blocks|key|821628|text|您可以使用以下正则表达式：\b(\w)\1%2B\b。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|821629|解释：|821630|至少一次相同的单词分隔符和单个字符重复(至少一次相同的(\b)|821631|A+a|unordered-list-item|821632|821633|821634|821635|821636|编辑：如果有更多的细节，我会说你可以去掉第一个\b。因此，它变成了：(\w)\1%2B\b|BOLD|821637|entityMap^0|D|B|0|0|S|2|0|0|0|0|0|0|0|3|N|2|Y|9|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Y|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|J|3|K|5|L|7|12|8|@]|D|@]|E|$]]|$1|M|3|-4|5|L|7|13|8|@]|D|@]|E|$]]|$1|N|3|-4|5|L|7|14|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|15|8|@]|D|@]|E|$]]|$1|P|3|-4|5|6|7|16|8|@]|D|@]|E|$]]|$1|Q|3|R|5|6|7|17|8|@$9|18|A|19|B|S]|$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|T|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]]|U|$]]

You can use the following regex: <code>\b(\w)\1+\b</code>.

Explanation:

<ul>
<li>A word break (<code>\b</code>)</li>
<li>A single character</li>
<li>The repetition (at least one time the same character)</li>
<li>Again, a word break</li>
</ul>

EDIT: With more details, I would say you can get rid of the first <code>\b</code>. So, it becomes: <code>(\w)\1+\b</code>

blocks|key|4235587|text|$text+=+"aaa+bbb+cc+dddd";
$replacedText+=+preg_replace('{(\w)\1%2B}','$1',$text);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4235588|如果您也不想要重复的空格，请尝试以下操作：|unstyled|4235589|$replacedText+=+preg_replace('{(.)\1%2B}','$1',$text);|4235590|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>$text = "aaa bbb cc dddd";
$replacedText = preg_replace('{(\w)\1+}','$1',$text);
</code></pre>

if you don't want repetitive whitespaces as well, try the following:

<pre><code>$replacedText = preg_replace('{(.)\1+}','$1',$text);
</code></pre>

blocks|key|827837|text|尝试如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|827838|preg_replace('/(\b)(\w)\2%2B(\b)/',+'$2',+$string);|code-block|syntax|javascript|827839|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Try something like:

<pre><code>preg_replace('/(\b)(\w)\2+(\b)/', '$2', $string);
</code></pre>

blocks|key|4235652|text|下面的正则表达式适用于任何语言中带有u-unicode标志的所有字母：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4235653|/([\p{L}\W])\1%2B(?=+)/u|code-block|syntax|javascript|4235654|解释：|4235655|(+++++++++++++++++#+beginning+of+1st+capturing+group
++++[+++++++++++++#+beginning+of+characters+class
++++++++\p{L}+++++#+any+letter+from+any+language
++++++++\W++++++++#+any+non-word+character
++++]+++++++++++++#+end+of+character+class
)+++++++++++++++++#+end+of+1st+capturing+group
\1++++++++++++++++#+back+reference+to+our+1st+capturing+group+for+repetition
%2B+++++++++++++++++#+one+or+more+character+repetition
(?=+)+++++++++++++#+using+positive+lookahead+to+be+sure+it's+followed+by+a+space|4235656|使用preg_replace来完成以下工作：|4235657|$string+=+preg_replace("/([\p{L}\W])\1%2B(?=+)/u",+"$1",+$string);|4235658|示例的输出：|4235659|"aaa+bbb+cc+dddd+"++=>++"a+b+c+d+"
"Google+is+the+best"++=>++"Google+is+the+best"
"this+is+tesaaat.+are+you+ook?"++=>++"this+is+tesaaat.+are+you+ook?"
"Good+morning+mmmm+yyyy+friendssss+"++=>++"Good+morning+m+y+friends+"
"Hello+friendd+okk"++=>++"Hello+friend+okk"|4235660|4235661|entityMap^0|I|1|0|0|0|0|2|C|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|19|8|@]|D|@]|E|$]]|$1|U|3|V|5|H|7|1A|8|@]|D|@]|E|$I|J]]|$1|W|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]|$1|X|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|Y|$]]

Following regex would work for all letters in any language with <code>u</code>-unicode flag:

<pre><code>/([\p{L}\W])\1+(?= )/u
</code></pre>

Explanations:

<pre><code>( # beginning of 1st capturing group
 [ # beginning of characters class
 \p{L} # any letter from any language
 \W # any non-word character
 ] # end of character class
) # end of 1st capturing group
\1 # back reference to our 1st capturing group for repetition
+ # one or more character repetition
(?= ) # using positive lookahead to be sure it's followed by a space
</code></pre>

Using <code>preg_replace</code> to achieve the job:

<pre><code>$string = preg_replace("/([\p{L}\W])\1+(?= )/u", "$1", $string);
</code></pre>

Output for your examples:

<pre><code>"aaa bbb cc dddd " =&gt; "a b c d "
"Google is the best" =&gt; "Google is the best"
"this is tesaaat. are you ook?" =&gt; "this is tesaaat. are you ook?"
"Good morning mmmm yyyy friendssss " =&gt; "Good morning m y friends "
"Hello friendd okk" =&gt; "Hello friend okk"
</code></pre>

<a href="http://regex101.com/r/pY7hA9" rel="nofollow">Live demo</a>

I am newbie in regular expression. 

I want to replace repeated characters from my string. here some example

<pre><code>$str1 = "aaa bbb cc dddd"; // output : a b c d

$str2 = "Google is the best"; // output : Google is the best
</code></pre>

I found lots of question related this question on stackoverflow. But it does not satisfy my requirement.

I tried this <code>(\w)\1</code> but this is not my solution

Any idea ? Thanks in advance

Edit : 

More examples

<pre><code> $str1 = "this is tesaaat. are you ook?"; // output : this is tesaaat. are you ook?

 $str2 = "Good morning mmmm yyyy friendssss "; // output : Good morning m y friends

 $str3 = "Hello friendd okk"; // output : Hello friend okk 
</code></pre>

Inshort I want to replace repeted charactor followed by space only.

Replace repeated characters regular expression

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我是正则表达式的新手。我想替换字符串中的重复字符。下面是一些例子$str1 = "aaa bbb cc dddd";  // output : a b c d$str2 = "Google is the best";  // output : Google is the best我在stackoverflow上发现了很...

问替换重复字符正则表达式
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