提取以字母开头的字母数字单词，忽略其他单词

Question

听起来很容易，但我无法弄明白。

/([a-z][a-z0-9]+)/gi

测试用例看起来像这样

Correct #123    2baZ   #1a2 thisToo  $bar  andTwo2

我正在与那些局部比赛作战。唯一有效的应该是：["Correct", "thisToo", "andTwo2"]。任何其他人应该保持无与伦比。

以下是测试人员的链接：http://regex101.com/r/qG7lU9/8

更新：

这是JS小提琴，它比测试者本身更好... http://jsfiddle.net/FredyCr/6hsgef82/

Answer 1

从组索引2中获取匹配的字符串。

(?:^| )([a-z][a-z0-9]+)(?: |$)

DEMO

Javascript代码将是，

> var re = /(?:^| )([a-z][a-z0-9]+)(?: |$)/gi
undefined
> var str = "    Correct #123 2baZ   #1a2 thisToo  $bar  andTwo2";
undefined
> var matches = [];
undefined
> while (match = re.exec(str))
... {
... matches.push(match[1]);
... }
3
> console.log(matches);
[ 'Correct', 'thisToo', 'andTwo2' ]

Answer 2

您可以使用基于前瞻和非捕获组的正则表达式：

(?:^| )([a-z][a-z0-9]+(?= |$))

并使用捕获的组＃1进行匹配：

Correct
thisToo
andTwo2

RegEx Demo

<强>代码：

var rx = /(?:^| )([a-z][a-z0-9]+(?= |$))/gi
var str = "    Correct #123 2baZ   #1a2 thisToo  $bar  andTwo2";

var matches = [];
while (match = rx.exec(str))
    matches.push(match[1]);

console.log(matches);
//=> ["Correct", "thisToo", "andTwo2"]

JsFiddle Demo

Answer 3

var regex = /(^|\s)[a-z][a-z0-9]+/gi;
var text = "Correct #123    2baZ   #1a2 thisToo  $bar  andTwo2";
var found;
while ((found = regex.exec(text)) !== null)
   console.log(found[0].trim());

输出

Correct
thisToo
andTwo2

Answer 4

解决方法：拆分并过滤所需的标记：

var rx = /^[a-zA-Z][a-zA-Z0-9]*$/;
var match = str.split(/\s+/).filter(rx.exec.bind(rx));

Example