I'm trying to grep(newbie to regular expressions) for the string on the right side of the following assignment in a source tree:
some_var = %1$s %2$s ID
I have tried:
grep -ri '[[:punct:]]1[[:punct:]]s [[:punct:]]2[[:punct:]]s ID' .
grep -ri "'[[:punct:]]1[[:punct:]]s'\|'[[:punct:]]2[[:punct:]]s ID'" .
I've ran:
grep -ri some_var .
And this returned some_var but having trouble figuring out how to return the other side of the assignment operator.
I've read through gnu grep character classes and bracket expressions but it's still not clear to me.
Use awk:
awk -F'[[:space:]]*=[[:space:]]*' '$1 == "some_var" { print $2 }'
or sed:
sed 's/^some_var[[:space:]]*=[[:space:]]*\(.*\)/\1/'
If you want to use grep you need to use GNU grep with perl regexes:
grep -oP '^some_var\s*=\s*\K.*'
But as I said, that works only with GNU grep and is not standard.
To remove the left side with GNU sed:
$ sed 's/^some_var\s=\s*//' <<< "some_var = %1$s %2$s ID"
%1 %2 ID
Related
I have some basic knowledge on using regular expressions with grep (bash).
But I want to use regular expressions the other way around.
For example I have a file containing the following entries:
line_one=[0-3]
line_two=[4-6]
line_three=[7-9]
Now I want to use bash to figure out to which line a particular number matches.
For example:
grep 8 file
should return:
line_three=[7-9]
Note: I am aware that the example of "grep 8 file" doesn't make sense, but I hope it helps to understand what I am trying to achieve.
Thanks for you help,
Marcel
As others haven pointed out, awk is the right tool for this:
awk -F'=' '8~$2{print $0;}' file
... and if you want this tool to feel more like grep, a quick bash wrapper:
#!/bin/bash
awk -F'=' -v seek_value="$1" 'seek_value~$2{print $0;}' "$2"
Which would run like:
./not_exactly_grep.sh 8 file
line_three=[7-9]
My first impression is that this is not a task for grep, maybe for awk.
Trying to do things with grep I only see this:
for line in $(cat file); do echo 8 | grep "${line#*=}" && echo "${line%=*}" ; done
Using while for file reading (following comments):
while IFS= read -r line; do echo 8 | grep "${line#*=}" && echo "${line%=*}" ; done < file
This can be done in native bash using the syntax [[ $value =~ $regex ]] to test:
find_regex_matching() {
local value=$1
while IFS= read -r line; do # read from input line-by-line
[[ $line = *=* ]] || continue # skip lines not containing an =
regex=${line#*=} # prune everything before the = for the regex
if [[ $value =~ $regex ]]; then # test whether we match...
printf '%s\n' "$line" # ...and print if we do.
fi
done
}
...used as:
find_regex_matching 8 <file
...or, to test it with your sample input inline:
find_regex_matching 8 <<'EOF'
line_one=[0-3]
line_two=[4-6]
line_three=[7-9]
EOF
...which properly emits:
line_three=[7-9]
You could replace printf '%s\n' "$line" with printf '%s\n' "${line%%=*}" to print only the key (contents before the =), if so inclined. See the bash-hackers page on parameter expansion for a rundown on the syntax involved.
This is not built-in functionality of grep, but it's easy to do with awk, with a change in syntax:
/[0-3]/ { print "line one" }
/[4-6]/ { print "line two" }
/[7-9]/ { print "line three" }
If you really need to, you could programmatically change your input file to this syntax, if it doesn't contain any characters that need escaping (mainly / in the regex or " in the string):
sed -e 's#\(.*\)=\(.*\)#/\2/ { print "\1" }#'
As I understand it, you are looking for a range that includes some value.
You can do this in gawk:
$ cat /tmp/file
line_one=[0-3]
line_two=[4-6]
line_three=[7-9]
$ awk -v n=8 'match($0, /([0-9]+)-([0-9]+)/, a){ if (a[1]<n && a[2]>n) print $0 }' /tmp/file
line_three=[7-9]
Since the digits are being treated as numbers (vs a regex) it supports larger ranges:
$ cat /tmp/file
line_one=[0-3]
line_two=[4-6]
line_three=[75-95]
line_four=[55-105]
$ awk -v n=92 'match($0, /([0-9]+)-([0-9]+)/, a){ if (a[1]<n && a[2]>n) print $0 }' /tmp/file
line_three=[75-95]
line_four=[55-105]
If you are just looking to interpret the right hand side of the = as a regex, you can do:
$ awk -F= -v tgt=8 'tgt~$2' /tmp/file
You would like to do something like
grep -Ef <(cut -d= -f2 file) <(echo 8)
This wil grep what you want but will not display where.
With grep you can show some message:
echo "8" | sed -n '/[7-9]/ s/.*/Found it in line_three/p'
Now you would like to transfer your regexp file into such commands:
sed 's#\(.*\)=\(.*\)#/\2/ s/.*/Found at \1/p#' file
Store these commands in a virtual command file and you will have
echo "8" | sed -nf <(sed 's#\(.*\)=\(.*\)#/\2/ s/.*/Found at \1/p#' file)
I have a string something like this
xsd:import schemaLocation="AppointmentManagementService.xsd6.xsd" namespace=
I want to extract the following from it :
AppointmentManagementService.xsd6.xsd
I have tried using regex, bash and sed with no success. Can someone please help me out with this?
The regex that I used was this :
/AppointmentManagementService.xsd\d{1,2}.xsd/g
Your string is:
nampt#nampt-desktop:$ cat 1
xsd:import schemaLocation="AppointmentManagementService.xsd6.xsd" namespace=
Try with awk:
cat 1 | awk -F "\"" '{print $2}'
Output:
AppointmentManagementService.xsd6.xsd
sed doesn't recognize \d, use [0-9] or [[:digit:]] instead:
sed 's/^.*schemaLocation="\([^"]\+[[:digit:]]\{1,2\}\.xsd\)".*$/\1/g'
## or
sed 's/^.*schemaLocation="\([^"]\+[0-9]\{1,2\}\.xsd\)".*$/\1/g'
You can use bash native regex matching:
$ in='xsd:import schemaLocation="AppointmentManagementService.xsd6.xsd" namespace='
$ if [[ $in =~ \"(.+)\" ]]; then echo "${BASH_REMATCH[1]}"; fi
Output:
AppointmentManagementService.xsd6.xsd
Based on your example, if you want to grant, at least, 1 or, at most, 2 digits in the .xsd... component, you can fine tune the regex with:
$ if [[ $in =~ \"(AppointmentManagementService.xsd[0-9]{1,2}.xsd)\" ]]; then echo "${BASH_REMATCH[1]}"; fi
using PCRE in GNU grep
grep -oP 'schemaLocation="\K.*?(?=")'
this will output pattern matched between schemaLocation=" and very next occurrence of "
Reference:
https://unix.stackexchange.com/a/13472/109046
Also we can use 'cut' command for this purpose,
[root#code]# echo "xsd:import schemaLocation=\"AppointmentManagementService.xsd6.xsd\" namespace=" | cut -d\" -f 2
AppointmentManagementService.xsd6.xsd
s='xsd:import schemaLocation="AppointmentManagementService.xsd6.xsd" namespace='
echo $s | sed 's/.*schemaLocation="\(.*\)" namespace=.*/\1/'
Everyone!!
I want to get specific substring from stdout of command.
stdout:
{"response":
{"id":"110200dev1","success":"true","token":"09ad7cc7da1db13334281b84f2a8fa54"},"success":"true"}
I need to get a hex string after token without quotation marks, the length of hex string is 32 letters.I suppose it can be done by sed or egrep. I don't want to use awk here. Because the stdout is being changed very often.
This is an alternate gnu-awk solution when grep -P isn't available:
awk -F: '{gsub(/"/, "")} NF==2&&$1=="token"{print $2}' RS='[{},]' <<< "$string"
09ad7cc7da1db13334281b84f2a8fa54
grep's nature is extracting things:
grep -Po '"token":"\K[^"]+'
-P option interprets the pattern as a Perl regular expression.
-o option shows only the matching part that matches the pattern.
\K throws away everything that it has matched up to that point.
Or an option using sed...
sed 's/.*"token":"\([^"]*\)".*/\1/'
With sed:
your-command | sed 's/.*"token":"\([^"]*\)".*/\1/'
YourStreamOrFile | sed -n 's/.*"token":"\([a-f0-9]\{32\}\)".*/\1/p'
doesn not return a full string if not corresponding
I'm trying to print,using grep, lines which contains vasile or line which contains ion . This is command but it doesn't work:
grep (vasile|ion) test.txt
I don't need this :
grep vasile test.txt | grep ion test.txt
try,
terminal$ grep -e vasile -e ion test.txt
Other way using OR operator | in grep
terminal$ grep 'vasile\|ion' test.txt
If you use awk, you can do:
awk '/vasile|ion/' test.txt
awk '/vasile/ || /ion/' test.txt
Try alternation with Grep's extended regex option:
grep -E 'vasile |ion' file
This should work with all Posix greps. \| is a GNU extension to BRE..
I have a string of text in a file that I am parsing out, I almost got it but not sure what I am missing
basic expression I am using is
cat cred.txt | grep -m 1 -o '&CD=[^&]*'
I am getting a results of
&CD=u8AA-RaF-97gc_SdZ0J74gc_SdZ0J196gc_SdZ0J211
I do not want the &CD= part in the resulting string, how would I do that.
The string I am parsing from is:
webpage.asp?UserName=username&CD=u8AA-RaF-97gc_SdZ0J74gc_SdZ0J196gc_SdZ0J211&Country=USA
If your grep knows Perl regex:
grep -m 1 -oP '(?<=&CD=)[^&]*' cred.txt
If not:
sed '1s/.*&CD=\([^&]*\).*/\1/' cred.txt
Many ways to skin this cat.
Extend your pipe:
grep -o 'CD=[^&]*' cred.txt | cut -d= -f2
Or do a replacement in sed:
sed -r 's/.*[&?]CD=([^&]*).*/\1/' cred.txt
Or get really fancy and parse the actual QUERY_STRING in awk:
awk -F'?' '{ split($2, a, "&"); for(i in a){split(a[i], kv, "="); out[kv[1]]=kv[2];} print out["CD"];}'