查找字符倍数包围的文本

Question

The problem:

Find pieces of text in a file enclosed by @

Input:

@abc@ abc @ABC@
cba @cba@ CBA

Output:

@abc@ @ABC@
@cba@

I've tried the following:

cat test.txt | perl -ne 'BEGIN { $/ = undef; } print $1 if(/(@.*@)/s)."\n"'

But this results in:

@abc@ abc @ABC@
cba @cba@

Additional: I was not complete. The goal of the above is the replace the characters between the @ with something else: a should become chr(0x430) b should become chr(0x431) c should become chr(0x446) A should become chr(0x410) B should become chr(0x411) C should become chr(0x426) so with the above input in mind it should result in: абц abc АБЦ cba цба CBA

Sorry for my imcompleteness. Thanks Kluther

Answer 1

The problem with (@.*@) is that * is greedy: it matches the largest amount possible. Thus it will match everything between the first @ in the string and the last one.

You can make it non-greedy with (@.*?@) . However, a better approach is to match everything that is not @ in between:

 (@[^@]*@)

If you want to match every occurrence instead of the first one, you also need to use the /g modifier and modify your code to use a loop:

perl -ne 'BEGIN { $/ = undef; } print $1 while(/(\@[^@]*\@)/gs); print "\n"'

Answer 2

使用这样的模式

@[a-zA-Z]+@

Answer 3

使用此正则表达式：

cat test.txt | perl -pe 's/(?:(@ )|^[^@]).*?(?: (@)|$)/$1$2/g'

Answer 4

Use non-greedy search .+? or /(\\@([^@]*)\\@)/gsm .

cat test.txt | perl -ne 'BEGIN { $/ = undef; } print $1." " while(/(\@([^@]*)\@)/gsm); print "\n";'

Answer 5

One way:

$ perl -pe '@a=$_=~/@[^@]+@/g; $_="@a";' file
@abc@ @ABC@ @cba@