Software :: Scripting : Using A Variable In A Command?
Jan 13, 2009
I need to create a zip file of jpg and bmp files. The zip file is named after the first file it finds which ends with .dat. Here is my script:
Code:
DAT_FILE= `find . -maxdepth 1 -iname "*.dat" | head -1 | sed 's/..(.*)..../1/'`
(cd pics; find . ( -name "*.bmp" -o -name "*.jpg" ) -print | zip ../$DAT_FILE -@ )
BTW my sed command cut off the first two chars and last four chars since find will return the filename is the form of "./filename.dat" and I just want to extract filename. When I run this script, it creates a zip file named ".zip". How do I fix this so the zip file is named after my dat file?
I did some searches and after a few hours was able to get what I needed. What I didn't find was a fully encompased means of what I'm used to in the windows world in working with delimted files. Hopefully this is helpful to others and if there is something better or leaner way, even better.We have an issue where managing printers, just viewing on RHEL w/ sys-conf-prtr we lose any number of, up to ~30 printers from lpadmin. Rather than stare and compare to find the missing ones, I wanted to make an intuitive script. This is what I came up with.
At my wit's end I can't find anything that I understand well enough to use. This is for a Unix class, we are working with shell scripting. File1 has 5 in it and File2 has 100 in it.The teacher wants us to read the values then do the math. This is what I have so far:#!/bin/bashvar1='cat File1'var2='cat File2'var3=`echo "scale=4; $var1 / $var2" | bc`echo The final result is: $var3
Now in my bash script, I want to get the output /home/user instead of $HOME once read. So far, I have managed to get the $HOME variable but I can't get it to echo the variable. All I get is the output $HOME.
How do i find out if a particular item is a file or a folder through the terminal ls -la gives 'd' before the permissions for every folder and '-' before every fileLike i want to write a script that backup data if it is a folder and deletes if it is a file
I'm using the linux 'script' command [URL]... to track some interactive sessions. The output files from that contain unprintable characters, including my backspace keystrokes.
Is there a way to tidy these output files up so they only contain what was displayed on screen?
Or is there another way to record an interactive shell session (input and output)?
I am attempting to create a script for Grid job submissions. I have most of it completed but when testing I've run into some problems.
How it works is that I have 2 wrapper files, one is the top of the script, the other the bottom, the original script imports commands for the middle.
Basically this is what is to happen: cat script-start.sh >> workerscript.sh echo commands >> workerscript.sh cat script-end.sh >> workerscript.sh
workerscript.sh gets created with no problems but when it is run I get: export: command not found
I think that the new script is not calling bash which gives the problem. I don't know how to fix the problem though because I have the shell set in the script-start.sh.
#!/bin/sh - tried /bin/bash, same result export PROJECT=/...... export PROJSOURCE=/......
Given that I want my shell script be invoked at the command line using the above parameters - where [these brackets] denote that they are optional - what is the best method to parse them?
I am very new to shell scripting.How does one pass a command-line parameter to a shell script?for the below program #/bin/bash mount -t cifs -o user=ramkannan,password=Linux123@ //10.200.1.125/ramkannan /MT cd /MT/test date=`/bin/date "+\%Y-\%m-\%d-\%H-\%M-\%S"` mysqldump -uroot -pram2@ employeedb > $date.sql gzip $date.sql
I want to pass parameter for everything,i tried in google and did but iam getting error while passing parameter to all
#/bin/bash mount -t cifs -o user=$1,password=$2 //10.200.1.125/ramkannan /MT cd /MT/test date=`/bin/date "+\%Y-\%m-\%d-\%H-\%M-\%S"` mysqldump -uroot -pram2@ employeedb > $date.sql gzip $date.sql
i was getting error while passing parameter to all.
I have written the following script in my linux server to add users for LDAP database.But i can't able to run this.
The script is as following
#!/bin/bash echo "Mention the username which you want to convert LDIF format" read username if ["$username" -e "/ldiffile/passwd"]; then echo "Username already exists" else cat /etc/passwd | grep -i "$username" >> /ldiffile/passwd fi The output which i got : . ldapadd.sh Mention the username which you want to convert LDIF format yal2361 -bash: [yal2361: command not found
i'm trying to execute a shell script, i'm trying to use the values in an array for use in a sed command:
sed -n '/Sales ID: ${array[$i]}/,/Totals:/p'
that command creates empty files. so my guess is that its not recognizing the array as an array but as text? how would i be able to utilize the array in the command? i got it, didnt think that if i doubled up the single quotes that it would work, but this worked:
I can print a specific line of a file with:$ sed -n '20p' myFileHow can I store it in a variable (in a shell script)?(I wasn't successful with "myVar=sed -n '20p' myFile" for example)
I was reading that if I want to do a one time scheduled command, I should use at, which I've never done, as opposed to cron, which i'm kinda familiar with. But what I want to do is reboot my server at 3am tomorrow and force it to check the file systems with a shutdown -rF. For this do I even need to use "at" or could I just say shutdown -rF 3:00.Will that also know that I mean 3am tomorrow and not say in 3 minutes from now or 3pm?
I tried using the tail command in my shell script and storing that value in a variable a but an error keeps coming. Is there any other way to store the output of a command into a variable. Cannot Read text from text file and store it in a variable using shell script. The thing is I need a number from the file new.txt and use that number in my script
#!/bin/bash a = `tail -1 new.txt|head -n 1` echo $a
I want to display 4 options using the case command and refresh the screen when options 1 and 2 are chosen (no changes to the options and you get asked again to chose option), but give a message for option 3 and exit on option 4. I set this up with the script below, but choosing option 1 works and choosing option 2 exits the script.
I am having all sorts of trouble trying to assign a variable within an awk script with the system command. I know there is a lot of ways around this problem, but for efficiency reasons, I would like to, within my awk script, do something like
system(x=3)
or
system(x=NR)
and, latter on the shell script which calls the awk script, use the variable $x. But nothing is passed to x. I have already tried things like
command = "x=3" system(command)
and also used a pipeline within the system to pipe it to /bin/sh In fact tried a lot of stuff like that, using $(( )) etc etc etc I can create directories e write to files (yes, i could write to a file and read from there, but I dont think it is efficient, plus I am puzzled).
When I run this command from shell, it runs ok export REVS=`svn info svn+ssh://svn.myone.ca/var/svn/story/trunk/lib |grep 'Last Changed Rev:'| awk -F: '{print $2}'` However when I save it into a file called test.sh (of course, I chmod it with +x), I got error "export: 2: bad variable name"
Here is the file: #!/bin/bash export REVS=`svn info svn+ssh://svn.myone.ca/var/svn/story/trunk/lib |grep 'Last Changed Rev:'| awk -F: '{print $2}'` I am using ubuntu.
!<number> to execute the Nth command(use history to see the list). Or you can use
Code:
cd !-2:1
to cd into the value in the first field that was executed 2 commands ago Anyhow, say I run a command and the output is a path. Any way to cd and then some variable where OUTPUT of the previous command was stored? A variable that always stores the OUTPUT of the last command.
