What is the value of ${1} in the regular function processing? How do you spell it?

$user _pattern = "/\@ (. +?) ([\s|:]|$)/";
$str = Preg_replace ($user _pattern, ' ${1} ', $str);

How to ${1} in Chinese UrlEncode function transcoding? I do not use the following as a string.

$str = Preg_replace ($user _pattern, ' ${1} ', $str);

Reply to discussion (solution)

With Preg_replace_callback (),
Such as:

Function CB ($matches) {
Return sprintf ('%1 $ S ', UrlEncode ($matches [1]));
}
$user _pattern = "/\@ (. +?) ([\s|:]|$)/";
$str = ' jacky@yyq.cn ';
$str = Preg_replace_callback ($user _pattern, ' CB ', $STR);
Echo $str;

With Preg_replace_callback (),
Such as:

Function CB ($matches) {
Return sprintf ('%1 $ S ', UrlEncode ($matches [1]));
}
$user _pattern = "/\@ (. +?) (......

As if, but the two values are all turned, I just change one, that is, click on the link inside, the link display text does not change, such as only change the regular value of this http://127.0.0.1/${1}, the other does not change.

See Preg_replace_callback 5th parameter, set to 1 to replace only once, and the first replacement is a tag in the

Reference to the 1 floor Varkychan reply: With Preg_replace_callback () bar,
Such as:

Function CB ($matches) {
Return sprintf ('%1 $ S ', UrlEncode ($matches [1]));
}
$user ...

Very simple, modify the parameters in the sprintf,
Such as:
Function CB ($matches) {
Return sprintf ('%s ', UrlEncode ($matches [1]), $matches [1]);
}

Your original plan is doable, just missing something.

$user _pattern = "/\@ (. +?) ([\s|:]|$)/E ";  $str = Preg_replace ($user _pattern, ' ${1} ', $str);

You can tell by comparing it with what you wrote.

This post was last edited by xuzuning on 2013-04-10 09:53:52
Your original plan is doable, just missing something.
PHP Code?12$user_pattern = "/\@ (. +?) ([\s|:]|$)/E "; $str = Preg_replace ($user _pattern, ' "
Starting with php5.5, the/E modifier starts discarding, and it is recommended to use Preg_replace_callback ()

This post was last edited by xuzuning on 2013-04-10 09:53:52
Your original plan is doable, just missing something.
PHP Code?12$user_pattern = "/\@ (. +?) ([\s|:]|$)/E "; $str = Preg_replace ($user _pattern, ' "
As if not, or as a character. There's no link.

I don't need to fool you!

$str = ' @ Chinese aa: '; $user _pattern = "/\@ (. +?) ([\s|:]|$)/E ";  $str = Preg_replace ($user _pattern, ' ${1} ', $str); Echo $str;

Chinese AA

Reference 2 floor Billssjone reply: Reference 1 floor Varkychan reply: With Preg_replace_callback () bar,
Such as:

Function CB ($matches) {
Return sprintf ('%1 $ S ', UrlEncode ($m ...

It doesn't seem to be good.

Function CB ($matches) {    return sprintf (' a href= "http://127.0.0.1/%1$s" >%1$s ', UrlEncode ($matches [1]), $ Matches[1]);} $STR = "Test @ My Weibo add link"; $user _pattern = "/\@ (. +?) ([\s|:]|$)/"; $str = Preg_replace_callback ($user _pattern, ' CB ', $STR); Echo $str;

I deliberately removed the link <符号，看看直接输出效果，还是两个值都会变，输出是：
<符号，看看直接输出效果，还是两个值都会变，输出是：>
Test a href= "Http://127.0.0.1/%E6%88%91%E7%9A%84%E5%BE%AE%E5%8D%9A" >%e6%88%91%e7%9a%84%e5%be%ae%e5%8d%9a add link

#9 code should be written

Function CB ($matches) {    return sprintf (' a href= "http://127.0.0.1/%s" >%s ', UrlEncode ($matches [1]), $matches [1] );} $STR = "Test @ My Weibo add link"; $user _pattern = "/\@ (. +?) ([\s|:]|$)/"; $str = Preg_replace_callback ($user _pattern, ' CB ', $STR); Echo $str;

Test a href= "http://127.0.0.1/%CE%D2%B5%C4%CE%A2%B2%A9" > My Weibo Add link

I don't need to fool you! PHP code?1234$str = ' @ Chinese aa: '; $user _pattern = "/\@ (. +?) ([\s|:]|$)/E "; $str = Preg_replace ($user _pattern, ' ${1} ', $str ...

Oh, thank you, this can, just now I test your code is not correct, that is, did not add that e, so the result of the test is the direct output:
A href=\ "HTTP://127.0.0.1/". UrlEncode ("Chinese AA"). " \ "> Chinese AA

#9 code should be written in PHP code?1234567function CB ($matches) {return sprintf (' a href= "http://127.0.0.1/%s" >%s ', UrlEncode ($ Matches[1]), $matches [1]);} $STR = "Test @ My Weibo add link"; $user _pattern = "/\@ (...)

Thank you so much, it's finally done, and I've been testing from yesterday to today.

