使用正则表达式,我想检测字符串中括号内是否存在特定单词,如果是,则删除括号及其内容。
我想要定位的词是:
picture
see
lorem
所以,这里有3个字符串示例:
$text1 = 'Hello world (see below).';
$text2 = 'Lorem ipsum (there is a picture here) world!';
$text3 = 'Attack on titan (is lorem) great but (should not be removed).';
我可以使用preg_replace()
使用什么正则表达式:
$text = preg_replace($regex, '' , $text);
要删除这些括号及其内容(如果它们包含这些字词)?
结果应该是:
$text1 = 'Hello world.';
$text2 = 'Lorem ipsum world!';
$text3 = 'Attack on titan great but (should not be removed).';
这是一个ideone用于测试。
答案 0 :(得分:6)
您可以使用以下方法(感谢@Casimir之前指出错误!):
<?php
$regex = '~
(\h*\( # capture groups, open brackets
[^)]*? # match everything BUT a closing bracket lazily
(?i:picture|see|lorem) # up to one of the words, case insensitive
[^)]*? # same construct as above
\)) # up to a closing bracket
~x'; # verbose modifier
$text = array();
$text[] = 'Hello world (see below).';
$text[] = 'Lorem ipsum (there is a picture here) world!';
$text[] = 'Attack on titan (is lorem) great but (should not be removed).';
for ($i=0;$i<count($text);$i++)
$text[$i] = preg_replace($regex, '', $text[$i]);
print_r($text);
?>
答案 1 :(得分:3)
您可以使用此正则表达式进行搜索:
\h*\([^)]*\b(?:picture|see|lorem)\b[^)]*\)
这意味着
\h* # match 0 or more horizontal spaces
\( # match left (
[^)]* # match 0 or more of any char that is not )
\b # match a word boundary
(?:picture|see|lorem) # match any of the 3 keywords
\b # match a word boundary
[^)]* # match 0 or more of any char that is not )
\) # match right )
并替换为空字符串:
<强>代码:强>
$re = '/\h*\([^)]*\b(?:picture|see|lorem)\b[^)]*\)/';
$result = preg_replace($re, '', $input);