[img]XXXXXXg;base64,iVQA/XXXXXXXXXXXXXXX[/img]
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如何把[img] 与[/img]之间的内容替换成空?
回复讨论(解决方案)
$s = '[img]XXXXXXg;base64,iVQA/XXXXXXXXXXXXXXX[/img]';echo preg_replace('/(?<=])[^[]+/', '', $s);登入後複製
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[img][/img]
$s = '[img]XXXXXXg;base64,iVQA/XXXXXXXXXXXXXXX[/img]';echo preg_replace('/(?<=])[^[]+/', '', $s);登入後複製
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[img][/img]
$s = '[a]XXXXXXg;base64,iVQA/XXXXXXXXXXXXXXX[/a]';echo preg_replace('/(?<=])[^[]+/', '', $s);登入後複製
[a][/a]
a标签或其它标签包含的内容也被替换了,怎么把它写死,仅针对[img][/img]
echo preg_replace('/(?<=\[img\])[^[]+/', '', $s);登入後複製
$str = '[img]XXXXX[/img]';$regx = '/\[img\](.*)\[\/img\]/';$rep_result = preg_replace_callback( $regx, function($match){ return str_replace($match[1],'',$match[0]); }, $str);登入後複製
