admin管理员组文章数量:1416082
Here I want to convert object properties to ma separated values like following- Join the elements of an array into a string:
var fruits = {"f1":"Banana", "f2":"Orange", "f3":"Apple","f4":"Mango"};
var energy = Object.keys(fruits).map(function(k){return fruits[k]}).join(",");
The result of energy will be:
Banana,Orange,Apple,Mango
But When I apply like this-
var fruits = {"f1":"Banana", "f2":"null", "f3":"Apple","f4":"Mango"};
var energy = Object.keys(fruits).map(function(k){return fruits[k]}).join(",");
The result of energy is like this:
Banana,,Apple,Mango
And I want result like this-
Banana,null,Apple,Mango
I have tried these links but No luck found.
Easy way to turn properties of Javascript object into ma-separated list?
Updated the Question
var fruits = {"f1":"Banana", "f2":"", "f3":"Apple","f4":"Mango"};
var energy = Object.keys(fruits).map(function(k){return fruits[k]}).join(",");
The result of energy is like this:
Banana,,Apple,Mango
I want result like this-
Banana,null,Apple,Mango
Here I want to convert object properties to ma separated values like following- Join the elements of an array into a string:
var fruits = {"f1":"Banana", "f2":"Orange", "f3":"Apple","f4":"Mango"};
var energy = Object.keys(fruits).map(function(k){return fruits[k]}).join(",");
The result of energy will be:
Banana,Orange,Apple,Mango
But When I apply like this-
var fruits = {"f1":"Banana", "f2":"null", "f3":"Apple","f4":"Mango"};
var energy = Object.keys(fruits).map(function(k){return fruits[k]}).join(",");
The result of energy is like this:
Banana,,Apple,Mango
And I want result like this-
Banana,null,Apple,Mango
I have tried these links but No luck found.
Easy way to turn properties of Javascript object into ma-separated list?
Updated the Question
var fruits = {"f1":"Banana", "f2":"", "f3":"Apple","f4":"Mango"};
var energy = Object.keys(fruits).map(function(k){return fruits[k]}).join(",");
The result of energy is like this:
Banana,,Apple,Mango
I want result like this-
Banana,null,Apple,Mango
Share
Improve this question
edited Nov 10, 2017 at 6:46
sapics
1,1248 silver badges22 bronze badges
asked Nov 10, 2017 at 6:20
Vikas GuptaVikas Gupta
1,2311 gold badge12 silver badges27 bronze badges
8
-
3
Is it
"null"
ornull
? ( and you may useObject.values(fruits)
...) – Jonas Wilms Commented Nov 10, 2017 at 6:22 - 1 It works for me. – castletheperson Commented Nov 10, 2017 at 6:24
-
If it’s actually
null
–return String(fruits[k])
. – Ry- ♦ Commented Nov 10, 2017 at 6:24 - 1 Possible duplicate of How to convert key-value pair object into an array of values in ES6? – Shubham Khatri Commented Nov 10, 2017 at 6:26
- 1 Sorry, I have updated the question. Please Check Jonas Ryan Killer Death – Vikas Gupta Commented Nov 10, 2017 at 6:31
3 Answers
Reset to default 6Using Object.values
would be easier.
// for "null"
var fruits = {"f1":"Banana", "f2":"null", "f3":"Apple","f4":"Mango"};
var energy = Object.values(fruits).join(",");
console.log(energy);
// for null
var fruits = {"f1":"Banana", "f2":null, "f3":"Apple","f4":"Mango"};
var energy = Object.values(fruits).map(String).join(",");
console.log(energy);
// for ""
var fruits = {"f1":"Banana", "f2":"", "f3":"Apple","f4":"Mango"};
var energy = Object.values(fruits).map(v=>v===""?"null":String(v)).join(",");
console.log(energy);
var fruits = {"f1":"Banana", "f2":"", "f3":"Apple","f4":"Mango"};
var energy = Object.keys(fruits).map(function(k){
if(fruits [k] ==="") return "null";
return fruits [k]
}).join(",");
about join()
,not show value null
as string "null"
, you can check it in map()
same as below :
const fruits = {"f1":"Banana", "f2":"", "f3":"Apple","f4":"Mango"};
const energy = Object.values(fruits).map(function(value){
return !value ? "null" : value;
}).join(",");
console.log(energy);
本文标签: converting object properties to comma seperated values in JavascriptStack Overflow
版权声明:本文标题:converting object properties to comma seperated values in Javascript - Stack Overflow 内容由网友自发贡献,该文观点仅代表作者本人, 转载请联系作者并注明出处:http://www.betaflare.com/web/1745244583a2649498.html, 本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌抄袭侵权/违法违规的内容,一经查实,本站将立刻删除。
发表评论