Bash script to pull out x number of starting comment lines into another file
I am looking for a one liner to pull out the first comment block in any file. The comment blocks look like this:
/*
* T开发者_Go百科his is a
* comment block
*/
I've been trying to play with sed, but just can't get it to work right. Help?
sed -n '/^\/\*/,/^ \*\//p;/^ \*\//q' file
Or equivalently:
sed -n '\|^/\*|,\|^ \*/|p;\|^ \*/|q' file
Edit:
Here is a version of the second one above that handle's the situation mentioned in ghostdog74's comment:
sed -n '\|^/\*.*\*/|{p;q};\|^/\*|,\|^ \*/|p;\|^ \*/|q' file
And if you want to handle whitespace at the beginning and end of lines:
sed -n '\|^[[:space:]]*/\*.*\*/|{p;q};\|^[[:space:]]*/\*|,\|^ \*/|p;\|^[[:space:]]*\*/|q'
$ cat file
one two
// comment.....
/*
* This is a
* comment block
*/
# asdgihj
/* adsjf */
$ awk -v RS="*/" -vFS="/[*]" 'NR==1{print "/*"$2RT}' file
/*
* This is a
* comment block
*/
$ cat file
/* * comment */
/*
* This is a
* comment block
*/
/* adsjf */
$ awk -v RS="*/" -vFS="/[*]" 'NR==1{print "/*"$2RT}' file
/* * comment */
Another way without using gawk RS
$ cat file
dsfsd
/*
* This is a
* comment block
*/ sdff
blasdf
/* adsjf */
$ awk '/\/\*/&&/\*\//{print;exit}f&&/\*\//{print "*/";exit}/\/\*/&&!/\*\//{f=1}f' file
/*
* This is a
* comment block
*/
I know it's not exactly what you are asking but it is ridiculously easy to do it with a little python code:
#!/usr/bin/env python
import sys
f = open(sys.argv[1])
incomment = False
for line in f:
line = line.rstrip("\n")
if "*/" in line:
print line
break
elif "/*" in line:
incomment = True
print line
elif incomment:
print line
To Run:
python code.py <filename>
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