替换字符串中所有出现的内容，将数组作为参数

Question

Is there a way to replace all occurrences in string giving an array as a parameter in replace() method.

For example:

Having this string: "ABCDEFG"

And having this array: ['A','D','F']

It is possible to replace the same letters in the string with something else? Something like:

"ABCDEFG".replace(['A','D','F'], '')

So the final result be: "BCEG"

Answer 1

You can loop through your array:

 var str = "ABCDEFG"; ['A','D','F'].forEach(c => str = str.replace(c, '*')) console.log(str); 

Answer 2

No. You will have to iterate through your array of replacements, calling replace() for every individual item in the array I'm afraid. Alternatively, you could try to formulate your array of individual strings as a regular expression, eg

"ABCDEFG".replace(/(A|D|F)/g, '')

Please note, though, that depending on your array and the length of strings in it, this may be considerably less efficient than a number of replace calls.

Answer 3

的种类：

 "ABCDEFG".replace(new RegExp(['A','D','F'].join("|")), '')

Answer 4

There is actually a way to do this from an Array.

You'll need to create a RegEx dynamically :

 let arr = ['A','D','F']; let expression = arr.join('|'); let rx = new RegExp(expression, 'g'); console.log("ABCDEFG".replace(rx,'')); 

Answer 5

If you want an array as input:

'ABCDEF'.replace(new RegExp(['A','D','F'].join('|'), 'g'), '')

By using the 'g' flag, it will replace all occurrences of 'A', 'D' or 'F' in the string.

You could also do this in a simpler way:

'ABCDEF'.replace(/A|D|F/g, '')

Answer 6

Here's a general function that uses regex similar to the other answers, but allows you to pass in whatever replacement string you like:

 const str = 'ABCDEFG'; const arr = ['A', 'D', 'F']; function replace(str, arr, r) { const regex = new RegExp(arr.join('|'), 'g'); return str.replace(regex, p => r); } console.log(replace(str, arr, '')); console.log(replace(str, arr, 'bob'));