blocks|key|86303|text|这可能适用于您(GNU+sed)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|86304|sed+-r+'s/[%5E:]*([%5E:]):%2B%7C:%2B/\1/g;s/\B/\//g'+file|code-block|syntax|javascript|86305|将零个或多个非:替换为整行中的单个字符，然后在每个字符之间插入一个:。|offset|length|style|CODE|86306|entityMap^0|0|0|7|1|X|1|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

This might work for you (GNU sed):

<pre><code>sed -r 's/[^:]*([^:]):+|:+/\1/g;s/\B/\//g' file
</code></pre>

Replace zero or more non <code>:</code>'s followed by a single character followed by a <code>:</code> or a lone <code>:</code> by the single character globally throughout the line. Then replace insert a <code>/</code> between each character.

blocks|key|2607738|text|下面是什么：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2607739|awk+'BEGIN{FS=":"}{for(i=1;i<NF;i%2B%2B){if(i>2)printf"/";printf+substr($i,length($i))}print""}'+input.txt|code-block|syntax|javascript|2607740|input.txt：|2607741|:hd:fg:kl:
:df:lkjh:
:glki:l:s:d:|2607742|输出：|2607743|d/g/l
f/h
i/l/s/d|2607744|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

What about:

<pre><code>awk 'BEGIN{FS=":"}{for(i=1;i&lt;NF;i++){if(i&gt;2)printf"/";printf substr($i,length($i))}print""}' input.txt
</code></pre>

input.txt:

<pre><code>:hd:fg:kl:
:df:lkjh:
:glki:l:s:d:
</code></pre>

Output:

<pre><code>d/g/l
f/h
i/l/s/d
</code></pre>

blocks|key|2607751|text|Solution+Could：你可以试着关注一下，让我知道这是否对你有帮助。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|2607752|awk+-F":"+'
{
++for(i=1;i<=NF;i%2B%2B){
++++if($i){+val=(val?val:"")substr($i,length($i))+}
++}
++print+val;
++val=""
}'+Input_file|code-block|syntax|javascript|2607753|输出如下所示。|2607754|dgl
fh
ilsd|2607755|解决方案2:输出字符串之间带有/的。|CODE|2607756|awk+'
BEGIN{
++OFS="/";
++FS=":"
}
{
++for(i=1;i<=NF;i%2B%2B){
++++if($i){
++++++val=(val?val+OFS:"")substr($i,length($i))
++++}}
++print+val;
++val=""
}'+Input_file|2607757|2607758|d/g/l
f/h
i/l/s/d|2607759|解决方案3：和awk的match实用程序。|2607760|awk+'
{
++while(match($0,/[a-zA-Z]:/)){
++++val=(val?val:"")substr($0,RSTART,RLENGTH-1)
++++$0=substr($0,RSTART%2BRLENGTH)
+++}
++print+val
++val=""
}'++Input_file|2607761|entityMap^0|0|F|0|0|0|0|0|H|F|1|0|0|0|0|0|6|7|3|B|5|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|16|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|Q]]|D|@]|E|$]]|$1|R|3|S|5|H|7|1D|8|@]|D|@]|E|$I|J]]|$1|T|3|L|5|6|7|1E|8|@]|D|@]|E|$]]|$1|U|3|V|5|H|7|1F|8|@]|D|@]|E|$I|J]]|$1|W|3|X|5|6|7|1G|8|@$9|1H|A|1I|B|C]|$9|1J|A|1K|B|Q]|$9|1L|A|1M|B|Q]]|D|@]|E|$]]|$1|Y|3|Z|5|H|7|1N|8|@]|D|@]|E|$I|J]]|$1|10|3|-4|5|6|7|1O|8|@]|D|@]|E|$]]]|11|$]]

Solution 1st: Could you please try following and let me know if this helps you.

<pre><code>awk -F":" '
{
 for(i=1;i&lt;=NF;i++){
 if($i){ val=(val?val:"")substr($i,length($i)) }
 }
 print val;
 val=""
}' Input_file
</code></pre>

Output will be as follows.

<pre><code>dgl
fh
ilsd
</code></pre>

Solution 2nd: With a <code>/</code> in between output strings.

<pre><code>awk '
BEGIN{
 OFS="/";
 FS=":"
}
{
 for(i=1;i&lt;=NF;i++){
 if($i){
 val=(val?val OFS:"")substr($i,length($i))
 }}
 print val;
 val=""
}' Input_file
</code></pre>

Output will be as follows.

<pre><code>d/g/l
f/h
i/l/s/d
</code></pre>

Solution 3rd: With <code>match</code> utility of <code>awk</code>.

<pre><code>awk '
{
 while(match($0,/[a-zA-Z]:/)){
 val=(val?val:"")substr($0,RSTART,RLENGTH-1)
 $0=substr($0,RSTART+RLENGTH)
 }
 print val
 val=""
}' Input_file
</code></pre>

blocks|key|2820045|text|Perl和负前视：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2820046|$+perl+-p+-e+'s/.(?!:)//g'+file
dgl
fh
ilsd|code-block|syntax|javascript|2820047|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|$]]

Perl and negative lookahead:

<pre><code>$ perl -p -e 's/.(?!:)//g' file
dgl
fh
ilsd
</code></pre>

blocks|key|88849|text|使用perl更容易做到这一点|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|88850|$+cat+ip.txt
:hd:fg:kl:
:df:lkjh:
:glki:l:s:d:

$+perl+-lne+'print+join+"/",+/.(?=:)/g'+ip.txt
d/g/l
f/h
i/l/s/d|code-block|syntax|javascript|88851|88852|/.(?=:)/g获取|unordered-list-item|88853|(?=:)为lookahead+construct|88854|88855|的:之前的所有字符|88856|​|88857|88858|然后使用/作为分隔符字符串|88859|打印结果匹配项|88860|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/1570896/what-does-mean-in-a-regular-expression^0|2|4|0|0|0|0|9|1|0|5|6|J|0|1|0|1|1|0|0|0|4|1|0|0^^$0|@$1|2|3|4|5|6|7|18|8|@$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]|$1|L|3|M|5|N|7|1D|8|@$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|O|3|P|5|N|7|1G|8|@$9|1H|A|1I|B|C]]|D|@$9|1J|A|1K|1|1L]]|E|$]]|$1|Q|3|-4|5|6|7|1M|8|@]|D|@]|E|$]]|$1|R|3|S|5|6|7|1N|8|@$9|1O|A|1P|B|C]]|D|@]|E|$]]|$1|T|3|U|5|6|7|1Q|8|@]|D|@]|E|$]]|$1|V|3|-4|5|6|7|1R|8|@]|D|@]|E|$]]|$1|W|3|X|5|N|7|1S|8|@$9|1T|A|1U|B|C]]|D|@]|E|$]]|$1|Y|3|Z|5|6|7|1V|8|@]|D|@]|E|$]]|$1|10|3|-4|5|6|7|1W|8|@]|D|@]|E|$]]]|11|$12|$5|13|14|15|E|$16|17]]]]

This is easier to do with <code>perl</code>

<pre><code>$ cat ip.txt
:hd:fg:kl:
:df:lkjh:
:glki:l:s:d:

$ perl -lne 'print join "/", /.(?=:)/g' ip.txt
d/g/l
f/h
i/l/s/d
</code></pre>

<ul>
<li><code>/.(?=:)/g</code> get all characters preceding <code>:</code>

<ul>
<li><code>(?=:)</code> is a <a href="https://stackoverflow.com/questions/1570896/what-does-mean-in-a-regular-expression">lookahead construct</a></li>
</ul></li>
<li>the resulting matches are then printed using <code>/</code> as delimiter string</li>
</ul>

blocks|key|86538|text|所有的sed都带着ERE|type|unstyled|depth|inlineStyleRanges|entityRanges|data|86539|sed+-E+'s#[%5E:]*(.):#\1/#g;s/%5E.%7C.$//g'+infile|code-block|syntax|javascript|86540|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|$]]

With all sed with ERE

<pre><code>sed -E 's#[^:]*(.):#\1/#g;s/^.|.$//g' infile
</code></pre>

blocks|key|88910|text|使用GNU+sed：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|88911|sed+-E+'s/[%5E:]*([%5E:]):/\1/g;+s/([%5E:])/\/\1/g;+s/%5E:\///'|code-block|syntax|javascript|88912|第一个命令s/[%5E:]*([%5E:]):/\1/g+matches去掉了多余的字符和冒号(第一个除外)，因此产生如下结果：|offset|length|style|CODE|88913|:dgl
:fh
:ilsd|88914|第二个命令s/([%5E:])/\/\1/g在每个字符之前插入一个/，结果如下：|88915|:/d/g/l
:/f/h
:/i/l/s/d|88916|最后一个命令s/%5E:\///只是从每一行的开头删除:/：|88917|d/g/l
f/h
i/l/s/d|88918|entityMap^0|0|0|5|J|0|0|5|F|V|1|0|0|6|8|P|2|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@$I|11|J|12|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|14|8|@$I|15|J|16|K|L]|$I|17|J|18|K|L]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|1A|8|@$I|1B|J|1C|K|L]|$I|1D|J|1E|K|L]]|9|@]|A|$]]|$1|U|3|V|5|D|7|1F|8|@]|9|@]|A|$E|F]]|$1|W|3|-4|5|6|7|1G|8|@]|9|@]|A|$]]]|X|$]]

Using GNU sed:

<pre><code>sed -E 's/[^:]*([^:]):/\1/g; s/([^:])/\/\1/g; s/^:\///'
</code></pre>

The first command, <code>s/[^:]*([^:]):/\1/g</code> matches strips out the extra characters and the colons (except the first one), so yields this:

<pre><code>:dgl
:fh
:ilsd
</code></pre>

The second command <code>s/([^:])/\/\1/g</code> inserts a <code>/</code> before each character, yielding:

<pre><code>:/d/g/l
:/f/h
:/i/l/s/d
</code></pre>

The last command <code>s/^:\///</code> simply removes the <code>:/</code> from the beginning of each line:

<pre><code>d/g/l
f/h
i/l/s/d
</code></pre>

blocks|key|89198|text|您可以使用gawk遍历从第二个字符开始的每一行。每次迭代器命中冒号时，打印前一个字符。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|89199|$+awk+<file.txt+'{for(i=2;i<=length($0);i%2B%2B)+{+\
++++++++++++++++++++if+(substr($0,i,1)==":")+printf+substr($0,i-1,1);}+printf+"\n";}'
dgl
fh
ilsd|code-block|syntax|javascript|89200|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|$]]

You could iterate across each line starting at the second character with gawk. Everytime the iterator hits a colon print the previous character.

<pre><code>$ awk &lt;file.txt '{for(i=2;i&lt;=length($0);i++) { \
 if (substr($0,i,1)==":") printf substr($0,i-1,1);} printf "\n";}'
dgl
fh
ilsd
</code></pre>

I'm trying to get the character that precede each occurrence of given character/pattern in a string using standard bash tools as grep, awk/gawk, sed ...
<code>Step I</code>: get the character that precede each occurrence of the character <code>:</code>
Example:
<blockquote>
String 1 =&gt; :hd:fg:kl:
String 2 =&gt; :df:lkjh:
String 3 =&gt; :glki:l:s:d:
</blockquote>
Expected results
<blockquote>
Result 1 =&gt; dgl
Result 2 =&gt; fh
Result 3 =&gt; ilsd
</blockquote>
I tried many times with awk but without success
<code>Step II</code>: Insert a given character between each character of the resulting string
Example with <code>/</code>
<blockquote>
Result 1 =&gt; d/g/l
Result 2 =&gt; f/h
Result 3 =&gt; i/l/s/d
</blockquote>
I have an awk expression for this step <code>awk -F '' -v OFS=&quot;/&quot; '{$1=$1;print}'</code>
I don't know if it is possible to do <code>Step I</code> with awk or sed and why not do <code>Step I</code> and <code>Step II</code> in once.
Kind Regards

Get the character that precede each occurrence of given character/pattern in a string

Perl

我正在尝试使用标准bash工具获取字符串中每次出现给定字符/模式之前的字符，如grep、awk/gawk、sed……Step I：获取每次出现字符:之前的字符示例：字符串1 => :hd:fg:kl：String 2 => :df:lkjh：String 3 => :glki:l:s:d：预期结果结果1 => dgl结...

问获取字符串中每次出现给定字符/模式之前的字符
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