删除括号前后的空格

Question

我试图在打开括号后和圆括号和方括号的括号之前删除一个或多个空格.

$s = "This is ( a sample ) [ string ] to play with"

预期结果:

"This is (a sample) [string] to play with"

我设法删除空格之前:

$s = preg_replace('/\s+(?=[\])])/', '', $s);

结果:

"This is ( a sample) [ string] to play with"

但不是空格后的括号!

Answer 1

试试这个正则表达式:

(?<=[([]) +| +(?=[)\]])

单击"演示"

用空白字符串替换匹配项

说明:

• (?<=[([]) +- 匹配1个出现的空格,前面有一个[或(
• | - 要么
• +(?=[)\]])- 匹配1次出现的空格,后面跟着)或]