从php字符串中删除一个有条件的href标签

Question

我有一个这样的字符串。

$str = "this is a bet and he is <a href=''>Mansoon</a> and he please search on <a href='http://www.google.com'>Google</a> and say he <a href=''>Yammy</a>";

我想从带有空白href的字符串中删除所有href标记。你能不能看看这个。

像这样输出

$str = "this is a bet and he is Mansoon and he please search on <a href='http://www.google.com'>Google</a> and say he Yammy";

非常感谢

Answer 1

试试这个： 的更新 它可能会填满所有异常情况：

$re = "/(<a[^href]*href=[\"']{2}[^>]*>)([^<>]*|.*)(<\\/a>)/m";
$str = "this is a bet and he is <a id=\"sss\" href='' dfsd fdg >Mansoon</a> and he please search on <a href='http://www.google.com'>Google</a> and say he <a href=''>Yammy</a> \n<a href=''> <i>Yammy </i> <br/> </a>\n\nthis is a bet and he is Mansoon and he please search on <a href='http://www.google.com'>Google</a> and say he Yammy";
$subst = "$2";

$result = preg_replace($re, $subst, $str);

<强> live demo

Answer 2

此正则表达式适用于您

<a[^>]*herf='*'[^>]*>[^>](.*?)<\/a> //works well even after space anywhere

Answer 3

此<a\h+href=(['"])\1>([^<>]*)<\/a>正则表达式将匹配具有空href值的所有锚标记。将匹配的<a>代码替换为$2将为您提供所需的输出。

DEMO

<?php
$str = "this is a bet and he is <a href=''>Mansoon</a> and he please search on <a href='http://www.google.com'>Google</a> and say he <a href=''>Yammy</a>";
echo preg_replace('~<a\h+href=([\'"])\1>([^<>]*)<\/a>~', '$2', $str);
?>

输出：

this is a bet and he is Mansoon and he please search on <a href='http://www.google.com'>Google</a> and say he Yammy

\h+匹配一个或多个水平空白字符。

Answer 4

您可以使用：

$str = "this is a bet and he is <a href=''>Mansoon</a> and he please search on <a href='http://www.google.com'>Google</a> and say he <a href=''>Yammy</a>";
$pattern = '/<a[^>]*?href=(?:\'\'|"")[^>]*?>(.*?)<\/a>/i';
$replacement = '$1';
echo preg_replace($pattern, $replacement, $str);

测试regex: here