正则表达式排除以哈希标记开头的单词(Regex to exclude words beginning with hash tag)

正则表达式排除以哈希标记开头的单词(Regex to exclude words beginning with hash tag)

我正在寻求帮助，以提出一个正则表达式，它将匹配票号字（任意数字后跟数字），而不是以#SN-开头的字符。 有任何想法吗？

示例字符串：

A string with has ref like #SN-INC0000058 and simple mention like INC0000059.

RegEx比赛： INC0000059

我目前停留在这个与'＃SN-'后面的数字匹配的分类，而不是排除整个单词/字符串。 我怀疑 - 被视为一个单词突破......任何想法？

/(?!#SN-)([A-Z]+[0-9]+)\b/g

https://www.regexpal.com/?fam=99443

I am looking for assistance in coming up with a regular expression which will match ticket number words (any number of characters followed by digits) but not ones that begin with #SN-. Any ideas?

Example String:

A string with has ref like #SN-INC0000058 and simple mention like INC0000059.

RegEx Matches: INC0000059

I am currently stuck at this breakdown which is matching the number after '#SN-' rather than excluding the entire word/string. I suspect - is being treated as a word break... any ideas?

/(?!#SN-)([A-Z]+[0-9]+)\b/g

https://www.regexpal.com/?fam=99443

最满意答案

由于Javascript正则表达式不具备lookbehind功能，因此无法通过简单模式捕捉您想要的内容。 但是，在替换上下文中，可以使用函数作为替换参数和模式来系统地尝试捕获不想要的部分，从而可以轻松地处理此问题：

var result = yourstr.replace(/(#SN-)?\b[A-Z]+[0-9]+\b/g, function(m,g1) { return g1 ? m : 'yourreplacement'; });

当捕获组1被定义时，函数返回整个匹配，否则返回替换字符串。

请注意，在函数内部，您无法使用$&或$1..$n占位符作为整个匹配或捕获组。 您必须使用函数参数来计算它们。

---

只有更严格，如果在目标之前描述所有可能性，也可以在没有回调函数的情况下完成，但是写入会很痛苦，并且效率不高，因为模式以替代方式开始：

var result = yourstring.replace(/(^|[^-]|(?:[^N]|^)-|(?:[^S]|^)N-|(?:^|[^#])SN-)\b[A-Z]+[0-9]+\b/g, '$1yourreplacement');

Since Javascript regex doesn't have the lookbehind feature, there's no way to catch what you want with a simple pattern. However, in a replacement context, you can easily handle this using a function as replacement parameter and a pattern that systematically tries to catch the part you don't want:

var result = yourstr.replace(/(#SN-)?\b[A-Z]+[0-9]+\b/g, function(m,g1) { return g1 ? m : 'yourreplacement'; });

When the capture group 1 is defined, the function returns the whole match, otherwise it returns the replacement string.

Note that inside the function you can't use the $& or $1..$n placeholders for the whole match or capture groups. You have to use the function parameters to figure them.

---

Only to be more rigorous, it can also be done without a callback function if you describe all possibilities before your target, but it's a pain to write and it isn't efficient since the pattern starts with an alternation:

var result = yourstring.replace(/(^|[^-]|(?:[^N]|^)-|(?:[^S]|^)N-|(?:^|[^#])SN-)\b[A-Z]+[0-9]+\b/g, '$1yourreplacement');正则表达式排除以哈希标记开头的单词(Regex to exclude words beginning with hash tag)

我正在寻求帮助，以提出一个正则表达式，它将匹配票号字（任意数字后跟数字），而不是以#SN-开头的字符。 有任何想法吗？

示例字符串：

A string with has ref like #SN-INC0000058 and simple mention like INC0000059.

RegEx比赛： INC0000059

我目前停留在这个与'＃SN-'后面的数字匹配的分类，而不是排除整个单词/字符串。 我怀疑 - 被视为一个单词突破......任何想法？

/(?!#SN-)([A-Z]+[0-9]+)\b/g

https://www.regexpal.com/?fam=99443

I am looking for assistance in coming up with a regular expression which will match ticket number words (any number of characters followed by digits) but not ones that begin with #SN-. Any ideas?

Example String:

A string with has ref like #SN-INC0000058 and simple mention like INC0000059.

RegEx Matches: INC0000059

I am currently stuck at this breakdown which is matching the number after '#SN-' rather than excluding the entire word/string. I suspect - is being treated as a word break... any ideas?

/(?!#SN-)([A-Z]+[0-9]+)\b/g

https://www.regexpal.com/?fam=99443

最满意答案

由于Javascript正则表达式不具备lookbehind功能，因此无法通过简单模式捕捉您想要的内容。 但是，在替换上下文中，可以使用函数作为替换参数和模式来系统地尝试捕获不想要的部分，从而可以轻松地处理此问题：

var result = yourstr.replace(/(#SN-)?\b[A-Z]+[0-9]+\b/g, function(m,g1) { return g1 ? m : 'yourreplacement'; });

当捕获组1被定义时，函数返回整个匹配，否则返回替换字符串。

请注意，在函数内部，您无法使用$&或$1..$n占位符作为整个匹配或捕获组。 您必须使用函数参数来计算它们。

---

只有更严格，如果在目标之前描述所有可能性，也可以在没有回调函数的情况下完成，但是写入会很痛苦，并且效率不高，因为模式以替代方式开始：

var result = yourstring.replace(/(^|[^-]|(?:[^N]|^)-|(?:[^S]|^)N-|(?:^|[^#])SN-)\b[A-Z]+[0-9]+\b/g, '$1yourreplacement');

Since Javascript regex doesn't have the lookbehind feature, there's no way to catch what you want with a simple pattern. However, in a replacement context, you can easily handle this using a function as replacement parameter and a pattern that systematically tries to catch the part you don't want:

var result = yourstr.replace(/(#SN-)?\b[A-Z]+[0-9]+\b/g, function(m,g1) { return g1 ? m : 'yourreplacement'; });

When the capture group 1 is defined, the function returns the whole match, otherwise it returns the replacement string.

Note that inside the function you can't use the $& or $1..$n placeholders for the whole match or capture groups. You have to use the function parameters to figure them.

---

Only to be more rigorous, it can also be done without a callback function if you describe all possibilities before your target, but it's a pain to write and it isn't efficient since the pattern starts with an alternation:

var result = yourstring.replace(/(^|[^-]|(?:[^N]|^)-|(?:[^S]|^)N-|(?:^|[^#])SN-)\b[A-Z]+[0-9]+\b/g, '$1yourreplacement');