#!/bin/bash
object=$1
tenant=$2
server=$(sshpass -p 'password' ssh -tt root#ipaddress "/opt/something/sh/mysql -A 'admin_site' -e 'select id from something_servers where tenantcode=$2' | grep -o '[0-9]*'")
http -b "http://ipaddress/?app=something&t=users&v=users&server=$server&apikey=apikey&action=something.$1.list"
Everything seems to work fine. The value of variable $server is "24" as it should be. But still, when I call upon it in the http request it simply won't work... I honestly have no clue. I tested the variable's value by echoing it and it's correct. But for some reason when using it in the request it doesn't.
Any ideas?
Your code creates a variable called object and you set its value to the value stored in a variable called 1.
It creates a variable called tenant and sets it to the value of what's stored in the variable called 2.
It then creates a variable called server and then calls a subshell and runs the routine between $().
In the subroutine:
You use single quotes around
-e 'select id from something_servers where tenantcode=$2'
But for variable interpolation to work, you would need to use double quotes
Maybe that is your issue?
Related
I wrote a script to create sqs resource on local stack. I wanted to pass a value that I get from one cli command to the next but inside an inline json. Following is the section of the script in question.
arn=$(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
I have the following function in makefile:
define INSTALL_SCRIPT
SRC_DIR = $(ROOT)\src
cd $(SRC_DIR)
$(SRC_DIR)\stage.bat
endef
I also echo the steps, so here's the output of the above snippet:
$SRC_DIR = C:\project_root\src
'SRC_DIR' is not recognized as an internal or external command,
operable program or batch file.
$cd
C:\project_root
\stage.bat
'\stage.bat' is not recognized as an internal or external command,
operable program or batch file.
It seems that in assignment statement the value is expanded correctly but then $(SRC_DIR) gives an error. Then cd goes to one directory up (and not src), then when I need to execute the batch file, $(SRC_DIR)'s value seems to be empty.
Assuming you're trying to do this from a recipe context, you would need to do it as follows:
define INSTALL_SCRIPT
set SRC_DIR=$(ROOT)\\src & \
cd %SRC_DIR% & \
%SRC_DIR%\\stage.bat
endef
sometarget:
#$(INSTALL_SCRIPT)
You need the \ at the end of each line to concatinate them into a single recipe line (other wise the variable you set will fall out of context when the first recipe line's shell terminates). You seem to be using windows so I believe you need to use the %varname% syntax to refer to the variables. Notice that $(ROOT) is a makefile variable in this case, so it still uses the $ syntax. (Note that if you were in bash you would need to use $$ to refer to shell variables). You also need to double the \\ in directory names, as make will interpret the first slash as an escape, and then pass a single slash to cmd.
Note that my windows machine doesn't have make installed on it, so I couldn't test the above, so it's quite possible I missed something.
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 = $SIZE;'
This should be an incredibly easy question but I am not very familiar with bash and I am taking way longer than I should to figure it out.
declare -a ids=( 1 2 3 )
for i in "${ids[#]}";
do
re= $(mysql -h .... "SELECT col_A FROM DBA WHERE id=$i")
if [ $re -eq 0 ]; then
echo sucess
fi
done
This is an example of what I am trying to do, I have an id array and I want to send a query to my db so I can get a flag in the row with a certain id and then do something based on that. But I keep getting unexpected token errors and I am not entirely sure why
Edit: While copying the code and deleting some private information somehow I deleted the then, it was present in the code I was testing.
Based on what you described and the partial script, I am not certain I can completely create what you are trying to do but the token error messages you are experiencing usually have to do with the way bash handles whitespace as a delimiter. A few comments based on what you posted:
You need to remove the space around the equal sign in declaring an variable, so the space after the equal sign in re= needs to removed.
Because bash will is sensitive to whitespace, you need to quote variables declarations that might contain a space. To be safe, quotes need to be around the sub-shell $( )
You were missing the then in the if statement
It is important that variables in the test brackets, that is single [ ]s, must be quoted. Using an unquoted string with -eq, or even just the unquoted string alone within test brackets normally works, however, this is an unsafe practice and can give unpredictable results.
So, taking into account the items noted, the updated script would look something like:
declare -a ids=( 1 2 3 )
for i in "${ids[#]}";
do
re="$(mysql -h .... "SELECT col_A FROM DBA WHERE id=$i")"
if [ "$re" -eq "0" ]; then
echo "success"
fi
done
Can you try working the edits mentioned into your script and see if you are able to get it working? Remember, it will be helpful for you to use a site like ShellCheck to learn more about potential pitfalls or the uniquenesses of bash syntax. This will help to ensure you are working toward a solution to your specific need rather then getting trapped by some tricky syntax.
After you have worked through those edits, can you report back your experience?
EDIT
Based on your comments there is a good chance you are not running your script with bash despite the including #!/bin/bash at the top of your script. When you run the script as sh scriptname.sh you are forcing the script to be run by sh not bash. Try running your script like this /bin/bash scriptname.sh then report back on your experience.
For more information on the differences between various shells, see Unix/Linux : Difference between sh , csh , ksh and bash Shell
Your problem with your if statement is that you do not have the then keyword. A simple fix is:
declare -a ids=( 1 2 3 )
for i in "${ids[#]}";
do
re= $(mysql -h .... "SELECT col_A FROM DBA WHERE id=$i")
if [ $re -eq 0 ]; then
echo sucess
fi
done
Also here is a great reference on if statements in bash
I am trying to write a GNU make call function (example below) which has multiple shell commands to execute, such that it can be called with different arguments.
shell_commands = $(shell echo $(1); ls -ltr $(2))
try:
$(call shell_commands,$(FILE1),$(FILE2))
1) Is above the correct way to write a call function with multiple commands? By using a semi-colon to separate them? To make it readable, I write my targets as shown below. Is there a similar way to write a call function?
shell_commands:
echo $(1)
ls -ltr $(2)
2) I get this error from make when I execute make -B try. It looks like it is trying to execute /home/user/file1. But why?
make: execvp: /home/user/file1: Permission denied
make: *** [try] Error 127
3) Is it possible to pass variable number of parameters to a call function? Like pass in just the second parameter and not the first one.
$(call shell_commands,,$(FILE2))
I tried google-ing, searching on SO, and looking on gnu.org but I did not get any solutions. Will appreciate any answers or pointers to any resources which document call function with multiple optional arguments and commands.
Question 1: No, this is not right. The shell make function should NEVER be used inside a recipe: the recipe is already running in the shell, so why would you run another shell? It's just confusing. Second, it's generally recommended to use && between multiple commands in a recipe, so that if the first command fails the entire command will immediately fail, rather than continuing on and perhaps succeeding. Of course, that is not always correct either, it depends on what you're trying to do.
Question 2: This happens because the shell make function is like backticks in the shell: it expands to the output printed by the shell command it runs. Your shell command that make runs is:
echo $(1); ls -ltr $(2)
(where, one assumes, $1 expands to /home/user/file1) which prints the string /home/user/file1. After the expansion, that string is substituted into the recipe and make tries to run that recipe, giving the error you see above.
You want this, most likely:
shell_commands = echo $(1) && ls -ltr $(2)
try:
$(call shell_commands,$(FILE1),$(FILE2))
Now the call expands to the actual text, not an invocation of make's shell function, and then that text is run as the recipe.
Question 3: Sure, just using empty parameters means that the variable $1 (in this case) expands to the empty string.