如何从文本中删除特定用户的所有bbcode引用块？

Question

我期待删除PHP中的BBCode作出引号，像这样的例子：

[quote=testuser] 
[quote=anotheruser]a sdasdsa dfv rdfgrgre gzdf vrdg[/quote] 
sdfsd fdsf dsf sdf[/quote] 
the rest of the post text 

我期待在做一个阻塞系统，所以用户不必看那些他们不”内容不想要。所以说“testuser”被封锁了，他们不希望整个引用部分，包括嵌套在里面的第二个引用，因为这是主引用的一部分。

所以职位将只剩下：

后文

我想知道的最佳方式的其余部分做这个。我希望正则表达式，但它更复杂，我想，我有这样的尝试：

/\[quote\=testuser\](.*)\[\/quote\]/is 

然而，然后捕获所有最终报价标记。

是否有另一种方法是快速的，或者是我的正则表达式的一个很好的解决方法？

总结：删除被阻止用户的初始报价以及该报价中的所有内容，但除此之外没有其他内容。

Answer 1

据我所知，这不是一个简单的过程。这里是我的步骤...

1. 使用preg_split()来划分输入字符串3种方式：打开报价标签，关闭报价标签和其他。我正在拆分开始和结束标记，但使用DELIM_CAPTURE将它们保留在输出数组中并保留原始位置/顺序。使用NO_EMPTY以便在foreach循环中没有无用的迭代。
2. 循环遍历生成的数组并搜索用户名称以省略。
3. 当发现由所述目标用户报价，存储元件的起始索引，并设置$open为1
4. 每当一个新的开引号标记发现$open递增。
5. 每当发现新的结束标记标记$open递减。
6. 只要$open达到0,$start和end索引被送到range()生成一个数组填充两点之间的数字。
7. array_flip()，当然，将值移到键。
8. array_diff_key()删除由preg_split()生成的数组中的点的范围。
9. 如果一切顺利，implode()会将子串重新粘合在一起，只保留所需的组件。

代码：（Demo）

/* 
This function DOES NOT validate the $bbcode string to contain a balanced number of opening & closing tags. 
This funcion DOES check that there are enough closing tags to conclude a targeted opening tag. 
*/ 
function omit_user_quotes($bbcode,$user){ 
    $substrings=preg_split('~(\[/?quote[^]]*\])~',$bbcode,NULL,PREG_SPLIT_DELIM_CAPTURE|PREG_SPLIT_NO_EMPTY); 
    $opens=0; // necessary declaration to avoid Notice when no quote tags in $bbcode string 
    foreach($substrings as $index=>$substring){ 
     if(!isset($start) && $substring=="[quote={$user}]"){ // found targeted user's first opening quote 
      $start=$index; // disqualify the first if statement and start searching for end tag 
      $opens=1; // $opens counts how many end tags are required to conclude quote block 
     }elseif(isset($start)){ 
      if(strpos($substring,'[quote=')!==false){ // if a nested opening quote tag is found 
       ++$opens; // increment and continue looking for closing quote tags 
      }elseif(strpos($substring,'[/quote]')!==false){ // if a closing quote tag is found 
       --$opens; // decrement and check for quote tag conclusion or error 
       if(!$opens){ // if $opens is zero ($opens can never be less than zero) 
        $substrings=array_diff_key($substrings,array_flip(range($start,$index))); // slice away unwanted elements from input array 
        unset($start); // re-qualify the first if statement to allow the process to repeat 
       } 
      } 
     } 
    } 
    if($opens){ // if $opens is positive 
     return 'Error due to opening/closing tag imbalance (too few end tags)'; 
    }else{ 
     return trim(implode($substrings)); // trims the whitespaces on either side of $bbcode string as feature 
    }  
} 

/* Single unwanted quote with nested innocent quote: */ 
/*$bbcode='[quote=testuser] 
[quote=anotheruser]a sdasdsa dfv rdfgrgre gzdf vrdg[/quote] 
sdfsd fdsf dsf sdf[/quote] 
the rest of the test'; */ 
/* output: the rest of the test */ 

/* Complex battery of unwanted, wanted, and nested quotes: */ 
$bbcode='[quote=mickmackusa]Keep this[/quote] 
[quote=testuser]Don\'t keep this because 
    [quote=mickmackusa]said don\'t do it[/quote] 
    ... like that\'s a good reason 
    [quote=NaughtySquid] It\'s tricky business, no?[/quote] 
    [quote=nester][quote=nesty][quote=nested][/quote][/quote][/quote] 
[/quote] 
Let\'s remove a second set of quotes 
[quote=testuser]Another quote block[/quote] 
[quote=mickmackusa]Let\'s do a third quote inside of my quote... 
[quote=testuser]Another quote block[/quote] 
[/quote] 
This should be good, but 
What if [quote=testuser]quotes himself [quote=testuser] inside of his own[/quote] quote[/quote]?'; 
/* output: [quote=mickmackusa]Keep this[/quote] 

Let's remove a second set of quotes 

[quote=mickmackusa]Let's do a third quote inside of my quote... 

[/quote] 
This should be good, but 
What if ? */ 

/* No quotes: */ 
//$bbcode='This has no bbcode quote tags in it.'; 
/* output: This has no bbcode quote tags in it. */ 

/* Too few end quote tags by innocent user: 
(No flag is raised because the targeted user has not quoted any text) */ 
//$bbcode='This [quote=mickmackusa] has not end tag.'; 
/* output: This [quote=mickmackusa] has not end tag. */ 

/* Too few end quote tags by unwanted user: */ 
//$bbcode='This [quote=testuser] has not end tag.'; 
/* output: Error due to opening/closing tag imbalance (too few end tags) */ 

/* Too many end quote tags by unwanted user: 
(No flag is raised because the function does not validate the bbcode text as fully balanced) */ 
//$bbcode='This [quote=testuser] has too many end[/quote] tags.[/quote]'; 
/* output: This tags.[/quote] */ 

$user='testuser'; 

echo omit_user_quotes($bbcode,$user); // omit a single user's quote blocks 

/* Or if you want to omit quote blocks from multiple users, you can use a loop: 
$users=['mickmackusa','NaughtySquid']; 
foreach($users as $user){ 
    $bbcode=omit_user_quotes($bbcode,$user); 
} 
echo $bbcode; 
*/ 

Answer 2

你想要的是一个正则表达式不可能的。我会建议扫描文件，直到找到[quote=testuser]。找到它后，设置一个布尔值以开始过滤，并将计数器设置为1.在布尔值为true后，为每个[quote=...]标记递增计数器。每遇到一个[/quote]标签，减少计数器。当计数器达到0时，将用于过滤的布尔值更改为false。

这是一些sudocode。您可能需要根据您的应用程序对其进行修改，但我认为它显示了使用的一般算法。

filtering = false 
counter = 0 
for each line: 
    if line contains "[quote=testuser]" 
     filtering = true 
     counter = 0 
    if line contains "[quote=" 
     counter += 1 
    if line contains "[/quote] 
     counter -= 1 
    if counter = 0 
     filtering = false 
    if not filtering 
     print line