Javascript 如何删除json对象键和值。？
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how to remove json object key and value.?
提问by krish kr
I have a json object as shown below. where i want to delete the "otherIndustry" entry and its value by using below code which doesn't worked.
我有一个 json 对象，如下所示。我想通过使用以下不起作用的代码删除“otherIndustry”条目及其值。
var updatedjsonobj = delete myjsonobj['otherIndustry'];
How to remove Json object specific key and its value. Below is my example json object where i want to remove "otherIndustry" key and its value.
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);
where the log still prints the same object without removing 'otherIndustry' entry from the object.
日志仍然打印相同的对象而不从对象中删除“otherIndustry”条目。
回答by Mihai Alexandru-Ionut
deleteoperator is used to removean object property.
delete运算符用于remove对象property。
deleteoperator does notreturns the new object, only returns a boolean: trueor false.
delete运算符不返回新对象，只返回一个boolean: true或false。
In the other hand, after interpreter executes var updatedjsonobj = delete myjsonobj['otherIndustry'];, updatedjsonobjvariable will store a booleanvalue.
另一方面，解释器执行后var updatedjsonobj = delete myjsonobj['otherIndustry'];，updatedjsonobj变量将存储一个boolean值。
How to remove Json object specific key and its value ?
如何删除 Json 对象特定的键及其值？
You just need to know the property name in order to delete it from the object's properties.
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);If you want to remove a keywhen you know the value you can use Object.keysfunction which returns an array of a given object's own enumerable properties.
如果您想key在知道值时删除 a ，您可以使用Object.keys返回给定对象自己的可枚举属性数组的函数。
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);回答by Mohammed Amir Ansari
There are several ways to do this, lets see them one by one:
有几种方法可以做到这一点，让我们一一看：
- deletemethod: The most common way
- 删除方法：最常用的方法
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);- By making key value undefined: Alternate & a faster way:
- 通过使键值未定义：替代和更快的方法：
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);- With es6 spreadOperator:
- 使用 es6扩展运算符：
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);Or if you can use omit()of underscorejs library:
或者，如果你可以用略（）的下划线JS库：
const filteredObject = _.omit(currentIndustry, 'myObject');
console.log(filteredObject);
When to use what??
什么时候用什么？？
If you don't wanna create a new filtered object, simply go for either option 1 or 2. Make sure you define your object with letwhile going with the second option as we are overriding the values. Or else you can use any of them.
如果您不想创建新的过滤对象，只需选择选项 1 或 2。确保在使用第二个选项时使用let定义对象，因为我们将覆盖值。否则，您可以使用其中任何一个。
hope this helps :)
希望这可以帮助 ：）
回答by Muayyad Ayesh
Follow this, it can be like what you are looking:
按照这个，它可以像你正在寻找的那样：
var obj = {
Objone: 'one',
Objtwo: 'two'
};
var key = "Objone";
delete obj[key];
console.log(obj); // prints { "objtwo": two}回答by Penny Liu
Here is one more example. (check the reference)
这里还有一个例子。（检查参考）
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;
}回答by Nick Gooch
I had issues with trying to delete a returned JSON object and found that it was actually a string. If you JSON.parse() before deleting you can be sure your key will get deleted.
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 JSON.parse() ，您可以确保您的密钥将被删除。
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}
