I'm trying to make a very small script but i run into a problem, i want to call a simple bash script, passing an IP addres, like this:
./bashScript 192.111.211.211
the script looks like this:
#!/bin/bash
curl https://www.xxx.com/api_json.html \
-d 'a=ban' \
-d 'tkn=xxxxxx' \
-d 'email=xxx#gmail.com' \
-d 'key=$1' \
but it isn't working, the $1 argument is not sending and i get an error from the web-service.
What i'm doing wrong?
Thanks a lot!
Use doubles quotes:
-d "key=$1"
Single quotes prevent variable expansion:
~$ foo=bar
~$ echo '$foo'
$foo
~$ echo "$foo"
bar
Related
I wrote a small bash script in this post: How to search for a string in a text file and perform a specific action based on the result
I noticed that when I ran the script and check the logs, everything appears to be working but when I look at the Nagios UI, almost half of the servers listed in my text file did not get their notifications disabled. A revised version of the script is below:
host=/Users/bob/wsus.txt
password="P#assw0rd123"
while read -r host; do
region=$(echo "$host" | cut -f1 -d-)
if [[ $region == *sea1* ]]
then
echo "Disabling host notifications for: $host"
curl -vs -o /dev/null -d "cmd_mod=2&cmd_typ=25&host=$host&btnSubmit=Commit" https://nagios.$region.blah.com/nagios/cgi-bin/cmd.cgi" -u "bob:$password" -k 2>&1
else
echo "Disabling host notifications for: $host"
curl -vs -o /dev/null -d "cmd_mod=2&cmd_typ=25&host=$host&btnSubmit=Commit" https://nagios.$region.blah02.com/nagios/cgi-bin/cmd.cgi" -u "bob:$password" -k 2>&1
fi
done < wsus.txt >> /Users/bob/disable.log 2>&1
If i run the command against the servers having the issue manually, it does get disabled in the Nagios UI, so I'm a bit confused. FYI, I'm not well versed in Bash either so this was my attempt at trying to automate this process a bit.
1 - There is a missing double-quote before the first https occurence:
You have:
curl -vs -o /dev/null -d "cmd_mod=2&cmd_typ=25&host=$host&btnSubmit=Commit" https://nagios.$region.blah.com/nagios/cgi-bin/cmd.cgi" -u "bob:$password" -k 2>&1
Should be:
curl -vs -o /dev/null -d "cmd_mod=2&cmd_typ=25&host=$host&btnSubmit=Commit" "https://nagios.$region.blah.com/nagios/cgi-bin/cmd.cgi" -u "bob:$password" -k 2>&1
2 - Your first variable host is never used (overwritten inside the while loop).
I'm guessing what you were trying to do was something like:
hosts_file="/Users/bob/wsus.txt"
log_file="/Users/bob/disable.log"
# ...
while read -r host; do
# Do stuff with $host
done < $hosts_file >> $log_file 2>&1
3 - This looks suspicious to me:
if [[ $region == *sea1* ]]
Note: I haven't tested it yet, so this is my general feeling about this, might be wrong.
The $region isn't double-quoted, so make sure there could be no spaces / funny stuff happening there (but this should not be a problem inside a double-bracket test [[).
The *sea* looks like it would be expanded to match your current directory files matching this globbing. If you want to test this as a regular expression, you should use ~= operator or (my favorite for some reason) grep command:
if grep -q ".*sea.*" <<< "$region"; then
# Your code if match
else
# Your code if no match
fi
The -q keeps grep quiet
There is no need for test like [ or [[ because the return code of grep is already 0 if any match
The <<< simply redirects the right strings as the standard input of the left command (avoid useless piping like echo "$region" | grep -q ".*sea.*").
If this doesn't solve your problem, please provide a sample of your input file hosts_file as well as some output logs.
You could also try to see what's really going on under the hood by enclosing your script with set -x and set +x to activate debug/trace mode.
I'm trying to decode my password through a script while it's being run but it seems like the script is being run with a literal and the password is not processed. Is there a better way of doing this?
#!/bin/bash
MYENC="Tk9UX1RIQVRfU1RVUElEX0xPTAo="
rdesktop -u FOO -d mgmt -p 'echo $(echo $MYENC) | base64 --decode' 192.0.0.0
I also tried to just pass in a variable but that failed as well.
Try this instead:
#!/bin/bash
MYENC="Tk9UX1RIQVRfU1RVUElEX0xPTAo="
rdesktop -u FOO -d mgmt -p $(echo $MYENC | base64 --decode) 192.0.0.0
Note that I wrapped the juicy stuff echo...base64... in $(...). This is called "command substitution" - basically you're telling bash that you want the code inside the $(...) to be executed before the rest of the line, with the result substituted in its place. More info here: http://www.tldp.org/LDP/abs/html/commandsub.html
Or this
#!/bin/bash
MYENC="Tk9UX1RIQVRfU1RVUElEX0xPTAo="
rdesktop -u FOO -d mgmt -p $(base64 --decode <<< "$MYENC") 192.0.0.0
I have a shell script test.sh as give below.
sudo -H sh -c '
echo $1;
'
But while I am running this script as ./test.sh abcd it is not echoing anything. Then I changed my script as given below.
sudo -H sh -c '
echo \$1;
'
But now, it is displaying output as $1. What modification shall I need to do here for getting output as abcd. Please advice as I am a very beginner in shell scripting.
Thanks.
Try this:
sudo -H sh -c "
echo $1;
"
sudo runs the command in a new shell that doesn't know anything about the argument(s) passed to its parent, so you need to expand them before running the command string in the new (sub)shell.
Try:
sudo -H sh -c '
echo "$1";
' argv0 "$1"
From the bash man page:
man bash | less -Ip '-c string'
# -c string If the -c option is present, then commands are read from
# string. If there are arguments after the string, they are
# assigned to the positional parameters, starting with $0.
I am writing a bash script to call curl from the command line. I sadly cannot figure out how to substitute a simple variable into the -d section of the curl request.
Why doesn't this work?
#!/bin/sh
name=$1
test -z $name && echo "Repo name required." 1>&2 && exit 1
curl -u 'metaraine' https://api.github.com/user/repos -d '{"name":"$name"}'
It doesn't actually substitute the value of $name into the data.
What about this?
curl -u 'metaraine' https://api.github.com/user/repos -d "{\"name\":\"$name\"}"
That is, escape the quotes and use double quotes around the {} instead of simple ones.
I have an sh script that contains the line
$PHP_COMMAND -r 'echo get_include_path();'
I can not edit this script, but I need the eventual command line to be (equivalent to)
php -d include_path='/path/with spaces/dir' -r 'echo get_include_path();'
How can I achieve this?
Below is a script that demonstrates the problem.
#!/bin/sh
# shell script quoting problem demonstration
# I need to be able to set a shell variable with a command with
# some options, like so
PHP_COMMAND="php -d 'include_path=/path/with spaces/dir'"
# then use PHP_COMMAND to run something in another script, like this:
$PHP_COMMAND -r 'echo get_include_path();'
# the above fails when executed. However, if you copy/paste the output
# from this line and run it in the CLI, it works!
echo "$PHP_COMMAND -r 'echo get_include_path();'"
php -d include_path='/path/with spaces/dir' -r 'echo get_include_path();'
# what's going on?
# this is also interesting
echo "\n--------------------"
# this works great, but only works if include_path doesn't need quoting
PHP_COMMAND="php -d include_path=/path/to/dir"
echo "$PHP_COMMAND -r 'echo get_include_path();'"
$PHP_COMMAND -r 'echo get_include_path();'
echo "\n--------------------"
# this one doesn't when run in the sh script, but again if you copy/paste
# the output it does work as expected.
PHP_COMMAND="php -d 'include_path=/path/to/dir'"
echo "$PHP_COMMAND -r 'echo get_include_path();'"
$PHP_COMMAND -r 'echo get_include_path();'
Script also available online: http://gist.github.com/276500
Reading your comments on other answers, I have tried to decipher what you really intended to ask:
I have an sh script that contains the line
$PHP_COMMAND -r 'echo get_include_path();'
I can not edit this script, but I need the eventual command line to be (equivalent to)
php -d include_path='/path/with spaces/dir' -r 'echo get_include_path();'
How can I achieve this?
If this accurately reflects your situation, then this is one way:
Write another script (I'll call it /some/path/to/bar.sh) to use as PHP_COMMAND. Since the variable is not quoted in the original script, you will have to make sure that the full pathname of bar.sh does not have any shell-special characters (like spaces).
/some/path/to/bar.sh
#!/bin/sh
exec php -d 'include_path=/path/with spaces/dir' ${1+"$#"}
Then, to run it, set PHP_COMMAND, and run the original script (/path/to/foo.sh):
env PHP_COMMAND=/some/path/to/bar.sh '/path/to/foo.sh'
There should be a simpler way, but one fix is to surround the entire command line with double quotes and then eval that:
PHP_COMMAND="php -d 'include_path=/path/with spaces/dir'"
eval "$PHP_COMMAND -r 'echo get_include_path();'"
Bash's printf has some special secret magic that might help. Try:
PHP_COMMAND="php -d $(printf "%q" "'include_path=/path/to/dir'")"
or some variation of that.
Edit:
I'm sorry, I should have included some explanation and an example. The %q flag causes printf to add escaping to the string so it can be reused in another command. The output of that printf would look like this (the single quotes get escaped):
\'include_path=/path/to/dir\'
This command illustrates some additional escaping:
$ printf "%q" "string ';\ text"
string\ \'\;\\\ text
what about this?
$PHP_COMMAND ' -r echo get_include_path();'