Question

我从PHP开始，我尝试删除所有出现的字符串。我的字符串是这样的。

'This is [a test] try [it]'

我想要做的是删除所有出现的[]和方括号内的文字。

我希望结果如下：

'This is try'

我如何实现这一目标？

Answer 1

您可以使用preg_replace功能。

preg_replace('~\[[^\]]*\]~', '', $string);

[^\]]*否定字符类，它匹配任何字符，但不匹配右括号]，零次或多次。

添加额外的修剪函数以从结果字符串中删除前导或尾随空格。

$string = 'This is [a test] try [it]';
$result =  preg_replace('~\[[^\]]*\]~', '', $string);
echo trim($result, " ");

Answer 2

你可以试试这个：

$myString = 'This is [a test] try [it]';
$myString = preg_replace('/\[[\w ]+\] */', '', $myString);
var_dump($myString);

说明：

/\[[\w ]+\]/g
 \[ matches the character [ literally
 [\w ]+ match a single character present in the list below
    Quantifier: + Between one and unlimited times, as many times as possible, giving back as needed [greedy]
    \w match any word character [a-zA-Z0-9_]
    ' ' the literal character ' '
 \] matches the character ] literally

在php中删除字符串的出现

2 个答案: