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PHP字符串模式替换

[英]PHP String Pattern Replace

I have a set of strings like this: 我有一组这样的字符串:

Pants [+$50]
Shirts [+$10]
Jeans [+$5]
Jackets [+$100]

How can I remove the ' [xxx]' in these lines and leaving just the item name (without the trailing space)? 如何删除这些行中的“ [xxx]”并仅保留商品名称(没有尾随空格)? I was told to define a regular expression, not sure how that works... 有人告诉我定义一个正则表达式,不知道它是如何工作的...

That's actually a bit of a confusing regex, since [ and ] are special characters: 实际上,这有点让人困惑,因为[]是特殊字符:

$str = 'Pants [+$50]';
$str = rtrim(preg_replace('/\[[^\]]*\]/', '', $str));

// 'Pants'

Basically the partern \\[[^\\]]*\\] means to match a literal [ followed by 0 or more characters that are not ] followed by a ] . 基本上,partern \\[[^\\]]*\\]装置以匹配字面[后跟0或多个字符不那]后跟一个] The second string in preg_replace is what it gets replaced with. preg_replace中的第二个字符串就是它所替换的字符串。 In this case the empty string since we want to remove it. 在这种情况下,因为我们要删除它,所以它是空字符串。 Then we use rtrim to trim any trailing whitespace. 然后,我们使用rtrim修剪任何尾随空白。

Try this one: 试试这个:

The RegEx 正则表达式

(?im)[ \t]*\[[^\]\[]+\][ \t]*$

Code

$result = preg_replace('/^(.+?)[ \t]*\[[^\][]+\][ \t]*$/im', '$1', $subject);

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