In Bash in the terminal, it is impossible to run a command that ends with and & (being sent to the background) followed by another command (obviously with a ; between them). Why is that? Why can't you run anything with &; or $ ; in it?
I'm trying to recreate a 502 error and was trying to DoS a specific page in a testing server. I was trying to run this:
while true; do curl -s https://some.site.com/someImage.jpg > /dev/null &; echo blah ; done
as a "one-liner" in the terminal. However, I got this error:
-bash: syntax error near unexpected token `;'
However the commands work individually, and when I run the curl command not in the background it works as a loop as well. It also works when it write a one line script, "/tmp/curlBack.sh" that includes only
curl -s https://some.site.com/someImage.jpg > /dev/null &
And then run
while true; do bash /tmp/curlBack.sh ; echo blah ; done
Why am I getting this error and how do I fix it?
The problem is with the semi-colon after the ampersand (&):
An ampersand does the same thing as a semicolon or newline in that it indicates the end of a command, but it causes Bash to execute the command asynchronously.
BashSheet
Basically, it's as if you were to put double semi-colons back to back, that would cause a syntax error too.
To fix this problem, you simply need to remove the semi-colon, I tested it that way and it seems to be working:
while true; do curl -s https://some.site.com/someImage.jpg > /dev/null & echo blah ; done
Related
I want to ssh to a node and run a program in a background using &. So, the script looks like
#!/bin/bash
ssh NODENAME 'lmgrd -c license.lic &;
exit;'
However, the bash interpreter complains for syntax error near unexpected token ;. It seems that &
& already finisheds a command. So after & you simply don't use any terminator.
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 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
i have got a bash file which i want to toggle via the sudo crontab list. Problem is, that it does not work, because when i run the script with sudo, there is a syntax error message on this line:
size=(`du -h $backupDir --summarize`)
If i run the same script without, i have to type the sudo pw, but it works without any problems.
I allready tried a few variations with brackets, with or without backticks, with or without spaces, etc but nothing helped. The error message is:
Syntax error: "(" unexpected (expecting ";;")
Any help?
The problem here is that you use bash-syntax, and the script (when it is executed from cron) is interpreted by /bin/sh (that known nothing about arrays and the () construction.).
You must either specify bash as an interpreter of the script using she-bang notation:
#!/bin/bash
or run the script explicitly with bash from cron:
bash /path/to/script
or rewrite script so, that it could run without arrays.
Fro readability use the $() form
size=$(du -h $backupDir --summarize)
If you want to stick to back quotes, then
size=`du -h $backupDir --summarize`
I would like to run a few instances of my bash script foo.bash in background.
When I write for i in {1..10}; do ~/bin/foo.bash & ; done in the command line I get an error: bash: syntax error near unexpected token ;
Could you explain why this error occurs and how to fix the command?
& and ; are both command separators; you don't need (and can't have) both.
for i in {1..10}; do ~/bin/foo.bash & done