당신이 같은 문자열을 가지고 있다면 다음 foo: 1, bar: 2을 사용하여 유효한 obj로 변환 할 수 있습니다 :
str
.split(',')
.map(x => x.split(':').map(y => y.trim()))
.reduce((a, x) => {
a[x[0]] = x[1];
return a;
}, {});
#javascript의 niggler에게 감사드립니다.
설명으로 업데이트하십시오.
const obj = 'foo: 1, bar: 2'
.split(',') // split into ['foo: 1', 'bar: 2']
.map(keyVal => { // go over each keyVal value in that array
return keyVal
.split(':') // split into ['foo', '1'] and on the next loop ['bar', '2']
.map(_ => _.trim()) // loop over each value in each array and make sure it doesn't have trailing whitespace, the _ is irrelavent because i'm too lazy to think of a good var name for this
})
.reduce((accumulator, currentValue) => { // reduce() takes a func and a beginning object, we're making a fresh object
accumulator[currentValue[0]] = currentValue[1]
// accumulator starts at the beginning obj, in our case {}, and "accumulates" values to it
// since reduce() works like map() in the sense it iterates over an array, and it can be chained upon things like map(),
// first time through it would say "okay accumulator, accumulate currentValue[0] (which is 'foo') = currentValue[1] (which is '1')
// so first time reduce runs, it starts with empty object {} and assigns {foo: '1'} to it
// second time through, it "accumulates" {bar: '2'} to it. so now we have {foo: '1', bar: '2'}
return accumulator
}, {}) // when there are no more things in the array to iterate over, it returns the accumulated stuff
console.log(obj)
혼란스러운 MDN 문서 :
데모: http://jsbin.com/hiduhijevu/edit?js,console
함수:
const str2obj = str => {
return str
.split(',')
.map(keyVal => {
return keyVal
.split(':')
.map(_ => _.trim())
})
.reduce((accumulator, currentValue) => {
accumulator[currentValue[0]] = currentValue[1]
return accumulator
}, {})
}
console.log(str2obj('foo: 1, bar: 2')) // see? works!