I have a string that looks something like this
我有一个看起来像这样的字符串
<a href="/team.php?team_id=521">@Arsenal Fc</a> and <a href="/profile.php?fid=50683">@Tester Alpha</a>
And I need to convert it to
我需要将其转换为
'#ArsenalFc and Tester Alpha'
'#ArsenalFc和Tester Alpha'
three things to keep in mind.
要记住三件事。
1) for links to the team.php page, The @ is converted to #
1)对于team.php页面的链接,@被转换为#
2) for links to the team.php page, string spaces are removed (Arsenal Fc to ArsenalFc)
2)对于team.php页面的链接,删除了字符串空格(Arsenal Fc to ArsenalFc)
3) for links to the profile.php page, the @ is removed
3)对于profile.php页面的链接,@被删除
Any ideas show to do this simply?
任何想法都表明要做到这一点?
1 个解决方案
#1
1
This works for me
这对我有用
function myPregCallback($res)
{
return '#'.str_replace(' ','',$res[1]);
}
$str = '<a href="/team.php?team_id=521">@Arsenal Fc</a> and <a href="/profile.php?fid=50683">@Tester Alpha</a>';
$newStr = preg_replace_callback('#^.+"/team\.php[^>]+>@?([^<]+)</a>#','myPregCallback',$str);
$newStr = preg_replace('#<a.+?"/profile\.php.+?>@?([^<]+)</a>#','$1',$newStr);
var_dump($newStr); //string(27) "#ArsenalFc and Tester Alpha"
This could of course be simpler if the explicit checks for "team.php" and "profile.php" weren't necessary.
如果不需要明确检查“team.php”和“profile.php”,这当然可以更简单。
#1
1
This works for me
这对我有用
function myPregCallback($res)
{
return '#'.str_replace(' ','',$res[1]);
}
$str = '<a href="/team.php?team_id=521">@Arsenal Fc</a> and <a href="/profile.php?fid=50683">@Tester Alpha</a>';
$newStr = preg_replace_callback('#^.+"/team\.php[^>]+>@?([^<]+)</a>#','myPregCallback',$str);
$newStr = preg_replace('#<a.+?"/profile\.php.+?>@?([^<]+)</a>#','$1',$newStr);
var_dump($newStr); //string(27) "#ArsenalFc and Tester Alpha"
This could of course be simpler if the explicit checks for "team.php" and "profile.php" weren't necessary.
如果不需要明确检查“team.php”和“profile.php”,这当然可以更简单。