I want to get the filename from a long string in shell script.After reading some example from likegeeks.com,I write a simple solution:
#/bin/bash
cdnurl="http://download.example.com.cn/download/product/vpn/rules/vpn_patch_20190218162130_sign.pkg?wsSecret=9cadeddedfr7bb85a20a064510cd3f353&wsABSTime=5c6ea1e7"
echo ${cndurl}
url=`echo ${cdnurl} | awk -F'/' '{ print $NF }'`
result=`echo ${url} | awk -F '?' '{ print $1}'`
echo ${url}
echo ${result}
I just want to get vpn_patch_20190218162130_sign.pkg,and the it does.I wonder is there any smart ways (may be one line).
If behind pkg it's not ?,how to use pkg to get the filename,I am not sure if always ? after pkg,but the filename always be *.pkg.
You can try : this is more robust as compare to second awk command:
echo "$cdnurl"|awk -v FS='/' '{gsub(/?.*/,"",$NF);print $NF}'
vpn_patch_20190218162130_sign.pkg
#less robust
echo "$cdnurl"|awk -vFS=[?/] '{print $(NF-1)}'
You should use sed :
sed -r 's|.*/(.*.pkg).*|\1|g'
Related
I have string lists :
./SolutionController.php core/app/Http/Controllers/Admin/SolutionController.php
./ContentController.php core/app/Http/Controllers/Frontpage/ContentController.php
./country-flag vendor/country-flag
I wish I could get the final value between the './' sign and the 'space'
Output:
SolutionController.php
ContentController.php
country-flag
This code with bash script:
#!/bin/bash
tanggal=$(date +%d-%m-%Y)
filename="./update/$tanggal/lists.md"
n=1
tanggalWaktu=$(date +"%d-%m-%Y %H:%M:%S")
mkdir -p ./logs
while read line; do
fileName=$(awk -F'[/ ]' '{print $2}' $line)
echo "file -> $fileName"
done < $filename
Output:
awk: can't open file ./SolutionController.php
source line number
Please help me
Using awk :
awk -F'[/ ]' '{print $2}' string.txt
Using gawk:
awk '{print gensub(/\.\/(.*) (.*)/,"\\1","g")}' string.txt
Test Results:
$ cat string.txt
./SolutionController.php core/app/Http/Controllers/Admin/SolutionController.php
./ContentController.php core/app/Http/Controllers/Frontpage/ContentController.php
./country-flag vendor/country-flag
$ awk -F'[/ ]' '{print $2}' string.txt
SolutionController.php
ContentController.php
country-flag
$ awk '{print gensub(/\.\/(.*) (.*)/,"\\1","g")}' string.txt
SolutionController.php
ContentController.php
country-flag
You can do it
echo "./SolutionController.php core/app/Http/Controllers/Admin/SolutionController.php" | sed -r 's/\.\/(.*) .*/\1/'
If you stored it in the file.
sed -r 's/\.\/(.*) .*/\1/' strings.txt
I want to print next sentence after match
My file content like this:
SSID:CoreFragment
Passphrase:WiFi1234
SSID:CoreFragment_5G
Passphrase:WiFi1234
SSID:Aleph_inCar
Passphrase:1234567890
As per my search,e.g. If I found WIFI-3(SSID) than, I want to print 1234ABCD. I used this command to search SSID:
grep -oP '^SSID:\K.+' file_name
After this search I want to print Passphrase of that particular match.
I'm working on Ubuntu 18.04
ssid=$(grep -oP '^SSID:\K.+' list_wifi.txt)
for ssid in $(sudo iwlist wlp2s0 scan | grep ESSID | cut -d '"' -f2)
do
if [ $ssid == $ssid_name ]; then
echo "SSID found...";
fi
done
I want to print next line after match.
another awk
$ awk -F: -v s="$ssid" '$0=="SSID:"s{c=NR+1} c==NR{print $2; exit}' file
1234ABCD
will only print the value if it's on the next line.
awk -F: '/WIFI-3/{getline;print $2; exit}' file
1234ABCD
Robustly (wont fail due to partial matches, etc.) and idiomatically:
$ awk -F':' 'f{print $2; exit} ($1=="SSID") && ($2=="WIFI-3"){f=1}' file
1234ABCD
Please try the following:
ssid="WIFI-3"
passphrase=$(grep -A 1 "^SSID:$ssid" file_name | tail -n 1 | cut -d: -f2)
echo "$passphrase"
which yields:
1234ABCD
Since code tags have changed the look of samples so adding this now.
var=$(awk '/SSID:[a-zA-Z]+-[0-9]+/{flag=1;next} flag{sub(/.*:/,"");value=$0;flag=""} END{print value}' Input_file)
echo "$var"
Could you please try following.
awk '/Passphrase/ && match($0,/WIFI-3 Passphrase:[0-9a-zA-Z]+/){val=substr($0,RSTART,RLENGTH);sub(/.*:/,"",val);print val;val=""}' Input_file
Using Perl
$ export ssid="WIFI-3"
$ perl -0777 -lne ' /SSID:$ENV{ssid}\s*Passphrase:(\S+)/ and print $1 ' yash.txt
1234ABCD
$ export ssid="Aleph_inCar"
$ perl -0777 -lne ' /SSID:$ENV{ssid}\s*Passphrase:(\S+)/ and print $1 ' yash.txt
1234567890
$
$ cat yash.txt
SSID:CoreFragment
Passphrase:WiFi1234
SSID:CoreFragment_5G
Passphrase:WiFi1234
SSID:Aleph_inCar
Passphrase:1234567890
SSID:WIFI-1
Passphrase:1234ABCD
SSID:WIFI-2
Passphrase:123456789
SSID:WIFI-3
Passphrase:1234ABCD
You can capture it in variables as
$ passphrase=$(perl -0777 -lne ' /SSID:$ENV{ssid}\s*Passphrase:(\S+)/ and print $1 ' yash.txt)
$ echo $passphrase
1234567890
$
Below is the output of my 'COMMAND' command .The output format is FILE:LINENO:PATTERN. I want to
take the below command output values in diff variables and use them in 'sed' command mentioned at the bottom.
<COMMAND>
./core.pkglist:16:package linux-release 6Server 9.0.3
./core.pkglist:18:package release-server 6Server 6.9.0.4.0.1.el6
./core.pkglist:32:package upstart 0.6.5 16.el6
./core.pkglist:33:package libnih 1.0.1 7.el6
I want to replace it with the command sed with inputs from the output of above command like:
sed "$var1 s/$var2/$c' $var3
that helps me run virtually the below command:
sed '16s/9.0.3/$c/' core.pkglist
1)The value 16 above should come from a variable like:
var1=$(COMMAND |awk -F':' '{print $2}')
2)9.0.3 should come from a variable.
var2=$(COMMAND |awk -F '{print $4}')
3)core.pkglist should come from a variable.
var3=$(COMMAND |awk -F '{print $1}')
4) $c is another command output
v3=$(echo Hello world); # Stupid examples of variables set from output commands
v1=$(wc -l <<< $v3);
v2=$(awk '{print $1}' <<< $v3);
c=$(echo Good morning);
sed "${v1}s/$v2/$c/" <<< $v3 # Use double quotes instead of singles
# echo $v3 | sed "${v1}s/$v2/$c/" works as well
Returns :
Good morning world
My KSH-Script should replace a String in a txt file from the same directory.
sed -i 's/"$original"/"$reversed"/' inputtext.txt
is what I'm using currently, but it doesn't work. There is no error in the code or things like that. It just doesn't work.
Here is my whole code:
#!/bin/ksh
original=$1
reversed=""
counter=0
echo $original | awk -v ORS="" '{ gsub(/./,"&\n") ; print }' | \
while read char
do
letters[$counter]+="$char"
((counter=counter+1))
done
length=${#original}
((length=length-1))
echo $original | awk -v ORS="" '{ gsub(/./,"&\n") ; print }' | \
while read char
do
reversed+=${letters[$length]}
((length=length-1))
done
echo $reversed
sed -i 's/"$original"/"$reversed"/' inputtext.txt
exit 0
I want, that in the file "inputtext.txt" (same dir as the .sh file) every word that equals "$original" gets changed to "$reversed".
What am I doing wrong?
I think single quotes prevent variable expansion. You can try this:
sed -i "s/$original/$reversed/" inputtext.txt
how to pass and compare an external variable in awk command? is it also dependent on unix shell that we are using
I am trying to do :
mgrid=`echo $file1 | awk -F'|' '{ print $40}' `
echo $mgrid
var=`/usr/bin/more $HOME/pwd_date_chk/file2.txt | awk -F'|' ' -v search="$mgrid" '{ $41 ~ search print $15}'`
echo $var
awk can read the input from file. No need to use more. You can try this,
mgrid=`echo $file1 | awk -F'|' '{ print $40}' `
echo $mgrid
var=`awk -F'|' -v search="$mgrid" '$41 ~ search {print $15}' $HOME/pwd_date_chk/file2.txt`
echo $var