I have the following code
#!/bin/bash
output= cat $1 | sed s/$2/$3/
if [ -f "$1" ]
then
echo $output > "$1"
echo "Done"
fi
Arguments:
1 is file
2 old word
3 new word to replace
File Permission is 777 and for some reason the code will replace the current file with a newline and that's it. Any possible reason for this issue?
Try:
#!/bin/bash
output=`cat $1 | sed s/$2/$3/`
if [ -f "$1" ]
then
echo $output > "$1"
echo "Done"
fi
Related
Im writing a simple script to list all files in a directory and whether each file is in fact a file or if it is a directory. If it is a directory then it outputs how many files are in the directory.
#!/bin/bash
for filename in $(ls)
do
if [ -f "$filename" ]
then
printf "$filename - file\n"
fi
if [ -d "$filename" ]
then
count=$(ls "$filename" | wc -l)
printf "$filename - directory $count files\n"
fi
done
This works perfectly fine. But if I try to pass a command line argument (directory name) to ls then the script doesn't work. Does anyone know what causes this to break. Example below.
#!/bin/bash
for filename in $(ls $1)
do
if [ -f "$filename" ]
then
printf "$filename - file\n"
fi
if [ -d "$filename" ]
then
count=$(ls "$filename" | wc -l)
printf "$filename - directory $count files\n"
fi
done
$filename exists in directory/$filename. And you are check $filename in current (./) directory.
You should check against directory/$filename
if [ -z $1 ]; then
echo USAGE: basename $0 directory
exit
else
directory=$1
fi
...
for ...
if [ -f "$directory/$filename" ]; then
...
...
When you are passing directory as argument you shouldn't check files $filename but $1/$filename.
If you are sure that there will always be an argument you should use something like:
#!/bin/bash
for filename in $(ls "$1")
do
if [ -f "$1/$filename" ]
then
printf "$1/$filename - file\n"
fi
if [ -d "$1/$filename" ]
then
count=$(ls "$1/$filename" | wc -l)
printf "$1/$filename - directory $count files\n"
fi
done
If dirname as argument is optional you should check if there is argument and process as you want. My suggestion:
#!/bin/bash
if [ -z ${1+x} ];
then
echo "No argument";
else
echo "There is argument";
cd $1;
fi
for filename in $(ls)
do
if [ -f "$filename" ]
then
printf "$filename - file\n"
fi
if [ -d "$filename" ]
then
count=$(ls "$filename" | wc -l)
printf "$filename - directory $count files\n"
fi
done
I am trying to create a bash dictionary script that accepts first argument and creates file named after that, then script accepts next arguments (which are files inside same folder) and outputs their content into file (first argument). It also sorts, deletes symbols etc., but main problem is, that sometimes ouptut file is empty (I am passing one non empty file and one non existing file), after deleting and running script few more times it is sometimes empty sometimes not.
#!/bin/bash
numberoffileargs=$(( $# - 1 ))
exitstat=0
counterexit=0
acceptingstdin=0;
> "$1";
#check if we have given input files given
if [ "$#" -gt 1 ]; then
#for cycle going through input files
for i in "${#:2}"
do
#check whether input file is readable
if [ -r "${i}" ]; then
cat "${i}" >> "$1"
#else redirect to standard output
else
exitstat=2
counterexit=$((counterexit + 1))
echo "file does not exist" 1>&2
fi
done
else
echo "stdin code to be done"
acceptingstdin=1
#stdin input to output file
#stdin=$(cat)
fi
#one word for each line, alphabetical sort, alphabet only, remove duplicates
#all lowercase
#sort -u >> "$1"
if [ "$counterexit" -eq "$numberoffileargs" ] && [ "$acceptingstdin" -eq 0 ]; then
exitstat=3
fi
cat "$1" | sed -r 's/[^a-zA-Z\-]+/ /g' | tr A-Z a-z | tr ' ' '\n' | sort -u | sed '/^$/d' > "$1"
echo "$numberoffileargs"
echo "$counterexit"
echo "$exitstat"
exit $exitstat
Here is your script with some syntax improvement. Your trouble came from the fact that the dictionary was both on input and output on your pipeline; I added a temp file to fix it.
#!/bin/bash
(($# >= 1)) || { echo "Usage: $0 dictionary file ..." >&2 ; exit 1;}
dict="$1"
shift
echo "Creating $dict ..."
>| "$dict" || { echo "Failed." >&2 ; exit 1;}
numberoffileargs=$#
exitstat=0
counterexit=0
acceptingstdin=0
if (($# > 0)); then
for i ; do
#check whether input file is readable
if [ -r "${i}" ]; then
cat "${i}" >> "$dict"
else
exitstat=2
let counterexit++
echo "file does not exist" >&2
fi
done
else
echo "stdin code to be done"
acceptingstdin=1
fi
if ((counterexit == numberoffileargs && acceptingstdin == 0)); then
exitstat=3
fi
sed -r 's/[^a-zA-Z\-]+/ /g' < "$dict" | tr '[:upper:]' '[:lower:]' | tr ' ' '\n' |
sort -u | sed '/^$/d' >| tmp$$
mv -f tmp$$ "$dict"
echo "$numberoffileargs"
echo "$counterexit"
echo "$exitstat"
exit $exitstat
The pipeline might be improved.
currently I'm working on a code :
egrep '("$1"|"$2")' Cities.txt > test.txt
if [ $# -eq 1] && grep -q "$1" test.txt ; then
grep $1 Cities.txt
elif [ $# -eq 2 ] && egrep -q '("$1"|"$2")' test.txt ; then
egrep '("$1"|"$2")' Cities.txt > $2.txt
else $1 not found on Cities.txt
fi
exit
basically, it lets user to enter 1 or 2 arguments and the argument(s) is/are used as a grep pattern in Cities.txt and redirect the output to a file named test.txt
If the user entered 1 argument and the argument matched the content of the test.txt , then it display the lines that contain argument 1 on file Cities.txt.
If the user entered 2 argument and both argument matched the content of the file test.txt, then it matched both argument in Cities.txt and redirect the output to the file named by the user's second argument.
I couldn't seem to get the code to work, may be some of you guys could help me inspect the error.
thanks
egrep "($1|$2)" Cities.txt > test.txt # change single quote to double quote
if [ $# -eq 1 ] && grep -q -- "$1" test.txt ; then
grep -- "$1" Cities.txt
elif [ $# -eq 2 ] && egrep -q -- "($1|$2)" test.txt ; then
egrep -- "($1|$2)" Cities.txt > $2.txt
else
$1 not found on Cities.txt
fi
This greatly changes the semantics, but I believe is what you are trying to do. I've added -- to try to make this slightly robust, but if either argument contains metacharacters for the regex this will fail. But you could try:
if test $# -eq 1 && grep -F -q -e "$1" test.txt ; then
grep -F -e "$1" Cities.txt
elif [ $# -eq 2 ] && grep -q -F -e "$1" -e "$2" test.txt; then
grep -F -e "$1" -e "$2" Cities.txt > $2.txt
else
$1 not found on Cities.txt >&2
fi
I have a script which is checking a key in one file against a key in another to see if it exists in both. However in the script the grep never returns anything has been found but on the command line it does.
#!/bin/bash
# First arg is the csv file of repo keys separated by line and in
# this manner 'customername,REPOKEY'
# Second arg is the log file to search through
log_file=$2
csv_file=$1
while read line;
do
customer=`echo "$line" | cut -d ',' -f 1`
repo_key=`echo "$line" | cut -d ',' -f 2`
if [ `grep "$repo_key" $log_file` ]; then
echo "1"
else
echo "0"
fi
done < $csv_file
The CSV file is formatted as follows:
customername,REPOKEY
and the log file is as follows:
REPOKEY
REPOKEY
REPOKEY
etc
I call the script by doing ./script csvfile.csv logfile.txt
Rather then checking output of grep command use grep -q to check its return status:
if grep -q "$repo_key" "$log_file"; then
echo "1"
else
echo "0"
fi
Also your script can be simplified to:
log_file=$2
csv_file=$1
while IFS=, read -r customer repo_key; do
if grep -q "$repo_key" "$log_file"; then
echo "1"
else
echo "0"
fi
done < "$csv_file"
use the exit status of the grep command to print 1 or 0
repo_key=`echo "$line" | cut -d ',' -f 2`
grep -q "$repo_key" $log_file
if [ $? -eq 1 ]; then
echo "1"
else
echo "0"
fi
-q supresses the output so that no output is printed
$? is the exit status of grep command 1 on successfull match and 0 on unsuccessfull
you can have a much simpler version as
grep -q "$repo_key" $log_file
echo $?
which will produce the same output
My assignment is to write a Unix shell script that asks the user for the name of a directory, and then works exactly like find.
Here is what I have so far:
#!/bin/bash
dir_lister()
{
cd "$1"
echo "$1"
list=$(ls -l ${1})
nolines=$(echo "$list" | awk 'END{printf "%d",NF}')
if [ $nolines -eq 2 ]
then
echo "$1"
return
fi
filelist=$(echo "$list" | grep ^-.*)
dirlist=$(echo "$list" | grep ^d.*)
filename=$(echo "$filelist"| awk '{printf "%s\n",$NF}')
present=$(pwd)
echo "$filename"| awk -v pres=$present '{printf "%s/%s\n",pres,$0}'
dirlist2=$(echo "$dirlist" | awk '{printf "%s\n",$NF}')
echo "$dirlist2" | while IFS= read -r line;
do
nextCall=$(echo "$present/$line");
dir_lister $nextCall;
cd ".."
done
cd ".."
}
read -p "Enter the name of the direcotry: " dName
dir_lister $dName
The problem is, after a depth of three directories, this script gets into an infinite loop, and I don't see why.
EDIT:
Here is the code i came up with after looking at your answer, it still doesn't go more than 1 directory depth:
#!/bin/bash
shopt -s dotglob # don't miss "hidden files"
shopt -s nullglob # don't fail on empty directories
list_directory()
{
cd "$2"
cd "$1"
##echo -e "I am called \t $1 \t $2"
for fileName in "$1/"*
do
##echo -e "hello \t $fileName"
if [ -d "$fileName" ];
then
echo "$fileName"
list_directory $fileName $2
else
echo "$fileName"
fi
done
}
read -p "Enter the direcotory Name: " dirName
var=$(pwd)
list_directory $dirName $var
Okay, that is completely the wrong way to list files in a directory (see ParsingLs). I'll give you the pieces and you should be able to put them together into a working script.
Put this at the top of your script:
shopt -s dotglob # don't miss "hidden files"
shopt -s nullglob # don't fail on empty directories
Then you can easily loop over directory contents with:
for file in "$directory/"* ; do
#...
done
Test if you have a directory:
if [ -d "$file" ] ; then
# "$file" is a directory, recurse...
fi