I need to extract path from a string. I found examples in another post, but missing additional steps.
I have a string as below:
title="test test good dskgkdh hdfyr /rlsmodules/svnrepo/SOURCE/CBL/MQ/BASELINE/MQO000.CBL kdlkfg nsfgf trhrnrt"
cobsrc=$(awk '{match($0,/\/[^"]*/,a);print a[0]}' <<< $title)
echo $cobsrc
Output is
/rlsmodules/svnrepo/SOURCE/CBL/MQ/BASELINE/MQO000.CBL kdlkfg nsfgf trhrnrt
I need only
/rlsmodules/svnrepo/SOURCE/CBL/MQ/BASELINE/MQO000.CBL
What modification is required?
An existing post on similar query:
how to extract path from string in shell script
Four solutions, in order of my own preference.
First option would be simple parameter expansion, in two steps:
$ title="/${title#*/}"
$ title="${title%% *}"
$ echo "$title"
/rlsmodules/svnrepo/SOURCE/CBL/MQ/BASELINE/MQO000.CBL
The first line removes everything up to the first slash (while prepending a slash to replace the one that's stripped", the second line removes everything from the first bit of whitespace that remains.
Or, if you prefer, use a regex:
$ [[ $title =~ ^[^/]*(/[^ ]+)\ ]]
$ echo ${BASH_REMATCH[1]}
/rlsmodules/svnrepo/SOURCE/CBL/MQ/BASELINE/MQO000.CBL
The regex translates as:
null at the beginning of the line,
a run of zero or more non-slashes,
an atom:
a slash followed by non-space characters
a space, to end the previous atom.
The $BASH_REMATCH array contains the content of the bracketed atom.
Next option might be grep -o:
$ grep -o '/[^ ]*' <<<"$title"
(Result redacted -- you know what it'll be.)
You could of course assign this output to a variable using command substitution, which you already know about.
Last option is another external tool...
$ sed 's:^[^/]*::;s/ .*//' <<<"$title"
This is the same functionality as is handled by the parameter expansion (at the top of the answer) only in a sed script, which requires a call to an external program. Included only for pedantry. :)
Could you please try following.
echo "$title" | awk 'match($0,/\/.*\/[^ ]*/){print substr($0,RSTART,RLENGTH)}'
Output will be as follows.
/rlsmodules/svnrepo/SOURCE/CBL/MQ/BASELINE/MQO000.CBL
Solution 2nd: Considering that your variable don't have space in between its value then following may help you too.
echo "$title" | awk '{sub(/[^/]* /,"");sub(/ .*/,"")} 1'
Related
I have something like this in a file named file.txt
AA.201610.pancake.Paul
AA.201610.hello.Robert
A.201610.hello.Mark
Now, i ONLY get the first three fields in 3 variables like:
field1="A"
field2="201610"
field3='hello'.
I'd like to remove a line, if it contains exactly the first 3 fields, like , in the case described above, i want only the third line to be removed from the file.txt . Is there a way to do that? And is there a way to do that in the same file?
I tried with:
sed -i /$field1"."$field2"."$field3"."/Id file.txt
but of course this removes both the second and the third line
I suggest using awk for this as sed can only do regex search and that requires escaping all special meta-chars and anchors, word boundaries etc to avoid false matches.
Suggested awk with non-regex matching:
awk -F '[.]' -v f1="$field1" -v f2="$field2" -v f3="$field3" '
!($1==f1 && $2==f2 && $3==f3)' file
AA.201610.pancake.Paul
AA.201610.hello.Robert
Use ^ to anchor the pattern at the beginning of the line. Also note that . in a regex means "any character" and not a literal peridio. You have to escape it: either \. (be careful with shell escaping and the difference between single and double quotes) or [.]
Sed cannot do string matches, only regexp matches which becomes horrendously complicated to work around when you simply want to match a literal string (see Is it possible to escape regex metacharacters reliably with sed). Just use awk:
$ awk -v str="${field1}.${field2}.${field3}." 'index($0,str)!=1' file
AA.201610.pancake.Paul
AA.201610.hello.Robert
The question was about bash so in bash:
#!/usr/bin/env bash
field1="A"
field2="201610"
field3='hello'
IFS=
while read -r i
do
case "$i" in
"${field1}.${field2}.${field3}."*) ;;
*) echo -E "$i"
esac
done < file.txt
I was trying to do these few operations/commands on a single line and assign it to a variable. I have it working about 90% of the way except for one part of it.
I was unaware you could do this, but I read that you can nest $(..) inside other $(..).... So I was trying to do that to get this working, but can't seem to get it the rest of the way.
So basically, what I want to do is:
1. Take the output of a file and assign it to a variable
2. Then pre-pend some text to the start of that output
3. Then append some text to the end of the output
4. And finally remove newlines and replace them with "\n" character...
I can do this just fine in multiple steps but I would like to try and get this working this way.
So far I have tried the following:
My 1st attempt, before reading about nested $(..):
MY_VAR=$(echo -n "<pre style=\"display:inline;\">"; cat input.txt | sed ':a;N;$!ba;s/\n/\\n/g'; echo -n "</pre>")
This one worked 99% of the way except there was a newline being added between the cat command's output and the last echo command. I'm guessing this is from the cat command since sed removed all newlines except for that one, maybe...?
Other tries:
MY_VAR=$( $(echo -n "<pre style=\"display:inline;\">"; cat input.txt; echo -n "</pre>") | sed ':a;N;$!ba;s/\n/\\n/g')
MY_VAR="$( echo $(echo -n "<pre style=\"display:inline;\">"; cat input.txt; echo "</pre>") | sed ':a;N;$!ba;s/\n/\\n/g' )"
MY_VAR="$( echo "$(echo -n "<pre style=\"display:inline;\">"; cat input.txt; echo "</pre>")" | sed ':a;N;$!ba;s/\n/\\n/g' )"
*Most these others were tried with and without the extra double-quotes surrounding the different $(..) parts...
I had a few other attempts, but they didn't have any luck either... On a few of the other attempts above, it seemed to work except sed was NOT inserting the replacement part of it. The output was correct for the most part, except instead of seeing "\n" between lines it just showed each of the lines smashed together into one line without anything to separate them...
I'm thinking there is something small I am missing here if anyone has any idea..?
*P.S. Does Bash have a name for the $(..) structure? It's hard trying to Google for that since it doesn't really search symbols...
You have no need to nest command substitutions here.
your_var='<pre style="display:inline;">'"$(<input.txt)"'</pre>'
your_var=${your_var//$'\n'/'\n'}
"$(<input.txt)" expands to the contents of input.txt, but without any trailing newline. (Command substitution always strips trailing newlines; printf '%s' "$(cat ...)" has the same effect, albeit less efficiently as it requires a subshell, whereas cat ... alone does not).
${foo//bar/baz} expands to the contents of the shell variable named foo, with all instances of bar replaced with baz.
$'\n' is bash syntax for a literal newline.
'\n' is bash syntax for a two-character string, beginning with a backslash.
Thus, tying all this together, it first generates a single string with the prefix, the contents of the file, and the suffix; then replaces literal newlines inside that combined string with '\n' two-character sequences.
Granted, this is multiple lines as implemented above -- but it's also much faster and more efficient than anything involving a command substitution.
However, if you really want a single, nested command substitution, you can do that:
your_var=$(printf '%s' '<pre style="display:inline;">' \
"$(sed '$ ! s/$/\\n/g' <input.txt | tr -d '\n')" \
'</pre>')
The printf %s combines its arguments without any delimiter between them
The sed operation adds a literal \n to the end of each line except the last
The tr -d '\n' operation removes literal newlines from the file
However, even this approach could be done more efficiently without the nesting:
printf -v your_var '%s' '<pre style="display:inline;">' \
"$(sed '$ ! s/$/\\n/g' <input.txt | tr -d '\n')" \
'</pre>')
...which has the printf assign its results directly to your_var, without any outer command substitution required (and thus saving the expense of the outer subshell).
I want to add the string %%% to the beginning of some specific lines in a text file.
This is my script:
#!/bin/bash
a="c:\Temp"
sed "s/$a/%%%$a/g" <File.txt
And this is my File.txt content:
d:\Temp
c:\Temp
e:\Temp
But nothing changes when I execute it.
I think the 'sed' command is not finding the pattern, possibly due to the \ backslashes in the variable a.
I can find the c:\Temp line if I use grep with -F option (to not interpret strings):
cat File.txt | grep -F "$a"
But sed seems not to implement such '-F` option.
Not working neither:
sed 's/$a/%%%$a/g' <File.txt
sed 's/"$a"/%%%"$a"/g' <File.txt
I have found similar threads about replacing with sed, but they don't refer to variables.
How can I replace the desired lines by using a variable adding them the %%% char string?
EDIT: It would be fine that the $a variable could be entered via parameter when calling the script, so it will be assigned like:
a=$1
Try it like this:
#!/bin/sh
a='c:\\Temp' # single quotes
sed "s/$a/%%%$a/g" <File.txt # double quotes
Output:
Johns-MacBook-Pro:sed jcreasey$ sh x.sh
d:\Temp
e:\Temp
%%%c:\Temp
You need the double slash '\' to escape the '\'.
The single quotes won't expand the variables.
So you escape the slash in single quotes and pass it into the double quotes.
Of course you could also just do this:
#!/bin/sh
sed 's/\(.*Temp\)/%%%&/' <File.txt
If you want to get input from the command line you have to allow for the fact that \ is an escape character there too. So the user needs to type 'c:\\' or the interpreter will just wait for another character. Then once you get it, you will need to escape it again. (printf %q).
#!/bin/sh
b=`printf "%q" $1`
sed "s/\($b\)/%%% &/" < File.txt
The issue you are having has to do with substitution of your variable providing a regular expression looking for a literal c:Temp with the \ interpreted as an escape by the shell. There are a number of workarounds. Seeing the comments and having worked through the possibilities, the following will allow the unquoted entry of the search term:
#!/bin/bash
## validate that needed input is given on the command line
[ -n "$1" -a "$2" ] || {
printf "Error: insufficient input. Usage: %s <term> <file>\n" "${0//*\//}" >&2
exit 1
}
## validate that the filename given is readable
[ -r "$2" ] || {
printf "Error: file not readable '%s'\n" "$2" >&2
exit 1
}
a="$1" # assign a
filenm="$2" # assign filename
## test and fix the search term entered
[[ "$a" =~ '/' ]] || a="${a/:/:\\}" # test if \ removed by shell, if so replace
a="${a/\\/\\\\}" # add second \
sed -e "s/$a/%%%$a/g" "$filenm" # call sed with output to stdout
Usage:
$ bash sedwinpath.sh c:\Temp dat/winpath.txt
d:\Temp
%%%c:\Temp
e:\Temp
Note: This allows both single-quoted or unquoted entry of the dos path search term. To edit in place use sed -i. Additionally, the [[ operator and =~ operator are limited to bash.
I could have sworn the original question said replace, but to append, just as you suggest in the comments. I have updated the code with:
sed -e "s/$a/%%%$a/g" "$filenm"
Which provides the new output:
$ bash sedwinpath.sh c:\Temp dat/winpath.txt
d:\Temp
%%%c:\Temp
e:\Temp
Remember: If you want to edit the file in place use sed -i or sed -i.bak which will edit the actual file (and if -i.bak is given create a backup of the original in originalname.bak). Let me know if that is not what you intended and I'm happy to edit again.
Creating your script with a positional parameter of $1
#!/bin/bash
a="$1"
cat <file path>|sed "s/"$1"/%%%"$1"/g" > "temporary file"
Now whenever you want sed to find "c:\Temp" you need to use your script command line as follows
bash <my executing script> c:\\\\Temp
The first backslash will make bash interpret any backslashes that follows therefore what will be save in variable "a" in your executing script is "c:\\Temp". Now substituting this variable in sed will cause sed to interpret 1 backlash since the first backslash in this variable will cause sed to start interpreting the other backlash.
when you Open your temporary file you will see:
d:\Temp
%%%c:\Temp
e:\Temp
I have a string variable x=tmp/variable/custom-sqr-sample/test/example
in the script, what I want to do is to replace all the “-” with the /,
after that,I should get the following string
x=tmp/variable/custom/sqr/sample/test/example
Can anyone help me?
I tried the following syntax
it didnot work
exa=tmp/variable/custom-sqr-sample/test/example
exa=$(echo $exa|sed 's/-///g')
sed basically supports any delimiter, which comes in handy when one tries to match a /, most common are |, # and #, pick one that's not in the string you need to work on.
$ echo $x
tmp/variable/custom-sqr-sample/test/example
$ sed 's#-#/#g' <<< $x
tmp/variable/custom/sqr/sample/test/example
In the commend you tried above, all you need is to escape the slash, i.e.
echo $exa | sed 's/-/\//g'
but choosing a different delimiter is nicer.
The tr tool may be a better choice than sed in this case:
x=tmp/variable/custom-sqr-sample/test/example
echo "$x" | tr -- - /
(The -- isn't strictly necessary, but keeps tr (and humans) from mistaking - for an option.)
In bash, you can use parameter substitution:
$ exa=tmp/variable/custom-sqr-sample/test/example
$ exa=${exa//-/\/}
$ echo $exa
tmp/variable/custom/sqr/sample/test/example
I have long file with the following list:
/drivers/isdn/hardware/eicon/message.c//add_b1()
/drivers/media/video/saa7134/saa7134-dvb.c//dvb_init()
/sound/pci/ac97/ac97_codec.c//snd_ac97_mixer_build()
/drivers/s390/char/tape_34xx.c//tape_34xx_unit_check()
(PROBLEM)/drivers/video/sis/init301.c//SiS_GetCRT2Data301()
/drivers/scsi/sg.c//sg_ioctl()
/fs/ntfs/file.c//ntfs_prepare_pages_for_non_resident_write()
/drivers/net/tg3.c//tg3_reset_hw()
/arch/cris/arch-v32/drivers/cryptocop.c//cryptocop_setup_dma_list()
/drivers/media/video/pvrusb2/pvrusb2-v4l2.c//pvr2_v4l2_do_ioctl()
/drivers/video/aty/atyfb_base.c//aty_init()
/block/compat_ioctl.c//compat_blkdev_driver_ioctl()
....
It contains all the functions in the kernel code. The notation is file//function.
I want to copy some 100 files from the kernel directory to another directory, so I want to strip every line from the function name, leaving just the filename.
It's super-easy in python, any idea how to write a 1-liner in the bash prompt that does the trick?
Thanks,
Udi
cat "func_list" | sed "s#//.*##" > "file_list"
Didn't run it :)
You can use pure Bash:
while read -r line; do echo "${line%//*}"; done < funclist.txt
Edit:
The syntax of the echo command is doing the same thing as the sed command in Eugene's answer: deleting the "//" and everything that comes after.
Broken down:
"echo ${line}" is the same as "echo $line"
the "%" deletes the pattern that follows it if it matches the trailing portion of the parameter
"%" makes the shortest possible match, "%%" makes the longest possible
"//*" is the pattern to match, "*" is similar to sed's ".*"
See the Parameter Expansion section of the Bash man page for more information, including:
using ${parameter#word} for matching the beginning of a parameter
${parameter/pattern/string} to do sed-style replacements
${parameter:offset:length} to retrieve substrings
etc.
here's a one liner in (g)awk
awk -F"//" '{print $1}' file
Here's one using cut and rev
cat file | rev | cut -d'/' -f2-| rev