[英]regular expression to match exactly 5 digits
testing= testing.match(/(\d{5})/g);
I'm reading a full html into variable. 我正在阅读完整的html变量。 From the variable, want to grab out all numbers with the pattern of exactly 5 digits. 从变量中,要截取所有具有精确5位数字的数字。 No need to care of whether before/after this digit having other type of words. 无需关心此数字之前/之后是否具有其他类型的单词。 Just want to make sure whatever that is 5 digit numbers been grabbed out. 只是要确保抓取到5位数字。
However, when I apply it, it not only pull out number with exactly 5 digit, number with more than 5 digits also retrieved... 但是,当我应用它时,它不仅提取出准确的5位数字,而且还检索了超过5位的数字...
I had tried putting ^
in front and $
behind, but it making result come out as null. 我曾尝试将^
放在前面,将$
放在后面,但是结果却为null。
I am reading a text file and want to use regex below to pull out numbers with exactly 5 digit, ignoring alphabets. 我正在阅读一个文本文件,并希望使用下面的正则表达式来提取5位数字,而忽略字母。
Try this... 尝试这个...
var str = 'f 34 545 323 12345 54321 123456',
matches = str.match(/\b\d{5}\b/g);
console.log(matches); // ["12345", "54321"]
The word boundary \\b
is your friend here. 边界\\b
是您的朋友。
My regex will get a number like this 12345
, but not like a12345
. 我的正则表达式将得到一个像这样的数字12345
,而不是像a12345
这样的a12345
。 The other answers provide great regexes if you require the latter. 如果您需要后者,其他答案也提供了很好的正则表达式。
My test string for the following: 我的测试字符串如下:
testing='12345,abc,123,54321,ab15234,123456,52341';
If I understand your question, you'd want ["12345", "54321", "15234", "52341"]
. 如果我理解您的问题,则需要["12345", "54321", "15234", "52341"]
。
If JS engines supported regexp lookbehinds, you could do: 如果JS引擎支持正则表达式后退,则可以执行以下操作:
testing.match(/(?<!\d)\d{5}(?!\d)/g)
Since it doesn't currently, you could: 由于目前还没有,您可以:
testing.match(/(?:^|\D)(\d{5})(?!\d)/g)
and remove the leading non-digit from appropriate results, or: 并从适当的结果中删除前导的非数字,或:
pentadigit=/(?:^|\D)(\d{5})(?!\d)/g;
result = [];
while (( match = pentadigit.exec(testing) )) {
result.push(match[1]);
}
Note that for IE, it seems you need to use a RegExp stored in a variable rather than a literal regexp in the while
loop, otherwise you'll get an infinite loop. 请注意,对于IE,似乎您需要使用存储在变量中的RegExp而不是while
循环中的文字regexp,否则您将获得无限循环。
This should work: 这应该工作:
<script type="text/javascript">
var testing='this is d23553 test 32533\n31203 not 333';
var r = new RegExp(/(?:^|[^\d])(\d{5})(?:$|[^\d])/mg);
var matches = [];
while ((match = r.exec(testing))) matches.push(match[1]);
alert('Found: '+matches.join(', '));
</script>
what is about this? 这是什么 \\D(\\d{5})\\D
This will do on: 这将在:
f 23 23453 234 2344 2534 hallo33333 "50000" f 23 23453 234 2344 2534 hallo33333“ 50000”
23453, 33333 50000 23453,33333 50000
No need to care of whether before/after this digit having other type of words 无需担心此数字之前/之后是否还有其他类型的单词
To just match the pattern of 5 digits number anywhere in the string, no matter it is separated by space or not, use this regular expression (?<!\\d)\\d{5}(?!\\d)
. 要仅匹配字符串中任意位置的5位数字的模式,无论是否用空格将其分隔,请使用此正则表达式(?<!\\d)\\d{5}(?!\\d)
。
Sample JavaScript codes: 示例JavaScript代码:
var regexp = new RegExp(/(?<!\d)\d{5}(?!\d)/g);
var matches = yourstring.match(regexp);
if (matches && matches.length > 0) {
for (var i = 0, len = matches.length; i < len; i++) {
// ... ydo something with matches[i] ...
}
}
Here's some quick results. 这是一些快速的结果。
abc12345xyz (✓) abc12345xyz(✓)
12345abcd (✓) 12345abcd(✓)
abcd12345 (✓) abcd12345(✓)
0000aaaa2 (✖) 0000aaaa2(✖)
a1234a5 (✖) a1234a5(✖)
12345 (✓) 12345(✓)
<space>
12345 <space>
12345 (✓✓) <space>
12345 <space>
12345(✓✓)
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