[英]perl oneliner search replace pattern
I have thousands of rows where some contain the following: 我有数千行,其中一些包含以下内容:
id_s,title_dk
KKS2826,"Søslag ved Øland og Gulland, 1564",12312,2x2
KKS935,"Vignet til Edvard Brandes, afhandling om Johan Wiehe", 1233, 4x4
I'm looking for a Perl one-liner where I can delete any comma that might occur within quotations (the second column). 我正在寻找一种Perl线性代码,可以删除引号(第二列)中可能出现的任何逗号。 But obviously not the others, wince they are delimiters.
但是显然其他人不是,因为它们是分隔符。
So desired output would be: 因此,期望的输出将是:
id_s,title_dk
KKS2826,"Søslag ved Øland og Gulland 1564",12312,2x2
KKS935,"Vignet til Edvard Brandes afhandling om Johan Wiehe", 1233, 4x4
I have been playing with this: perl -ne 's/(?<!,),//g; print;'
我一直在玩这个游戏:
perl -ne 's/(?<!,),//g; print;'
perl -ne 's/(?<!,),//g; print;'
But I can't figure out how to keep the other commas. 但是我不知道如何保留其他逗号。
Easy using Text::CSV_XS : 易于使用Text :: CSV_XS :
perl -CS -MText::CSV_XS=csv -we '
my $aoa = csv(in => shift, allow_whitespace => 1);
$_->[1] =~ s/,//g for @$aoa;
csv(in => $aoa, out => *STDOUT, always_quote => 0);
' input.csv > output.csv
Try this one liner 尝试这个衬垫
perl -p -e 's/"([^"]*)"/my $m=$1;$m=~ s:,::g; $m /eg' file.txt
As per Borodin comment script updated. 根据Borodin评论脚本的更新。 Because above script will remove the
"
also. 因为上面的脚本还将删除
"
。
perl -p -e 's/ (?<=") ([^"]*) (?<=")/$1=~ s:,::rg; /xeg' file.txt
In second one, I used positive look ahead and look behind. 在第二篇文章中,我使用正面的眼光看待后面。 With non-destructive modifier
(r)
. 带有非破坏性修饰符
(r)
。 Non-destructive modifier will work on only > 5.14. 非破坏性修改器仅在> 5.14上有效。
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