obj = JSON.parse(this.getBody());
delete obj["BRL"];
console.log(obj) // {"AED":3.6729,"AZN":1.69805}
如果您想重新分配一个值(或2个值),您可以使用@Mohammed Amir Ansari的答案并像这样修改它:
const myCarObject= { manufacturer: 'BMW',model: "M4", hp:503 };
const { model, hp, ...manufacturer} = myCarObject;
console.log(manufacturer)
//OUTPUT:
//{ manufacturer: 'BMW'}
console.log(model)
//OUTPUT:
//"M4"
console.log(hp)
//OUTPUT:
//503
这将从汽车对象中提取值并将它们分配给其他变量。
聚会有点晚了,希望对某人有帮助。