I am new to git and github. I am working on a project where I need to commit my changes to github repository in a specific branch.
But I am getting the error
$ git commit
3.5.0.1
s/3.5.0.1/3.5.1.1/g
sed: can't read ../../build.gradle: No such file or directory
I have also attached the pre-commit file code here.
#!/bin/sh
## finding the exact line in the gradle file
#ORIGINAL_STRING=$(cat ../../build.gradle | grep -E '\d\.\d\.\d\.\d')
## extracting the exact parts but with " around
#TEMP_STRING=$(echo $ORIGINAL_STRING | grep -Eo '"(.*)"')
## the exact numbering scheme
#FINAL_VERSION=$(echo $TEMP_STRING | sed 's/"//g') # 3.5.0.1
#Extract APK version
v=$(cat build.gradle | grep rtVersionName | awk '{print $1}')
FINAL_VERSION=$(echo ${v} | cut -d"\"" -f2)
echo ${FINAL_VERSION}
major=0
minor=0
build=0
assets=0
regex="([0-9]+).([0-9]+).([0-9]+).([0-9]+)"
if [[ $FINAL_VERSION =~ $regex ]]; then
major="${BASH_REMATCH[1]}"
minor="${BASH_REMATCH[2]}"
build="${BASH_REMATCH[3]}"
assets="${BASH_REMATCH[4]}"
fi
# increment the build number
build=$(echo $build + 1 | bc)
NEW_VERSION="${major}.${minor}.${build}.${assets}"
SED_ARGUMENT=$(echo "s/${FINAL_VERSION}/${NEW_VERSION}/g")
echo $SED_ARGUMENT
sed -i -e `printf $SED_ARGUMENT` ../../build.gradle
The error comes in the last line of this file basically. I am using windows.
Things I tried:
sed -i -e `printf $SED_ARGUMENT` ../../build.gradle
sed -i ' ' -e `printf $SED_ARGUMENT` ../../build.gradle
I am unable to understand where am I actually doing wrong. Kindly help me out.
sed: can't read ../../build.gradle: No such file or directory
This one is rather simple. Your build.gradle file is not at ../../build.gradle.
The solution is to determine actual path to the build.gradle file relative to the script, and change the path in the script.
To debug this, do echo Current Directory: $PWD in the script to see what the actual working directory is, then you should be able to determine the correct path to use.
Related
I have a bash script which downloads some files from a url and stores them into a folder named "data1". Since these files are downloaded as .zip then the next step is to unzip them. After that, the variables extension and encoding are get from every file, where extension is the type of file (txt, csv, docx) and encoding is the encoding format of each file (ISO, utf-8). Since the files that this script downloads are not in utf-8 format i have to perform this transformation. This is the line which performs the encoding:
iconv -f $encoding -t UTF-8//TRANSLIT $name2.$extension -o conversion_$name2.$extension;
As you can see, I have to pass two parameters, the file to be encoded to utf-8 format and the name of the output file which will be: conversion_(name of the original file).(extension of the original file). However, I'm getting the following error:
iconv: illegal input sequence at position 1234704
This error is affecting the datos_abiertos_covid19.zip file which after the unzipping process is named as 200715COVID19MEXICO.csv (but it changes depending on the day this script is run). Does anyone knows how I can avoid this error? I specifically need all of the files downloaded to be in utf-8 format. I would really appreciate your help.
Here is the script I'm using:
! /usr/bin/bash
# creating folders
mkdir data1
cd data1
# downloading data
wget http://187.191.75.115/gobmx/salud/datos_abiertos/datos_abiertos_covid19.zip
wget http://187.191.75.115/gobmx/salud/datos_abiertos/diccionario_datos_covid19.zip
# unziping data
for i in `ls | grep .zip`; do unzip $i; done
# this for will iterate over all the files contained on the data1 folder
for name in `ls -F -1 | grep -v / | grep -v zip`; do
# getting extension of current file
extension=`echo $name | sed 's/\./ /g' | awk '{print $2}'`;
# getting encoding format of current file
encoding=`file -i $name | sed 's/=/ /g' |awk '{print $4}'`;
# echo $encoding
query="s/\.$extension//g"
# echo $query
name2=`echo $name | sed -e $query`;
# echo $name2
# echo $name" "$extension" "$encoding" "$name2
# encoding current file
iconv -f $encoding -t UTF-8//TRANSLIT $name2.$extension -o conversion_$name2.$extension;
done
mkdir old
mv `ls | grep -v "conversion_" | grep -v "old"` old
Since this script is intended to be run automatically every 24 hours, then I need the old data (data from a day before) to be stored in another place. That's why at the end of the script a new folder is created and all the "old files" are moved to the folder named "old".
I try to search for files and seperate path and version as variable because each will be needed later for creating a directory and to unzip a .jar in desired path.
file=$(find /home/user/Documents/test/ -path *.jar)
version=$(echo "$file" | grep -P -o '[0-9].[0-9].[0-9].[0-9]')
path=$(echo "$file" | sed 's/\(.*\)[/].*/\1/')
newpath=$(echo "${path}/${version}")
echo "$newpath"
result
> /home/user/Documents/test/gb0500
> /home/user/Documents/test/gb0500 /home/user/Documents/test/gb0500
> /home/user/Documents/test /home/user/Documents/test/1.3.2.0
> 1.3.2.1
> 1.3.2.2
> 1.2.0.0
> 1.3.0.0
It's hilarious that it's only working at one line.
what else I tried:
file=$(find /home/v990549/Dokumente/test/ -path *.jar)
version=$(grep -P -o '[0-9].[0-9].[0-9].[0-9]')
path=$(sed 's/\(.*\)[/].*/\1/')
while read $file
do
echo "$path$version"
done
I have no experience in scripting. Thats what I figured out some days ago. I am just practicing and trying to make life easier.
find output:
/home/user/Documents/test/gb0500/gb0500-koetlin-log4j2-web-1.3.2.0-javadoc.jar
/home/user/Documents/test/gb0500/gb0500-koetlin-log4j2-web-1.3.2.1-javadoc.jar
/home/user/Documents/test/gb0500/gb0500-koetlin-log4j2-web-1.3.2.2-javadoc.jar
/home/user/Documents/test/gb0500-co-log4j2-web-1.2.0.0-javadoc.jar
/home/user/Documents/test/gb0500-commons-log4j2-web-1.3.0.0-javadoc.jar
As the both variables version and path are newline-separated, how about:
file=$(find /home/user/Documents/test/ -path *.jar)
version=$(echo "$file" | grep -P -o '[0-9].[0-9].[0-9].[0-9]')
path=$(echo "$file" | sed 's/\(.*\)[/].*/\1/')
paste -d "/" <(echo "$path") <(echo "$version")
Result:
/home/user/Documents/test/gb0500/1.3.2.0
/home/user/Documents/test/gb0500/1.3.2.1
/home/user/Documents/test/gb0500/1.3.2.2
/home/user/Documents/test/1.2.0.0
/home/user/Documents/test/1.3.0.0
BTW I do not recommend to store multiple filenames in a single variable
as a newline-separated variable due to several reasons:
Filenames may contain a newline character.
It is not easy to manipulate the values of each line.
For instance you could simply say
the third line as path=${file%/*} if file contains just one.
Hope this helps.
How do you use a command line argument as a file path and check for file existence in Bash?
I have the simple Bash script test.sh:
#!/bin/bash
set -e
echo "arg1=$1"
if [ ! -f "$1" ]
then
echo "File $1 does not exist."
exit 1
fi
echo "File exists!"
and in the same directory, I have a data folder containing stuff.txt.
If I run ./test.sh data/stuff.txt I see the expected output:
arg1=data/stuff.txt
"File exists!"
However, if I call this script from a second script test2.sh, in the same directory, like:
#!/bin/bash
fn="data/stuff.txt"
./test.sh $fn
I get the mangled output:
arg1=data/stuff.txt
does not exist
Why does the call work when I run it manually from a terminal, but not when I run it through another Bash script, even though both are receiving the same file path? What am I doing wrong?
Edit: The filename does not have spaces. Both scripts are executable. I'm running this on Ubuntu 18.04.
The filename was getting an extra whitespace character added to it as a result of how I was retrieving it in my second script. I didn't note this in my question, but I was retrieving the filename from folder list over SSH, like:
fn=$(ssh -t "cd /project/; ls -t data | head -n1" | head -n1)
Essentially, I wanted to get the filename of the most recent file in a directory on a remote server. Apparently, head includes the trailing newline character. I fixed it by changing it to:
fn=$(ssh -t "cd /project/; ls -t data | head -n1" | head -n1 | tr -d '\n' | tr -d '\r')
Thanks to #bigdataolddriver for hinting at the problem likely being an extra character.
I want to copy only 20180721 files from Outgoing to Incoming folder. I also want to remove the first numbers from the file name and want to rename from -1 to -3. I want to keep my commands to minimum so I am using pax command below.
Filename:
216118105741_MOM-09330-20180721_102408-1.jar
Output expected:
MOM-09330-20180721_102408-3.jar
I have tried this command and it's doing most of the work apart from removing the number coming in front of the file name. Can anyone help?
Command used:
pax -rw -pe -s/-1/-3/ ./*20180721*.jar ../Incoming/
Try this simple script using just parameter expansion:
for file in *20180721*.jar; do
new=${file#*_}
cp -- "$file" "/path/to/destination/${new%-*}-3.jar"
done
You can try this
In general
for i in `ls files-to-copy-*`; do
cp $i `echo $i | sed "s/rename-from/rename-to/g"`;
done;
In your case
for i in `ls *_MOM*`; do
cp $i `echo $i | sed "s/_MOM/MOM/g" | sed "s/-1/-3/g"`;
done;
pax only applies the first successful substitution even if the -s option is specified more than once. You can pipe the output to a second pax instance, though.
pax -w -s ':^[^_]*_::p' *20180721*.jar | (builtin cd ../Incoming; pax -r -s ':1[.]jar$:3.jar:p')
I have written a bash script that finds any executable files in our scripts directory, then performs a grep on the resulting files to display a description, if it was included in the file.
A "description" is identified in each file as a line beginning with "# DESC:"
For some reason, the script also includes the grep command that is being run (but only once). Does anyone know why this is?
Script and output shown below. Why does the second line in the output happen?
Script
#!/bin/bash
# Find any FILES that are EXECUTABLE in the SCRIPTS
# directory and display any description, if there is one
find /opt/scripts/. -perm -111 -type f -maxdepth 1 | while read line ;
do
file=$(basename "$line")
printf "\033[1m%10s\033[0m : " $file
grep "# DESC:" "$line" | cut -c 9-
done
Output
desc : Displays all the scripts and their descriptions
DESC:" "$line" | cut -c 9-
showhelp : Displays the script help file
test : Script to perform system testing
Reason
Presumably your grepping script is also in /opt/scripts?
So it finds itself, and finds the grep subject '#DESC' and prints that.
You could fix that by adding a # DESC line to the top of your grep script, and just outputting the first result found by each grep using 'grep -m1'
grep -m1 '# DESC' "$line" | cut -c 9-
<humour>Otherwise its just turtles all the way down... ;-) </humour>
Alternative Fix
You could also improve the grep by using a regular expression and anchoring to the beginning of the line:
egrep -m1 '^# DESC' "$line" | cut -c 9-