Question

附录你们怎样! 页: 1

我希望尽可能消除背后图,如果出现类似于“!”的情况,则保留问题标记。过滤体应如下:

<编码> Hello. How? 页: 1

解决该问题的最简明的解决办法是最佳的。我期望,定期表达将是解决这一问题的最佳途径,但我对定期表达的了解是非常有限的,而且我不知道如何解决这一问题。然而,用非常规解决办法,我比罚款还要高! 对你提供的无耻的简单易懂的法典的解释也是空洞的。

谢谢!

Answer 1

$str = preg_replace( ~[?!]{2,}~ ,  ? , preg_replace( ~([.,!?])(\1+)~ ,  \1 , $str));

或preg_replace calls (thanks to Alix Axel)

$str = preg_replace(array( ~([.,!?])(\1+)~ ,  ~[?!]{2,}~ ), array( \1 ,  ? ), $str);

Just enumerate all the punctuation you care of in the braces

<>UPD: 处理!>?

Explanation of what it all means:

preg_replace( ~([.,!?])(\1+)~ ,  \1 , $str)

<代码>([,!])(1+)系指-发现任何<代码>,!>,条件是在此之前至少有一个相同的特性<代码>/1+,即<代码>/1-指的是以前的匹配,+ > 至少有一个。

仅用单一果园取代所有这些。

外部表述<代码>[!]{2,} 系指find all ? 如果在浏览中至少有2个,将其改为?。

Answer 2

您可以使用前言:

$a="Hello... how are you!!?? Im bored!!!!!!"; echo preg_replace("/([.!?])+/iS","$1",$a);

=> Hello. how are you? Im bored!

Answer 3

$string = "Hello... how are you!!?? I m bored!!"
$new_string = $string;
foreach(array( . , , , ? , ! ) as $value) {
  $i = ;
  do {
    $prev_string = $new_string;
    $string = str_replace($value . $value,$value,$string;
    $i++;
  } while ($string !== $prev_string && $i<100)
}

这消除了重复,但不是?

I think this second solution will work, keeping the FIRST of your "bad_chars". If you want to keep the LAST one, there are solutions to that as well.

<?php
$string = str_split($string);
$new_string = array();
$i = 0;
foreach($string as $key => $char) {
    echo  Processing:   . $char .  <br /> ;
    $prev_key = $key - 1;
    $prev_char = $string[$prev_key];
  if($i!== 0) {
    if(in_array($char,$bad_chars) && in_array($prev_char,$bad_chars) ) {
      // do nothing
    } else {
      $new_string[] = $char;
    }
  } else {
    $prev_char = $char;
    $new_string[] = $prev_char;
  }
  $i++;
}
$string = implode(  ,$string);
$new_string = implode(  ,$new_string);
?><br />

<?php echo $string; ?><br />
<?php echo $new_string; ?><br />

友情链接