Question

假设我有这一系列的推文：

var arr = ['@userone Hey man, whats popping', 'Shoutout to @usertwo haha']

我如何摆脱提及并仅包含消息？所以喜欢：

var newArr = ['Hey man, whats popping', 'Shoutout to haha']

这是我能想到的

if (tweet.includes('@')) {
  var atIndex = tweet.indexOf('@');
  var spaceIndex = // index of the nearest space after @
  var strToReplace = tweet.substring(atIndex, spaceIndex);
  tweet = tweet.replace(strToReplace, '');

}

请帮忙。

Answer 1

＆＃13;

var arr = ['@userone Hey man, whats popping', 
           'Shoutout to @usertwo haha'
          ];
var i = 0, at_pos, sp_pos, str_rep;
while(i<arr.length) {
  at_pos  = arr[i].indexOf('@');
  sp_pos  = arr[i].indexOf(' ', at_pos);
  str_rep = arr[i].substring(at_pos, sp_pos+1);
  arr[i]  = arr[i].replace(str_rep, '');
  console.log(arr[i]);
  i++;
}

＆＃13;

或者只是你可以使用像

这样的正则表达式

＆＃13;

var arr = ['@userone Hey man, whats popping', 
           'Shoutout to @usertwo haha'
          ];
var i=0;
while(i<arr.length){
  arr[i] = arr[i].replace(/\@[^\s]*\s/g, '');
  console.log(arr[i]);
  i++;
}

＆＃13;

Answer 2

arr.map(e => e.replace(/(?:^|\W)@(\w+)(?!\w)/g,"") )

在一行中完成:) ...演示是here

Answer 3

正则表达式是你的朋友。

   var arr = ['@userone Hey man, whats popping', 'Shoutout to @usertwo haha'];

    arr.forEach(function(element,index) {
        arr[index] = element.replace(/@[A-Za-z0-9]*\s/, "");
    }

    console.log(arr);

Answer 4

建立在@Abdennour ...

var newArr = arr.map(e => e.replace(/@\w+\s+/,"") );

JavaScript：如何获取紧跟在另一个特定字符之后的某个字符的出现的索引？

4 个答案: