I'm trying to perform a shell script from a groovy function loaded by a jenkins-pipeline to retrieve a zip file from an external location. I am building the address out in the function and passing it into the shell script via $. But I am getting a syntax error and I'm not sure why.
I've tried escaping the $ but dont think thats the correct approach here and my code has been coverted from triple single quotes (''') to triple double (""") so I can pass the variable in.
def DownloadBaseLineFromNexus(groupID, artifactID){
//add code for this method
def nexusLink = "${GetNexusLink()}/${GetNexusProdRepo()}/${groupID}/${artifactID}/"
sh """
# retrieving all available version from release repo to versionFile.xml
curl ${nexusLink} | grep "<a href=.*</a>" | grep "http" | cut -d'>' -f3 |cut -d'/' -f1 > versionFile.xml
# creating array from versionFile.xml
fileItemString=$(cat versionFile.xml |tr "\n" " ")
fileItemArray=($fileItemString)
# Finding maximum of array element
maxValue=`printf "%d\n" "${fileItemArray[#]}" | sort -rn | head -1`
# Download latest version artifact from nexus
curl -o ${(artifactID)}.zip ${(nexusLink)}/${(artifactID)}-$maxValue.zip
# Unzip the tool
unzip ${(artifactID)}.zip
"""
}
the results I get are:
Script1.groovy: 28: illegal string body character after dollar sign;
solution: either escape a literal dollar sign "\$5" or bracket the value expression "${5}" # line 28, column 22.
curl "${nexusLink}" | grep "" | grep "http" | cut -d'>' -f3 |cut -d'/' -f1 > versionFile.xml
You have to add escape characters like below:-
curl ${nexusLink} | grep \"<a href=.*</a>\" | grep \"http\" | cut -d'>' -f3 |cut -d'/' -f1 > versionFile.xml
So I have a bash script which is below, and whenever I try to execute it, it says only one type of list may be specified, any clue as to whats wrong with this code? I interpret this as we use grep to find the word (-word) of the first character ($1) typed by the user in the file femalenames.txt and then using the cut command we print the 2nd field of characters 16-20? sort of confused there.
#!/bin/bash
grep -w $1 femalenames.txt | cut -f2 -c16-20
Just pipe it ( | ) again & it'll work.
Command guide here
#!/bin/bash
grep -w $1 femalenames.txt | cut -f2 | cut -c16-20
Need to ignore grep if the line starts with ; or # for a specific string in a file. file.ini contains below line
output_partition_key=FILE_CREATED_DATE
doing a grep as below returns values FILE_CREATED_DATE
grep -w "output_partition_key" file.ini | cut -d= -f2
but if say the line starts with ; or # then it should not grep anything
;output_partition_key=FILE_CREATED_DATE
I tried solutions from other posts but its not working.Can anyone tell me how to achieve the expected result
It seems like what you really want is to find lines that start with output_partition_key=. The simplest way to do that is:
grep ^output_partition_key= file.ini | cut -d= -f2
(where ^ means "the start of a line").
The command 'grep -c blah *' lists all the files, like below.
% grep -c jill *
file1:1
file2:0
file3:0
file4:0
file5:0
file6:1
%
What I want is:
% grep -c jill * | grep -v ':0'
file1:1
file6:1
%
Instead of piping and grep'ing the output like above, is there a flag to suppress listing files with 0 counts?
SJ
How to grep nonzero counts:
grep -rIcH 'string' . | grep -v ':0$'
-r Recurse subdirectories.
-I Ignore binary files (thanks #tongpu, warlock).
-c Show count of matches. Annoyingly, includes 0-count files.
-H Show file name, even if only one file (thanks #CraigEstey).
'string' your string goes here.
. Start from the current directory.
| grep -v ':0$' Remove 0-count files. (thanks #LaurentiuRoescu)
(I realize the OP was excluding the pipe trick, but this is what works for me.)
Just use awk. e.g. with GNU awk for ENDFILE:
awk '/jill/{c++} ENDFILE{if (c) print FILENAME":"c; c=0}' *
I have a file that possibly contains bad formatting (in this case, the occurrence of the pattern \\backslash). I would like to use grep to return only the line numbers where this occurs (as in, the match was here, go to line # x and fix it).
However, there doesn't seem to be a way to print the line number (grep -n) and not the match or line itself.
I can use another regex to extract the line numbers, but I want to make sure grep cannot do it by itself. grep -no comes closest, I think, but still displays the match.
try:
grep -n "text to find" file.ext | cut -f1 -d:
If you're open to using AWK:
awk '/textstring/ {print FNR}' textfile
In this case, FNR is the line number. AWK is a great tool when you're looking at grep|cut, or any time you're looking to take grep output and manipulate it.
All of these answers require grep to generate the entire matching lines, then pipe it to another program. If your lines are very long, it might be more efficient to use just sed to output the line numbers:
sed -n '/pattern/=' filename
Bash version
lineno=$(grep -n "pattern" filename)
lineno=${lineno%%:*}
I recommend the answers with sed and awk for just getting the line number, rather than using grep to get the entire matching line and then removing that from the output with cut or another tool. For completeness, you can also use Perl:
perl -nE 'say $. if /pattern/' filename
or Ruby:
ruby -ne 'puts $. if /pattern/' filename
using only grep:
grep -n "text to find" file.ext | grep -Po '^[^:]+'
You're going to want the second field after the colon, not the first.
grep -n "text to find" file.txt | cut -f2 -d:
To count the number of lines matched the pattern:
grep -n "Pattern" in_file.ext | wc -l
To extract matched pattern
sed -n '/pattern/p' file.est
To display line numbers on which pattern was matched
grep -n "pattern" file.ext | cut -f1 -d: