仅在两个字符串之间移除空白。

I have a string like this: l' opp lop and I need to remove the white space between l' and o, so it has to return l'opp lop.

我有一个像这样的字符串:l'opp lop我需要删除l'和o之间的空白，所以它必须返回l'opp lop。

But I also need to handle strings with different cases like: L' opp LOP, L' Opp lop, l' Opp lop and return the correct string without the extra space between the those string, so the result has to be: L'opp LOP, L'Opp lop, l'Opp lop.

但我也需要处理不同情况下的字符串，比如L'opp LOP, L'opp LOP, L'opp LOP，返回正确的字符串，而不需要这些字符串之间的额外空间，所以结果必须是:L'opp LOP, L'opp LOP, L'opp LOP。

I have tried $new = str_ireplace( "l' o", "l'o", $string ); but it always returns a string that includes l'o so if the string is L'Opp lop the result is always in lower case: l'opp lop.

我尝试过$new = str_ireplace(“l'o”、“l'o”、$string);但是它总是返回一个包含l'o的字符串，所以如果字符串是l 'Opp lop，结果总是在较低的情况下:l 'Opp lop。

Is it possible to have a solution using regex because I also will need to use it to match l' a, l' i, l' u, l' e?

使用regex是否可能有一个解决方案，因为我还需要使用它来匹配l' a, l' I, l' u, l' e?

Thanks

谢谢

1 个解决方案

#1

You can use

您可以使用

~\b(l)'\s+([a-z])~i

And replace with $1$2.

和替换为1美元2美元。

See regex demo

查看演示正则表达式

The point is to use capturing groups with the subpatterns that you need to get back after replacement.

重点是使用捕获组和替换后需要返回的子模式。

Regex explanation:

正则表达式的解释:

\b - starting word boundary
\b -起始字边界
(l)- match and capture into Group 1 a letter l or L (as /i case insensitive modifier is used)
(l)-将字母l或l匹配并捕获到第1组(使用不区分大小写的修饰符)
' - a literal apostrophe
——字面意义上的撇号
\s+ - 1 or more whitespaces (you may use \h+ to limit to horizontal spaces only)
\s+ - 1或更多的白色空间(你可以只使用\h+来限制水平空间)
([a-z]) - match and capture into Group 2 any ASCII letter (lower- and uppercase).
([a-z]) -匹配并捕获到组2中的任何ASCII字母(小写和大写)。

In the replacement pattern, Group 1 and 2 are backreferences with $1 and $2 respectively.

在替换模式中，第1和第2组分别为$1和$2的反向引用。

1 个解决方案

#1

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