Question

我有这个字符串：

Hey I love #apple and #orange and also #banana

我想提取以#符号开头的每个单词。

目前我正在使用此代码实现它：

var last = 0;
var n = 0;
var str = "Hey I love #apple and #orange and also #banana";
do{
    n = str.indexOf("#", last);
    if(n != -1){
        //The code found the # char at position 'n'
        last = n+1; //saving the last found position for next loop

        //I'm using this to find the end of the word
        var suffixArr = [' ', '#'];
        var e = -1;
        for(var i = 0; i < suffixArr.length;i++){
            if(str.indexOf(suffixArr[i], n) != -1){
               e = str.indexOf(suffixArr[i], n+1);
               break;
            }
        }
        if(e == -1){
            //Here it could no find banana because there isn't any white space or # after
            e = str.length; //this is the only possibility i've found
        }

        //extracting the word from the string
        var word = str.substr(n+1, (e-1)-n);
   }
}
while (n != -1);

如何才能设法找到以＃开头且仅以a-Z characters开头的字词。如果我有#apple!我应该能够提取apple 而且，正如我在代码中提到的，如果它出现在字符串的末尾，我如何设法获取该字

Answer 1

(?:^|[ ])#([a-zA-Z]+)

试试这个。抓住捕获。参见演示。

https://regex101.com/r/wU7sQ0/18

    var re = /(?:^|[ ])#([a-zA-Z]+)/gm;
var str = 'Hey I love #apple and #orange and #apple!@ also #banana';
var m;

while ((m = re.exec(str)) != null) {
if (m.index === re.lastIndex) {
re.lastIndex++;
}
// View your result using the m-variable.
// eg m[0] etc.
}

Answer 2

您可以使用正则表达式/(^|\s)#[a-z]+/i和match，然后使用Array.join（此处正在执行+ ""时在内部使用）并替换所有{来自形成的字符串的{1}}并在#

上拆分

如果您还想在var arr = (str.match(/(^|\s)#[a-z]+/i)+"").replace(/#/g,"").split(",");中匹配test，请从正则表达式中删除Some#test。

Javascript：从字符串中的特定字符开始提取单词

2 个答案: