sed replace two double quotes and keep text

Text to be processed :

""text""
"text"
""

output desired :

"text"
"text"
""

Tried with :

 echo -e '""text""\n"text"\n""' | sed -e 's/"".*""/".*"/g'

But obviously no luck.

Cheers,

You want to use a backreference (something that refers to a previously matched part) in your sed command :

echo -e '""text""\n"text"\n""' | sed -E -e 's/""(.*)""/"\1"/g'

Here are the modifications I did :

I grouped what was inside the double-quote in the matching pattern with (...)
I referenced that matching group in the replacement pattern with \1
I told sed to use -Extended regex so I wouldn't have to escape the grouping parenthesis

Based on your sample input, you could just use this:

sed '/^""$/! s/""/"/g' file

On lines which don't only contain two double quotes, globally replace all pairs of double quotes with one double quote.

$ sed 's/"\("[^"]*"\)"/\1/' file
"text"
"text"
""

Could you please try following awk command and let me know if this helps you.

awk -v s1="\"" '!/^\"\"$/{gsub(/\"\"/,s1)} 1'  Input_file

Another approach

cat file | sed 's/"//g' | sed 's/^/"/g' | sed 's/$/"/g'

Details

sed 's/"//g' remove all double quote

sed 's/^/"/g' put a double quote at the beginning of each line

sed 's/$/"/g' put a double quote at the end of each line

Note

This approach will work only if there is one word per line, like in your example.

Answer by Aaron is also fine but back-referencing is costlier in general. Why not use a simpler command:

echo -e '""text""\n"text"\n""' | sed 's/""*/"/g'

Regex to replace more than one double quote to single one.

echo -e '""text""\n"text"\n""' |awk '/""text""/{gsub(/""/,"\42")}1'

"text"
"text"
""

来源：https://stackoverflow.com/questions/46488778/sed-replace-two-double-quotes-and-keep-text

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