for d in */ ; do
cd $d
NUM = $(echo ${PWD##*/} | grep -q "*abc*");
if [[ "$NUM" -ne "0" ]]; then
pwd
fi
cd ..
done
Here I'm trying to match a folder name to some substring 'abc' in the name of the folder and check if the output of the grep is not 0. But it gives me an error which reads that NUM: command not found
An error was addressed in comments.
NUM = $(echo ${PWD##*/} | grep -q "*abc*"); should be NUM=$(echo ${PWD##*/} | grep -q "*abc*");.
To clarify, the core problem would be to be able to match current directory name to a pattern.
You can probably simply the code to just
if grep -q "*abc*" <<< "${PWD##*/}" 2>/dev/null; then
echo "$PWD"
# Your rest of the code goes here
fi
You can use the exit code of the grep directly in a if-conditional without using a temporary variable here ($NUM here). The condition will pass if grep was able to find a match. The here-string <<<, will pass the input to grep similar to echo with a pipeline. The part 2>/dev/null is to just suppress any errors (stderr - file descriptor 2) if grep throws!
As an additional requirement asked by OP, to negate the conditional check just do
if ! grep -q "*abc*" <<< "${PWD##*/}" 2>/dev/null; then
Related
I am trying to execute the command git diff | grep pkg/client/clientset | wc -l and check if the output is more than 0 or not. I have the following script
if [ "$(git diff | grep pkg/client/clientset | wc -l "$i")" -gt "0" ]; then
echo "Hello"
fi
I am getting the following error while executing the script. The error I am getting is
line 29: [: : integer expression expected
Any idea of what can be going wrong?
Comparing the number of output lines to zero is almost always an antipattern. diff and grep both already tell you whether there was a difference (exit code 1) or a match (exit code 0) precisely so you can say
if diff old new; then
echo "There were differences"
fi
if git diff --exit-code; then
echo "There were differences"
fi
if git diff --exit-code pkg/client/clientset; then
echo "There were differences in this specific file"
fi
if git diff | grep -q pkg/client/clientset; then
echo "Hello"
fi
Notice that git diff requires an explicit option to enable this behavior.
-- EDIT --
There were some incorrect statements in the answer, pointed-out by commentators Gordon Davisson and iBug. They have been corrected in this version of the answer. The final conclusion (remove the "$i") remains the same though.
wc -l "$i" will count the lines in the file $i. If you never used i as a variable, then i will be empty and the command will be wc -l "". The output of that will be empty on STDOUT en contain wc: invalid zero-length file name on STDERR. If the variable i is used, wc will most likely complain about a non-existing file. The point is, that wc will not read STDIN.
I also made some incorrect statements about the quoting. As pointed out, between the ( and ), it is a different quoting context. This can be shown as follows:
$ a="$(/usr/bin/echo "hop")"
$ echo $a
hop
$ b=hop
$ a="$(/usr/bin/echo "$b")"
$ echo $a
hop
Just removing "$i" from the wc-l will solve your issue.
if [ "$(git diff | grep pkg/client/clientset | wc -l)" -gt "0" ]; then
echo "Hello"
fi
Only a note, that is to long for a comment:
if [ "$(git diff | grep pkg/client/clientset | wc -l "$i")" -gt "0" ]; then
I think you will test the existence of string pkg/client/clientset to enter the then part. In this case you can use:
if git diff | grep -q pkg/client/clientset; then
grep will only returns a status because option -q. The status is true after the first occurrence of the string. At this point grep stops. And this status is used by if.
I'm trying to use an if [grep] using the exit value to trigger it
to test, from the command line
$ ls ~/dir | grep txt
$ echo $?
0
However when i use it in an if statement in a script, i.e.
if [ ls /some/dir | grep -q pattern ]; then
echo y
fi
It says
line 1: [ missing `]' (the line the if statement was written on)
Why is this happening and is there a way to fix it?
That is because [ is and conditional expression, check the details.
You should use () instead:
if (ls /some/dir | grep -q pattern); then
echo y
fi
I wrote what I thought was a quick script I could run on a bunch of machines. Instead it print what looks like might be directory contents in a recursive search:
version=$(mysql Varnish -B --skip-column-names -e "SELECT value FROM sys_param WHERE param='PatchLevel'" | sed -n 's/^.*\([0-9]\.[0-9]*\).*$/\1/p')
if [[ $(echo "if($version == 6.10) { print 1; } else { print 0; }" | bc) -eq 1 ]]; then
status=$(dpkg-query -l | awk '{print $2}' | grep 'sg-status-polling');
cons=$(dpkg-query -l | awk '{print $2}' | grep 'sg-consolidated-poller');
if [[ "$status" != "" && "$cons" != "" ]]; then
echo "about to change /var/www/Varnish/lib/Extra/SG/ObjectPoller2.pm"; echo;
cp /var/www/Varnish/lib/Extra/SG/ObjectPoller2.pm /var/www/Varnish/lib/Extra/SG/ObjectPoller2.pm.bkup;
sed -ir '184s!\x91\x93!\x91\x27--timeout=35\x27\x93!' /var/www/Varnish/lib/Extra/SG/ObjectPoller2.pm;
sed -n 183,185p /var/www/Varnish/lib/Extra/SG/ObjectPoller2.pm; echo;
else
echo "packages not found. Assumed to be not applicable";
fi
else
echo "This is 4.$version, skipping";
fi
The script is supposed to make sure Varnish is version 4.6.10 and has 2 custom .deb packages installed (not through apt-get). then makes a backup and edits a single line in a perl module from [] to ['--timeout=35']
it looks like its tripping up on the sed replace one liner.
There are two major problems (minor ones addressed in comments). The first is that you use the decimal code for [] instead of the hexa, so you should use \x5b\x5d instead of \x91\x93. The second problem is that if you do use the proper codes, sed will still interpret those syntactically as []. So you can't escape escaping. Here's what you should call:
sed -ri'.bkup' '184s!\[\]![\x27--timeout=35\x27]!' /var/www/Varnish/lib/Extra/SG/ObjectPoller2.pm
And this will create the backup for you (but you should double check).
I need to make a for loop that loops for every item in a directory.
My issue is the for loop is not acting as I would expect it to.
cd $1
local leader=$2
if [[ $dOpt = 0 ]]
then
local items=$(ls)
local nitems=$(ls |grep -c ^)
else
local items=$(ls -l | egrep '^d' | awk '{print $9}')
local nitems=$(ls -l | egrep '^d' | grep -c ^)
fi
for item in $items;
do
printf "${CYAN}$nitems\n${NONE}"
let nitems--
if [[ $nitems -lt 0 ]]
then
exit 4
fi
printf "${YELLOW}$item\n${NONE}"
done
dOpt is just a switch for a script option.
The issue I'm having is the nitems count doesn't decrease at all, it's as if the for loop is only going in once. Is there something I'm missing?
Thanks
Goodness gracious, don't rely on ls to iterate over files.
local is only useful in functions.
Use filename expansion patterns to store the filenames in an array.
cd "$1"
leader=$2 # where do you use this?
if [[ $dOpt = 0 ]]
then
items=( * )
else
items=( */ ) # the trailing slash limits the results to directories
fi
nitems=${#items[#]}
for item in "${items[#]}" # ensure the quotes are present here
do
printf "${CYAN}$((nitems--))\n${NONE}"
printf "${YELLOW}$item\n${NONE}"
done
Using this technique will safely handle files with spaces, even newlines, in the name.
Try this:
if [ "$dOpt" == "0" ]; then
list=(`ls`)
else
list=(`ls -l | egrep '^d' | awk '{print $9}'`)
fi
for item in `echo $list`; do
... # do something with item
done
Thanks for all the suggestions. I found out the problem was changing $IFS to ":". While I meant for this to avoid problems with whitespaces in the filename, it just complicated things.
I have used the find command for this, but it doesnt return any message when a file is not found.
And I want the search to be recursive and return a message "not found" when a file is not found.
Here's the code I have done so far. Here "input.txt" contains the list of files to be searched.
set `cat input.txt`
echo $#
for i in $#
do
find $HOME -name $i
done
Try this:
listfile=input.txt
exec 3>&1
find | \
grep -f <( sed 's|.*|/&$|' "$listfile" ) | \
tee /dev/fd/3 | \
sed 's|.*/\([^/]*\)$|\1|' | \
grep -v -f - "$listfile" | \
sed 's/$/ Not found/'
exec 3>&-
open file descriptor 3
find the files
see if they're on the list (use sed to
send a copy of the found ones to file descriptor 3
strip off the directory name
get a list of the ones that don't appear
add the "Not found" message
close file descriptor 3
Output looks like:
/path/to/file1
/path/somewhere/file2
foo Not found
bar Not found
No loops necessary.
Whats wrong with using a script. I hope this will do.
#!/bin/bash -f
for i in $#
do
var=`find $HOME -name $i`
if [ -z "$var"]
then
var="File not found"
fi
echo $var
done
You can use the shell builtin 'test' to test the existence of a file. There is also an alternative syntax using square brackets:
if [ -f $a ]; then # Don't forget the semicolon.
echo $a
else
echo 'Not Found'
fi
Here is one way - create a list of all the files to grep against. If your implementation supports
grep -q otherwise use grep [pattern] 2&>1 >/dev/null....
find $HOME -type f |
while read fname
do
echo "$(basename $fname) $fname"
done > /tmp/chk.lis
while read fname
do
grep -q "^$fname" /tmp/chk.lis
[ $? -eq 0 ] && echo "$fname found" || echo "$fname not found"
done < /tmp/chk.lis
All of this is needed because POSIX find does not return an error when a file is not found
perl -nlE'say-f$_?$_:"not found: $_"' file