The action which I'm configuring is used to sending email notification. The problem is that if I use the macros in action's name and message the email will not be send out, if using just plain text the email can be sent out succesfully.
I'm using the media type of Script and the script do the job of sending email. The script as below:
#!/bin/sh
export smtpemailfrom=zabbix#yourdomain.com
export zabbixemailto=$1
export zabbixsubject=$2
export zabbixbody=$3
export smtpserver=yoursmtpserver.com
export smtplogin=smtpuser
export smtppass=smtppassword
/usr/bin/sendEmail -f $smtpemailfrom -t $zabbixemailto -u $zabbixsubject -m $zabbixbody -s $smtpserver:25 -xu $smtplogin -xp $smtppass
I was wondering if the reason is that the are some special chars in message, but even if I add quotation around the subject and body, it still not work.
It is advisable to put input variables in double quotes.
For instance, an email subject is likely to consist of multiple words. The following does not work as might be intended:
$ export subject=hello world
$ echo $subject
hello
Similarly, curly braces that are used in Zabbix macros have a special meaning to the shell, see Bash subshell creation with curly braces question.
Therefore, you should double quote variable expansion, at least for those variables where you are not sure about the content:
#!/bin/sh
export smtpemailfrom=zabbix#yourdomain.com
export zabbixemailto="$1"
export zabbixsubject="$2"
export zabbixbody="$3"
export smtpserver=yoursmtpserver.com
export smtplogin=smtpuser
export smtppass=smtppassword
/usr/bin/sendEmail -f $smtpemailfrom -t "$zabbixemailto" -u "$zabbixsubject" \
-m "$zabbixbody" -s $smtpserver:25 -xu $smtplogin -xp "$smtppass"
Related
I have a Perl script that uses some local variables as per below:
my $cool_variable="Initial value";
COOLVAR="Initial value for COOLVAR"
I would like to replace the content between the quotes using a bash script.
I got it to work for a non-variable like below:
#!/bin/sh
dummy_var="Replaced value"
sed -i -r "s#^(COOLVAR=).*#\1$dummy_var#" perlscript.pl
But if I replace it with cool_variable or $cool_variable:
sed -i -r "s#^($cool_variable=).*#\1$dummy_var#" perlscript.pl
It does not work..
The are multiple code injection bugs in that snippet. You shouldn't be generating code from the shell or sed.
Say you have
var=COOLVAR
val=coolval
As per How can I process options using Perl in -n or -p mode?, you can use any of
perl -spe's{^$var=\K.*}{"\Q$val\E";};' -- -var="$var" -val="$val" perlscript.pl
var=var val=val perl -pe's{^$ENV{var}=\K.*}{"\Q$ENV{val}\E";};' perlscript.pl
export var
export val
perl -pe's{^$ENV{var}=\K.*}{"\Q$ENV{val}\E";};' perlscript.pl
to transform
COOLVAR="dummy";
HOTVAR="dummy";
into
COOLVAR="coolvar";
HOTVAR="dummy";
The values are passed to the program using arguments to avoid injecting them into the fixer, and the fixer uses Perl's quotemeta (aka \Q..\E) to quote special characters.
Note that $var is assumed to be a valid identifier. No validation checks are performed. This program is absolutely unsafe using untrusted input.
Use -i to modify the file in place.
I wrote a script to create sqs resource on local stack. I wanted to pass a value that I get from one cli command to the next but inside an inline json. Following is the section of the script in question.
arn=$(aws --endpoint-url=http://localhost:4576 sqs get-queue-attributes \
--queue-url http://localhost:4576/my_dead_letter_queue_url \
--query 'Attributes.QueueArn' \
--output text)
aws --endpoint-url=http://localhost:4576 sqs create-queue \
--queue-name my_queue \
--attributes \
'{"RedrivePolicy":"{\"deadLetterTargetArn\":\"$arn\", \"maxReceiveCount\":\"5\"}"}'
So I'm trying to pass that "arn" variable but the cli is taking that as a string and trying to find a sqs with url "$arn" and fails. I also tried removing the quote. In that case, the error is malformed string.
Instead of the arn variable, if I use the arn value as string there, it works.
Can someone please show me how to pass that variable inside that inline json if it is possible?
Thank you for reading :)
Shahed
I was able to do the following with successful results, grant it it doesn't process the json (for that I'm just replacing tokens via sed), but I updated my example and tested it at least in bash with what I was doing:
#!/bin/bash
export awscmd="aws --region us-east-1 iam"
function setArn() {
${awscmd} list-policies --query 'Policies[?PolicyName==`'${1}'`].{ARN:Arn}' --output text
}
arn=$(setArn "some-policy-name")
echo '{"RedrivePolicy":"{"deadLetterTargetArn":"'$arn'", "maxReceiveCount":"5"}"}'
$ ./somearntest.sh
{"RedrivePolicy":"{"deadLetterTargetArn":"arn:aws:iam::############:policy/some-policy-name", "maxReceiveCount":"5"}"}
Notice the use of single tics to concatenate the output result outside of the string. This is in bash 4 and I removed the escaped \"s as I think that was added in error; ymmv.
The problem here is you are trying to expand a bash variable inside single quotes. Using single quotes like this is usually to pass a bunch of strings and unqoutable stuff as one argument. If you can't replace them with double quotes you'll have to resort to dirty eval hacks, which I do not recommend.
Here is an example:
$ arn=foobar
$ echo '{"RedrivePolicy":"{\"deadLetterTargetArn\":\"$arn\", \"maxReceiveCount\":\"5\"}"}'
{"RedrivePolicy":"{\"deadLetterTargetArn\":\"$arn\", \"maxReceiveCount\":\"5\"}"}
$ eval echo '{"RedrivePolicy":"{\"deadLetterTargetArn\":\"$arn\", \"maxReceiveCount\":\"5\"}"}'
{RedrivePolicy:{"deadLetterTargetArn":"foobar", "maxReceiveCount":"5"}}
For more information I suggest to check How eval works and Expansion of variables inside single quotes
Hello I am having trouble with a very specific line in a bash script.
Here is the code:
ssh $SOURCEIP "/usr/bin/time -f \"%e\" bash -c \"seq $ITER | parallel -n0 \"mysql --silent -h $TARGET -uroot -ppass -e 'SELECT * FROM dbname.tablename WHERE size = $SIZE;' >> out.txt\""
The problem is I ran out of quotes. The opening and escaped double quotes at the beginning of "mysql" are closing those from "bash -c". I have to put the mysql statement in double quotes and the query in single quotes, otherwise i get an error and I can't figure out how to proceed. I know that I should not pass the password like that and it will be changed later, I get this warning "$ITER"-times everytime i test this because --silent doesn't suppress this.
The problematic code is part of a small shell script that is supposed to just perform this data transfer.
I want to change to the other machine with ssh first and not via parallel because of consistency with other scripts.
So basically I need the double quotes around the bash -c command to get this whole parallel operation to work, which are already escaped because of the opening ssh doublequotes and also I need to put the mysql command inside quotes as well but they are closing each other somehow.
Any help will be greatly appreciated.
Thanks in advance.
Largio
Edit: (SOLUTION)
As suggested by #ole-tange the following command worked for me.
parallel --shellquote | parallel --shellquote
After invoking in a shell, i pasted my string in question into the prompt and got the masked string back. I still had troubles with finding out what exactly to paste but in the end it is just logical.
What exactly i pasted into the quoter was:
sql mysql://root:pass#$TARGET/ 'SELECT data FROM db_name.tablename WHERE size = ${SIZE};' >> out.txt
But still i had some problems with my variables inside my query. The problem here was that i had to de-mask the masking of the 2 variables $TARGET and $SIZE after everything got masked by the parallel quoter. Maybe my thinking has a too laborious manner but i could not get it to work in another way. Also note that i did not put quotes around the whole sql statement, as my plan was before, because now the quoter compensated for that. For consistency reasons i paste the final string that i got working in the end (with my changes afterwards):
ssh $SOURCEIP "/usr/bin/time -f \"%e\" bash -c \"seq $ITER | parallel -n0 sql\\\ mysql://root:pass#$TARGET/ \\\'SELECT\\\ data\\\ FROM\\\ db_name.tablename\\\ WHERE\\\ size\\\ =\\\ ${SIZE}\\\;\\\'\\\ \\\>\\\>\\\ out.txt\""
GNU Parallel has a quoter:
$ parallel --shellquote
"*\`$
[CTRL-D]
\"\*\\\`\$
And you can do it twice:
$ parallel --shellquote | parallel --shellquote
"*\`$
[CTRL-D]
\\\"\\\*\\\\\\\`\\\$
So just paste the string you want quoted.
But you might want to consider using functions and use env_parallel to copy the function:
myfunc() {
size=$1
target=$2
sql mysql://root:pass#$target/ "SELECT data FROM db_name.tablename WHERE size = $size;" >> out.txt
}
env_parallel --env myfunc -S $SOURCEIP --nonall myfunc $SIZE $TARGET
Also: Instead of mysql try sql mysql://root:pass#/ 'SELECT * FROM dbname.tablename WHERE size = $SIZE;'
I have a query that sends the results to an email. I would like not to send an email if the query has NO results. How can i do that ?
heres the code
mysql -umy -hmysql1.com -P2 -pmysq <<<" Select * from Data.data "| mail -aFrom:test#test.com -s 'test' test#gmail.com
Not every task can be done easily in a single command pipeline. Trying to force it into a one-liner can make it hard to code and hard to maintain.
Feel free to write some statements in a script:
result=`mysql -umy -hmysql1.com -P2 -pmysq -e " Select * from Data.data "`
if [ -n "$result" ]
then
echo "$result" | mail -aFrom:test#test.com -s 'test' test#gmail.com
fi
The -n test is for strings being nonzero length. Read http://linuxcommand.org/lc3_man_pages/testh.html for more details on that.
Re your comment:
The statements I showed above are things you could type at the command-line in bash. Bash supports variables and "if/then/else" constructs and a lot more.
Writing a bash script is easy. Anything you can type at the command-line can be in a file. Open a text editor and write the lines I showed above. Save the file. For example it could be called "mailmyquery.sh" (the .sh extension is only customary, it's not required).
Exit the text editor. Then run:
bash mailmyquery.sh
And it runs the statements in the file as if you had written them yourself at the command-line.
Voilà! You are now a shell script programmer!
I'm inserting a git diff of changed files into a JSON object to send using a curl request.
The problem is it doesn't like the new-line characters being inserted into the JSON but I'm not sure how to get around that. Translate tool didn't work, this perl solution I'm using is close but just replaces with spaces:
changedfiles=$(git diff --name-only $3..$4 | perl -p -e 's/\n/ /')
and changing it to this didn't help:
changedfiles=$(git diff --name-only $3..$4 | perl -p -e 's/\n/\\n/')
Can anyone point me in the right direction? It doesn't need to use perl, it just needs to work
(...being simple would be nice too)
Instead of trying to do ad-hoc escaping for characters that your immediate testing finds problematic, how about using an actual JSON library that handles all of them in a solid way?
Here's an example in bash using inlined python:
python -c '
import json
import sys
print(json.dumps({"data": sys.argv[1]}))
' "$(git diff --name-only $3..$4)"
It prints the json object { "data": "your command output here" } with standards compliant escaping.
This is what I think you want to do to get a quoted list of files separated by commas (i.e. for inserting into a JSON string):
git diff --name-only $3..$4 | perl -p -e 's/(.*)/"$1",/;s/\n//;s/""/","/'
This works if your files don't contain double quotes or special characters that need to be JSON escaped.
First, we put the files in quotes followed by a comma, then remove newlines, then change the "" between files to ",". Although, this is kind of a hack. Somewhat better might be:
git diff --name-only $3..$4 | perl -p -e '$/="";s/(.*)\n/"$1",/g;s/,$//'
Here we read in the whole input, newlines and all, do our substitution and remove the final comma.