So currently i need to grep a "." from a variable and if it is successful than do some work. So currently i am using this code
if echo "$arg3" | grep -iv '\.' ;
then
FS="'$arg4.$arg3'"
fi
Here i will pass the value in $arg3 like test and look for the "." in the variable and if it is not there than add a substring $arg4 and store it to the another variable.
SO my problemis, it is working but when i run my script it will print the $arg3 value as well.
Thanks in regards,
Vivek
You can redirect the grep output to null as below:
if echo "$arg3" | grep -iv '\.' > /dev/null 2>&1
Related
Is it possible in bash to filter out a part of a string with another given string ?
I have a fixed list of motifs defined in a string. The order IS important and I want to keep only the parts that are passed as a parameter ?
myDefaultList="s,t,a,c,k" #order is important
toRetains="k,t,c,u" #provided by the user, order is not enforced
retained=filter $myDefaultList $toRetains # code to filter
echo $retained # will print t,c,k"
I can write an ugly method that will use IFS, arrays and loops, but I wonder if there's a 'clever' way to do that, using built-in commands ?
here is another approach
tolines() { echo $1 | tr ',' '\n'; }
grep -f <(tolines "$toRetains") <(tolines "$myDefaultList") | paste -sd,
will print
t,c,k
assign to a variable as usual.
Since you mention in your comments that you are open to sed/awk , check also this with GNU awk:
$ echo "$a"
s,t,a,c,k
$ echo "$b"
k,t,c,u
$ awk -v RS=",|\n" 'NR==FNR{a[$1];next}$1 in a{printf("%s%s",$1,RT)}' <(echo "$b") <(echo "$a")
t,c,k
#!/bin/bash
myDefaultList="s,t,a,c,k"
toRetains="s,t,c,u"
IFS=","
for i in $myDefaultList
do
echo $toRetains | grep $i > /dev/null
if [ "$?" -eq "0" ]
then
retained=$retained" "$i
fi
done
echo $retained | sed -e 's/ /,/g' -e 's/,//1'
I have checked it running for me. Kindly check.
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
I'm trying to save the pgrep -f "instance" output inside a variable in a bash script. For some reason it doesn't work:
here is the code:
function check_running() {
local component_identifier=$1
local process_check=`get_component_param_value $component_identifier $process_check_col | tr -d '\n'`
if [ "$process_check" != "n/a" ]; then
log "process to check: ****""$process_check""****"
pid=`pgrep -f $process_check`
log "pid: " $pid
fi
}
I have tried with different ways, in single and double quotes.
Also, neither this works:
pid=$(pgrep -f "$process_check")
Please note that the process_check variable returns correctly and is definitely working.
I believe the problem is that this field is at the end of the line and may contain a \n char, that is why I've added a tr in the process_check var.
Any idea?
this is the output of the logs:
process to check: ****"instance"****
pid:
I found a way to answer this question:
echo `ps -ef | grep "$process_check" | grep -wv 'grep\|vi\|vim'` | awk '{print $2}'
I hope other people will find this useful
Well, in my case it didn't like the arguments I was trying to pass.
pgrep -f "example" # working
pgrep -f "-f -N example" # not working
I need to check if the recipient username is in file /etc/passwd which contains all the users in my class, but I have tried a few different combinations of if statements and grep without success. The best I could come up with is below, but I don't think it's working properly.
My logic behind it is that if the grep is null, the user is invalid.
send_email()
{
message=
address=
attachment=
validuser=1
until [ "$validuser" = "0" ]
do
echo "Enter the email address: "
read address
if [ -z grep $address /etc/passwd ]
then
validuser=0
else
validuser=1
fi
echo -n "Enter the subject of the message: "
read message
echo ""
echo "Enter the file you want to attach: "
read attachment
mail -s "$message" "$address"<"$attachment"
done
press_enter
}
Just do a simple if like this:
if grep -q $address /etc/passwd
then
echo "OK";
else
echo "NOT OK";
fi
The -q option is used here just to make grep quiet (don't output...)
Use getent and check for grep's exit code. Avoid using /etc/passwd. Equivalent in the shell:
getent passwd | grep -q valid_user
echo $?
Output:
0
And:
getent passwd | grep -q invalid_user
echo $?
Output:
1
Your piece of code:
if [ -z grep $address /etc/passwd ]
You haven't saved the results of grep $address /etc/passwd in a variable. Before putting it in the if statement and then testing the variable to see if it is empty.
You can try it like this:
check_address=`grep $address /etc/passwd`
if [ -z "$check_address" ]
then
validuser=0
else
validuser=1
fi
The -z check is for variable strings, which your grep isn't giving. To give a value from your grep command, enclose it in $():
if [ -z $(grep $address /etc/passwd) ]
The easiest one will be this:
cat test123
# Output: 12345678
cat test123 | grep 123 >/dev/null && echo "grep result exist" || echo "grep result doesn't exist"
# Output: grep result exist
cat test123 | grep 999 >/dev/null && echo "grep result exist" || echo "grep result doesn't exist"
# Output: grep result doesn't exist
My problem was that the file I was trying to grep was a binary file. On windows, the first two characters in the file were little squares. On mac, the first two characters were question marks. When I used more or less on the file, I could see it was binary and when I used diff, it responded that the "Binary files foo.log and requirements.txt differ".
I used cat to display the contents of the file, highlighted and copied the text (minus the two question marks at the top, deleted the file, created a new file with touch and then used vi to paste the text back into the new file.
After that, grep worked fine.
Shorter:
until ! grep $address /etc/passwd ; do {
do_your_stuff
}
I took the process ID and stored it in a variable. And I navigate to another directory. Then I should pass that process ID to get results in another command. how to achieve this? I tried the below with no luck... Please help me
pid=$(ps -ef | grep j[a]va | grep wmip | awk '{ print $2 }' | sed -n 1p)
echo $pid
#query=$(echo "cd /wmip/webMethods7/jvm/linux150/bin/ ; ./jstat -gcutil $pid")
#echo $query
result=$(cd /wmip/webMethods7/jvm/linux150/bin/ ; ./jstat -gcutil $pid)
echo $result
You should use the full path of ./jstat
Could this work? 'pid' could be a reserved word, change it
jpid=$(pgrep -f 'java .*wmip')
result=$(/wmip/webMethods7/jvm/linux150/bin/jstat -gcutil $jpid)