I need some help...
I'm creating a script of unit tests using shellscripts. That script, stores all the beeline calls from all files inside a directory.
The script is doing it's purpose, but I don't wanna append the file name if grep does not return results.
That's my code:
for file in $(ls)
do
cat $file | egrep -on '^( +)?\bbeeline.*password=;"?' >> testa_scripts.sh
echo $file >> testa_scripts.sh
done
How can I do that?
Thanks
grep returns a falsy exit status (1) if it doesn't find any matching lines, so you can put in an if statement to test if it matched anything. Inverted with ! here:
for file in ./*; do
if ! egrep -on '...' "$file" >> somefile; then
echo 'grep did not match anything'
fi
done
(I don't think there's any need for the ls instead of just a shell glob here.)
Related
I'm having a trouble with my code.
grep looks for files that doesn't have a word 'code'
and I need to add 'doesn't have' as a last line in those files
By logic
echo 'doesnt have' >> grep -ril 'code' file/file
I'm using -ril to ignore the cases and get file names
Does anyone know how to append a text to each .txt files found from grep searches?
How's this for a novel alternative?
echo "doesn't have" |
tee -a $(grep -riL 'code' file/file)
I switched to the -L option to list the files which do not contain the search string.
This is unfortunately rather brittle in that it assumes your file names do not contain whitespace or other shell metacharacters. This can be fixed at the expense of some complexity (briefly, have grep output zero-terminated matches, and read them into an array with readarray -d ''. This requires a reasonably recent Bash, and probably GNU grep.)
The 'echo' command can append output to a single file, which must be specified by redirecting the standard output. To update multiple files, a loop is needed. The loop iterated over all the files found with 'grep'
for file in $(grep -ril 'code' file/file) ; do
echo 'doesnt have' >> $file
done
Using a while read loop and Process Substitution.
#!/usr/bin/env bash
while IFS= read -r files; do
echo "doesn't have" >> "$files"
done < <(grep -ril 'code' file/file)
As mentioned by #dibery
#!/bin/sh
grep -ril 'code' file/file | {
while IFS= read -r files; do
echo "doesn't have" >> "$files"
done
}
I'm sorry for my poor English, first.
I want to read a file (tel.txt) that contains many tel numbers (a number per line) and use that line to grep command to search about the specific number in the source file (another file)!
I wrote this code :
dir="/home/mujan/Desktop/data/ADSL_CDR_Text_Parts_A"
file="$dir/tel.txt"
datafile="$dir/ADSL_CDR_Like_Tct4_From_960501_to_97501_Part0.txt"
while IFS= read -r line
do
current="$line"
echo `grep -F $current "$datafile" >> output.txt`
done < $file
the tel file sample :
44001547
44001478
55421487
but that code returns nothing!
when I declare 'current' variable with literals it works correctly!
what happened?!
Your grep command is redirected to write its output to a file, so you don't see it on the terminal.
Anyway, you should probably be using the much simpler and faster
grep -Ff "$file" "$datafile"
Add | tee -a output.txt if you want to save the output to a file and see it at the same time.
echo `command` is a buggy and inefficient way to write command. (echo "`command`" would merely be inefficient.) There is no reason to capture standard output into a string just so that you can echo that string to standard output.
Why don't you search for the line var directly? I've done some tests, this script works on my linux (CentOS 7.x) with bash shell:
#!/bin/bash
file="/home/mujan/Desktop/data/ADSL_CDR_Text_Parts_A/tel.txt"
while IFS= read -r line
do
echo `grep "$line" /home/mujan/Desktop/data/ADSL_CDR_Text_Parts_A/ADSL_CDR_Like_Tct4_From_960501_to_97501_Part0.tx >> output.txt`
done < $file
Give it a try... It shows nothing on the screen since you're redirecting the output to the file output.txt so the matching results are saved there.
You should use file descriptors when reading with while loop.instead use for loop to avoid false re-directions
dir="/home/mujan/Desktop/data/ADSL_CDR_Text_Parts_A"
file="$dir/tel.txt"
datafile="$dir/ADSL_CDR_Like_Tct4_From_960501_to_97501_Part0.txt"
for line in `cat $file`
do
current="$line"
echo `grep -F $current "$datafile" >> output.txt`
done
I have prepared a bash script to get only the directory (not full path) with file name where file is present. It has to be done only when file is located in sub directory.
For example:
if input is src/email/${sub_dir}/Bank_Casefeed.email, output should be ${sub_dir}/Bank_Casefeed.email.
If input is src/layouts/Bank_Casefeed.layout, output should be Bank_Casefeed.layout. I can easily get this using basename command.
src/basefolder is always constant. In some cases (after src/email(basefolder) directory), sub_directories will be there.
This script will work. I can use this script (only if module is email) to get output. but script should work even if sub directory is present in other modules. Maybe should I count the directories? if there are more than two directories (src/basefolder), script should get sub directories. Is there any better way to handle both scenarios?
#!/bin/bash
filename=`basename src/email/${sub_dir}/Bank_Casefeed.email`
echo "filename is $filename"
fulldir=`dirname src/email/${sub_dir}/Bank_Casefeed.email`
dir=`basename $fulldir`
echo "subdirectory name: $dir"
echo "concatenate $filename $dir"
Entity=$dir/$filename
echo $Entity
Using shell parameter expansion:
sub_dir='test'
files=( "src/email/${sub_dir}/Bank_Casefeed.email" "src/email/Bank_Casefeed.email" )
for f in "${files[#]}"; do
if [[ $f == *"/$sub_dir/"* ]]; then
echo "${f/*\/$sub_dir\//$sub_dir\/}"
else
basename "$f"
fi
done
test/Bank_Casefeed.email
Bank_Casefeed.email
I know there might be an easier way to do this. But I believe you can just manipulate the input string. For example:
#!/bin/bash
sub_dir='test'
DIRNAME1="src/email/${sub_dir}/Bank_Casefeed.email"
DIRNAME2="src/email/Bank_Casefeed.email"
echo $DIRNAME1 | cut -f3- -d'/'
echo $DIRNAME2 | cut -f3- -d'/'
This will remove the first two directories.
I have a folder with lots of files having a pattern, which is some string followed by a date and time:
BOS_CRM_SUS_20130101_10-00-10.csv (3 strings before date)
SEL_DMD_20141224_10-00-11.csv (2 strings before date)
SEL_DMD_SOUS_20141224_10-00-10.csv (3 strings before date)
I want to loop through the folder and extract only the part before the date and output into a file.
Output
BOS_CRM_SUS_
SEL_DMD_
SEL_DMD_SOUS_
This is my script but it is not working
#!/bin/bash
# script variables
FOLDER=/app/list/l088app5304d1/socles/Data/LEMREC/infa_shared/Shell/Check_Header_T24/
LOG_FILE=/app/list/l088app5304d1/socles/Data/LEMREC/infa_shared/Shell/Check_Header_T24/log
echo "Starting the programme at: $(date)" >> $LOG_FILE
# Getting part of the file name from FOLDER
for file in `ls $FOLDER/*.csv`
do
mv "${file}" "${file/date +%Y%m%d HH:MM:SS}" 2>&1 | tee -a $LOG_FILE
done #> $LOG_FILE
Use sed with extended-regex and groups to achieve this.
cat filelist | sed -r 's/(.*)[0-9]{8}_[0-9][0-9]-[0-9][0-9].[0-9][0-9].csv/\1/'
where filelist is a file with all the names you care about. Of course, this is just a placeholder because I don't know how you are going to list all eligible files. If a glob will do, for example, you can do
ls mydir/*.csv | sed -r 's/(.*)[0-9]{8}_[0-9][0-9]-[0-9][0-9].[0-9][0-9].csv/\1/'
Assuming you wont have numbers in the first part, you could use:
$ for i in *csv;do str=$(echo $i|sed -r 's/[0-9]+.*//'); echo $str; done
BOS_CRM_SUS_
SEL_DMD_
SEL_DMD_SOUS_
Or with parameter substitution:
$ for i in *csv;do echo ${i%_*_*}_; done
BOS_CRM_SUS_
SEL_DMD_
SEL_DMD_SOUS_
When you use ${var/pattern/replace}, the pattern must be a filename glob, not command to execute.
Instead of using the substitution operator, use the pattern removal operator
mv "${file}" "${file%_*-*-*.csv}.csv"
% finds the shortest match of the pattern at the end of the variable, so this pattern will just match the date and time part of the filename.
The substitution:
"${file/date +%Y%m%d HH:MM:SS}"
is unlikely to do anything, because it doesn't execute date +%Y%m%d HH:MM:SS. It just treats it as a pattern to search for, and it's not going to be found.
If you did execute the command, though, you would get the current date and time, which is also (apparently) not what you find in the filename.
If that pattern is precise, then you can do the following:
echo "${file%????????_??-??-??.csv}" >> "$LOG_FILE"
using grep:
ls *.csv | grep -Po "\K^([A-Za-z]+_)+"
output:
BOS_CRM_SUS_
SEL_DMD_
SEL_DMD_SOUS_
My code is given below. Echo works fine. But, the moment I redirect output of echo to touch, I get an error "no such file or directory". Why ? How do i fix it ?
If I copy paste the output of only echo, then the file is created, but not with touch.
while read line
do
#touch < echo -e "$correctFilePathAndName"
echo -e "$correctFilePathAndName"
done < $file.txt
If you have file names in each line of your input file file.txt then you don't need to do any loop. You can just do:
touch $(<file.txt)
to create all the files in one single touch command.
You need to provide the file name as argument and not via standard input. You can use command substitution via $(…) or `…`:
while read line
do
touch "$(echo -e "$correctFilePathAndName")"
done < $file.txt
Ehm, lose the echo part... and use the correct variable name.
while read line; do
touch "$line"
done < $file.txt
try :
echo -e "$correctFilePathAndName" | touch
EDIT : Sorry correct piping is :
echo -e "$correctFilePathAndName" | xargs touch
The '<' redirects via stdin whereas touch needs the filename as an argument. xargs transforms stdin in an argument for touch.