如何使用preg replace在替换字符串中插入找到的字母

Question

我想替换下面写的字符串

AND ( (wp_usermeta.meta_key = 'first_name' AND CAST(wp_usermeta.meta_value AS CHAR) LIKE '%A%')
AND  (mt1.meta_key = 'wp_capabilities' AND CAST(mt1.meta_value AS CHAR) LIKE '%\"s2member\\_level2\"%') )

至

AND ( (wp_usermeta.meta_key = 'first_name' AND CAST(wp_usermeta.meta_value AS CHAR) LIKE 'A%')
AND  (mt1.meta_key = 'wp_capabilities' AND CAST(mt1.meta_value AS CHAR) LIKE '%\"s2member\\_level2\"%') )

在上面的代码中，我想要实现的是将%[a-zA-Z]%替换为[a-zA-Z]%

Answer 1

您需要捕获字母并再次插入替换字符串。

%([a-zA-Z])%

这将捕获字母，因为周围有括号。

并替换为

$1%

$1是该捕获组的内容

在php中，它看起来像这样

$SourceString = preg_replace('/%([a-zA-Z])%/', '$1%', $SourceString);

Answer 2

使用$string = str_replace("%A%", "%A", $string);

Answer 3

我不确定，但你可以试试这个。

$find ="/%A%/"; 
$replace ="A%"; 

echo preg_replace ($find, $replace, "AND ( (wp_usermeta.meta_key = 'first_name' AND CAST(wp_usermeta.meta_value AS CHAR) LIKE '%A%')");