I would like to pass parameters to a perl script using positional parameters inside a bash script "tablecheck.sh". I am using an alias "tablecheck" to call "tablecheck.sh".
#!/bin/bash
/scripts/tables.pl /var/lib/mysql/$1/ /var/mysql/$1/mysql.sock > /tmp/chktables_$1 2>&1 &
Perl script by itself works fine. But when I do "tablecheck MySQLinstance", $1 stays $1. It won't get replaced by the instance. So I get the output as follows:
Exit /scripts/tables.pl /var/lib/mysql/$1/ /var/mysql/$1/mysql.sock > /tmp/chktables_$1 2>&1 &
The job exits.
FYI: alias tablecheck='. pathtobashscript/tablecheck.sh'
I have a bunch of aliases in another bash script. Hence . command.
Could anyone help me... I have gone till the 3rd page of Google to find an answer. Tried so many things with no luck.
I am a noob. But may be it has something to do with it being a background job or $1 in a path... I don't understand why the $1 won't get replaced...
If I copy your exact set up (which I agree with other commenters, is some what unusual) then I believe I am getting the same error message
$ tablecheck foo
[1]+ Exit 127 /scripts/tables.pl /var/lib/mysql/$1/ /var/mysql/$1/mysql.sock > /tmp/chktables_$1 2>&1
In the /tmp/chktables_foo file that it makes there is an additional error message, in my case "bash: /scripts/tables.pl: No such file or directory"
I suspect permissions are wrong in your case
Related
Say I start with the following statement, which echo-s a string into the ether:
$ echo "foo" 1>/dev/null
I then submit the following pipeline:
$ echo "foo" | cat -e - 1>/dev/null
I then leave the process out:
$ echo "foo" | 1>/dev/null
Why is this not returning an error message? The documentation on bash and piping doesn't seem to make direct mention of may be the cause. Is there an EOF sent before the first read from echo (or whatever the process is, which is running upstream of the pipe)?
A shell simple command is not required to have a command name. For a command without a command-name:
variable assignments apply to the current execution environment. The following will set two variables to argument values:
arg1=$1 arg3=$3
redirections occur in a subshell, but the subshell doesn't do anything other than initialize the redirect. The following will truncate or create the indicated file (if you have appropriate permissions):
>/file/to/empty
However, a command must have at least one word. A completely empty command is a syntax error (which is why it is occasionally necessary to use :).
Answer summarized from Posix XCU§2.9.1
In Ubuntu 14.04, I created the following bash script:
flock -nx "$1" xdg-open "$1" &
The idea is to lock the file specified in $1 (flock), then open it in my usual editor (xdg-open), and finally return to prompt, so I can open other files in sequence (&).
However, the & isn't working as expected. I need to press Enter to make the shell prompt appear again. In simpler constructs, such as
gedit test.txt &
it works as it should, returning the prompt immediately. I think it has to do with the existence of two commands in the first line. What am I doing wrong, please?
EDIT
The prompt is actually there, but it is somehow "hidden". If I issue the command
sudo ./edit error.php
it replies with
Warning: unknown mime-type for "error.php" -- using "application/octet-stream"
Error: no "view" mailcap rules found for type "application/octet-stream"
Opening "error.php" with Geany (application/x-php)
__
The errors above are not related to the question. But instead of __ I see nothing. I know the prompt is there because I can issue other commands, like ls, and they work. But the question remains: WHY the prompt is hidden? And how can I make it show normally?
Why isn't this command returning to shell after &?
It is.
You're running a command in the background. The shell prints a new prompt as soon as the command is launched, without waiting for it to finish.
According to your latest comment, the background command is printing some message to your screen. A simple example of the same thing:
$ echo hello &
$ hello
The cursor is left at the beginning of the line after the $ hello.
As far as the shell is concerned, it's printed a prompt and is waiting a new command. It doesn't know or care that a background process has messed up your display.
One solution is to redirect the command's output to somewhere other than your screen, either to a file or to /dev/null. If it's an error message, you'll probably have to redirect both stdout and `stderr.
flock -nx "$1" xdg-open "$1" >/dev/null 2>&1 &
(This assumes you don't care about the content of the message.)
Another option, pointed out in a comment by alvits, is to sleep for a second or so after executing the command, so the message appears followed by the next shell prompt. The sleep command is executed in the foreground, delaying the printing of the next prompt. A simple example:
$ echo hello & sleep 1
hello
[1] + Done echo hello
$
or for your example:
flock -nx "$1" xdg-open "$1" & sleep 1
This assumes that the error message is printed in the first second. That's probably a valid assumption for you example, but it might not be in general.
I don't think the command is doing what you think it does.
Have you tried to run it twice to see if the lock cannot be obtained the second time.
Well, if you do it, you will see that it doesn't fail because xdg-open is forking to exec the editor. Also if it fails you expect some indication.
You should use something like this
flock -nx "$1" -c "gedit '$1' &" || { echo "ERROR"; exit 1; }
I am writing a script that among other things runs a shell command several times. This command doesn't handle exit codes very well and I need to know if the process ended successfully or not.
So what I was thinking is to analyze the stderr to find out the word error (using grep). I know this is not the best thing to do, I'm working on it....
Anyway, the only way I can imagine is to put the stderr of that program in a variable and then use grep to well, "grep" it and throw it to another variable. Then I can see if that variable is valorized, meaning that there was an error, and do my work.
The qustion is: how can I do this ?
I don't really want to run the program inside a variable, because it has got a lot of arguments (with special characters such as backslash, quotes, doublequotes...) and it's a memory and I/O intensive program.
Awaiting your reply, thanks.
Redirect the stderr of that command to a temporary file and check if the word "error" is present in that file.
mycommand 2> /tmp/temp.txt
grep error /tmp/temp.txt
Thanks #Jdamian, this was my answer too, in the end.
I asked my principal if I can write a temp file and it allowed, so this is the end result:
... script
command to be launched -argument -other "argument" -other "other" argument 2>&1 | tee $TEMPFILE
ERRORCODE=( `grep -i error "$TEMPFILE" `)
if [ -z $ERRORCODE ] ;
then
some actions ....
I didn't tested this yet because I got some other scripts involved that I need to write before.
What I'm trying to do is:
run the command, having its stderr redirected to stdout;
using tee, have the above result printed on screen and also to the temp file;
have grep to store the string error found on the temp file (if any) in a variable called ERRORCODE;
if that variable is populated (which mean if it has been created by grep), then the script stops, quitting with status 1, else it contiunes.
What do you think ?
If you don't need the standard output:
if mycommand 2>&1 >/dev/null | grep -q error; then
echo an error occurred
fi
I know this has been asked many times, but I can find a suitable answer in my case.
I croned a backup script using rsync and would like to see all output, errors or not, from the all script commands. I must write the command inside the script itself, and do not want to see output in my shell.
I have been trying with no success. Below part of the script.
#!/bin/bash
.....
BKLOG=/mnt/backup_error_$now.txt
# Log everything to log file
# something like
exec 2>&1 | tee $BKLOG
# OR
exec &> $BKLOG
I have been adding at the script beginig all kinds of exec | tee $BKLOG with adding &>, 2>&1at various part of the command line, but all failed. I either get an empty log file or incomplete. I need to see on log file what rsync has done, and the error if script failed before syncing.
Thank you for help. My shell is zsh, so any solution in zsh is welcomed.
To redirect all the stdout/stderr to a file place this line on top of your script:
BKLOG=/mnt/backup_error_$now.txt
exec &> "$BKLOG"
I am trying to implement something which my logic says can't be done. But I need your help to understand why can't it be.
Short Version of Question
Is it possible to log stdout+stderr of a script in csh without using file redirection ( >& or tee ).
Detailed Explanation of Question
I have a requirement with a csh script (script1) where I am not allowed to use file redirection.(I will give the reason in a while)
So that means I can't use something like
echo just checking >& logfile
hence I can't use this or tee to create my logfile.
I also have a another script (script2) which is a top level script.
I can either run script1 in standalone mode or through script2.
In either case i need to create a log(stdout+stderr) of script1 in logfile.
There are two possible(but not complete) option for that
write this line in script2
./script1 >& logfile
But then I can't log script1 in logfile when script1 is run in standalone mode.
Another option is to use file redirections in script1 like this:
echo test starting >> logfile
echo test over
In this case thee are two disadvantages:
1) "test over" prints before "test starting" , i.e. the order of occurring of command logs is not certain.
2) It's tedious to put >>& after every statement if I am intending to cover whole script.
Now is there any other way,I can get what I need. That is I can run script1 without file redirection and still get to log its stdout+stderr in logfile.
You mention csh, so this may not help you. On the other had, it may motivate you to stop using csh for scripts, a task for which it is notoriously inappropriate. In sh, you can simply do:
#!/bin/sh
exec > logfile 2>&1
echo foo
To write foo (and the output and errors of all subsequent commands) to the logfile