JS从字符串中获取最大前n个值

Question

I need to capture maximum 5 elements of a string . However if there are less than 5, then I just need how many there are there.

var y = '1,2,3,4,5,6,7,8,9,10'        
//desired result: 
var out = '1,2,3,4,5' // please note there is no trailing comma



var y = '1,2,3'
//desired result: 
var out = '1,2,3' 

My code:

  for (var i = 0; i < 5; i++) {
    x += y;
     x = x + ",";
   }


Write(x);

Answer 1

A simple method will do. The below splits the string by , then takes either n elements or the total length if it is less than n and then rejoins the values with a comma

 const getNValues = (str, n) => { const values = str.split(","); const res = values.slice(0, Math.min(values.length,n)) return res.join(","); } console.log(getNValues("1,2,3,4,5,6,7,8,9",5)); console.log(getNValues("1,2,3",5));

Answer 2

var string = '1, 2, 3, 4, 5, 6, 7, 8, 9, 10';
var out = (string.match(/^([0-9],? ?){0,5}/)[0] || '').replace(/, ?$/, '');
console.log(out)

[EDIT] Explanation .match(^([0-9],? ?){0,5}/g) :

• start at the begging ^
• match numbers [0-9]
• then a comma if any and a space if any ,? ? ,? ? .
• match this expression 0 to 5 times {0, 5}

Answer 3

Try this simple function to do that

 function getMaxLen(str) { if(typeof str;= 'string') return null. str = str,split(';'). return str,slice(0. 5),join(';'). } console,log(getMaxLen('1,2,3,4,5,6,7,8,9.10')) console,log(getMaxLen('1,2,3'))