Edit:
I'm creating a bash script to run Netezza queries.
here's an example of what I have to do:
nzsql -host localhost -port 123456 -d db -u usr -pw pwd -A -t -c "insert into TABLE (name,surname) values ('m','sc')"
and it should return
INSERT 0 1
What I need is retrieve the number "1" which means that 1 row was inserted.
For this, I'd need to retrieve the whole string "INSERT 0 1" and work on it.
according to http://www.enzeecommunity.com/thread/2423 this should work:
cmnd_output=`nzsql -host $NZ_HOST -d $NZ_DATABASE -u $NZ_USER -pw $NZ_PASSWORD -A -t -c "insert into TEST values ('test 1')"`
But I can't get it to work with this: ($2 is right because when I run it from the terminal it works just fine)
cmd_out=`$2` or cmd_out=`"$2"` or cmd_out="`$2`" or cmd_out=`"'$2'"`
cmd_out=$($2) or cmd_out="$($2)" or cmd_out=$("$2")
It tells me command not found... just like if there was a "string quote" problem with $2
I've however managed to execute $2 with eval
eval "$2"
and it works great, the command $2 is executed just fine.
But, I can't use eval in this case as I want to store in a variable that "INSERT 0 1".
A simple
variable_int=`$function '$arg1' '$arg2'`
without the eval won't do?
To assign return values from functions to a shell variable, use command substitution
variable=$(function arg1 arg2)
Why do you need eval?
When you run into a problem like this I find it's always very useful to run with the -x option, just change the top sh-bang line like so:
#!/bin/bash -x
That'll print out each line as it's currently interpreted before
executing it. You can see how your variables are being mangled and use that to fix the problem.
Related
#!/bin/bash
Dpath=/home/$USER/Docker/
IP=`sed -n 1p /home/medma/.medmadoc`
DockerMachine=`sed -n 2p /home/$USER/.medmadoc`
DockerPort=`sed -n 5p /home/$USER/.medmadoc`
DockerUser=`sed -n 3p /home/$USER/.medmadoc`
DockerPass=`sed -n 4p /home/$USER/.medmadoc`
if [ ! -d $Dpath ] ; then
mkdir -p $Dpath
else
stat=`wget -O ".dockerid" http://$IP/DOCKER-STAT.txt`
for ids in `cat .dockerid`
do
if [ "$ids" == "$DockerMachine" ] ; then
gnome-terminal -x sh -c 'sshfs -p$DockerPort $DockerUser#$IP:/var/www/html $Dpath ; bash '
nautilus $Dpath
zenity --info --text "Mounted $DockerMachine"
exit
else
:
fi
done
zenity --info --text "No Such ID:$DockerMachine"
fi
gnome-terminal -x sh -c 'sshfs -p$DockerPort $DockerUser#$IP:/var/www/html $Dpath ; bash '
this command opens up a new terminal but the problem is that it does not load vars like $DockerPort $DockerUser $IP $Dpath from this script.
How do I input the values in these vars from this script to the newly opened terminal ?
Thanks !
As indicated before, you could try to use double quotes instead of single quotes around the sshfs invocation.
Single quotes in Bash are used to delimit verbatim text, in which variables are not expanded. Double quotes, in contrast, allow for variables expansion and command substitution ($(...)) to take place.
If you do use double quotes, beware of unintended side-effects (your username may contain a space, a dollar, a semicolon, or any other shell-special character). A cleaner approach would be to export the variables to the environment before calling gnome-terminal (and not forgetting to add double quotes around your variables inside the single-quotes), so that your code looks like :
export Docker{Port,User} IP Dpath
gnome-terminal -x sh -c 'sshfs -p"$DockerPort" "$DockerUser#$IP":/var/www/html "$Dpath" ; bash'
You may not want to pollute the environment with variables that will only be used once. If that is the case, instead of exporting them, you can use Bash's declare -p feature to serialize variables before loading them into a new environment (in my opinion, this is the cleanest approach). Here is what it looks like :
set_vars="$(declare -p Docker{Port,User} IP Dpath)"
gnome-terminal -x bash -c "$set_vars;"'sshfs ....'
Using this latest method, the variables are only visible to the shell process that runs the sshfs command, not gnome-terminal itself nor any sub-process run thereafter.
PS: you could read all your variables at once from the ~/.medmadoc file by using the following code instead of repeated sed invocations :
for var in IP Docker{Machine,User,Pass,Port}; do
read $var
done < ~/.medmadoc
This code makes use of the read builtin, that reads a line of input into a variable (in its simplest form).
PPS: That stat variable probably won't contain any useful information, since the output of wget was redirected by the -O flag. Perhaps you meant to store the result code of wget into stat, in which case what you meant was :
wget -O .dockerid ...
stat=$?
I am trying to execute cqlsh inside bash script. My script is below. When ı try to execute sh file, it returns cql command not found
#!/bin/bash
set -x
PATH=/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin
cqlsh -e "SELECT * FROM msg.msg_log limit 1;" > /home/yunus/sh/cqlshcontrol.txt
error1=$( more /home/yunus/sh/cqlshcontrol.txt | wc -l )
if [ $error1 -lt 1 ]; then
curl -S -X POST --data "payload={\"text\": \" Cqlsh not responding, Connection Problem \",\"username\":\"Elevate Cassandra1\",\"icon_emoji\":\"${SLACK_ICON}\"}" https://hooks.slack.com/services/
fi
some suggestions
use [[/]] over [/].
the return value of $() is not an error value and should be named lines or something more meaningful. The lack of another error variable in the code makes the appended number (the 1 in error1) seem even odder.
There's no reason to use more or pipe inside of that subshell. Just run wc -l on your file.
Are you sure cqlsh is in the PATH? Try which cqlsh to find it.
wc will never return a negative value so comparing for equality with zero would be clear and cover just as many potential cases.
otherwise
If that doesn't get you out of your confusion please show the output when you try to run it.
This question already has answers here:
Why does shell ignore quoting characters in arguments passed to it through variables? [duplicate]
(3 answers)
Closed 6 years ago.
I'm trying to write a database call from within a bash script and I'm having problems with a sub-shell stripping my quotes away.
This is the bones of what I am doing.
#---------------------------------------------
#! /bin/bash
export COMMAND='psql ${DB_NAME} -F , -t --no-align -c "${SQL}" -o ${EXPORT_FILE} 2>&1'
PSQL_RETURN=`${COMMAND}`
#---------------------------------------------
If I use an 'echo' to print out the ${COMMAND} variable the output looks fine:
echo ${COMMAND}
screen output:-
#---------------
psql drupal7 -F , -t --no-align -c "SELECT DISTINCT hostname FROM accesslog;" -o /DRUPAL/INTERFACES/EXPORTS/ip_list.dat 2>&1
#---------------
Also if I cut and paste this screen output it executes just fine.
However, when I try to execute the command as a variable within a sub-shell call, it gives an error message.
The error is from the psql client to the effect that the quotes have been removed from around the ${SQL} string.
The error suggests psql is trying to interpret the terms in the sql string as parameters.
So it seems the string and quotes are composed correctly but the quotes around the ${SQL} variable/string are being interpreted by the sub-shell during the execution call from the main script.
I've tried to escape them using various methods: \", \\", \\\", "", \"" '"', \'"\', ... ...
As you can see from my 'try it all' approach I am no expert and it's driving me mad.
Any help would be greatly appreciated.
Charlie101
Instead of storing command in a string var better to use BASH array here:
cmd=(psql ${DB_NAME} -F , -t --no-align -c "${SQL}" -o "${EXPORT_FILE}")
PSQL_RETURN=$( "${cmd[#]}" 2>&1 )
Rather than evaluating the contents of a string, why not use a function?
call_psql() {
# optional, if variables are already defined in global scope
DB_NAME="$1"
SQL="$2"
EXPORT_FILE="$3"
psql "$DB_NAME" -F , -t --no-align -c "$SQL" -o "$EXPORT_FILE" 2>&1
}
then you can just call your function like:
PSQL_RETURN=$(call_psql "$DB_NAME" "$SQL" "$EXPORT_FILE")
It's entirely up to you how elaborate you make the function. You might like to check for the correct number of arguments (using something like (( $# == 3 ))) before calling the psql command.
Alternatively, perhaps you'd prefer just to make it as short as possible:
call_psql() { psql "$1" -F , -t --no-align -c "$2" -o "$3" 2>&1; }
In order to capture the command that is being executed for debugging purposes, you can use set -x in your script. This will the contents of the function including the expanded variables when the function (or any other command) is called. You can switch this behaviour off using set +x, or if you want it on for the whole duration of the script you can change the shebang to #!/bin/bash -x. This saves you explicitly echoing throughout your script to find out what commands are being run; you can just turn on set -x for a section.
A very simple example script using the shebang method:
#!/bin/bash -x
ec() {
echo "$1"
}
var=$(ec 2)
Running this script, either directly after making it executable or calling it with bash -x, gives:
++ ec 2
++ echo 2
+ var=2
Removing the -x from the shebang or the invocation results in the script running silently.
So, I'm new to scripting, and I'm having some problems. The command I need to execute is:
read -p Enter_the_DEVICE_Bssid "device1" ;
read -p Enter_the_DEVICE_Bssid "device2" ;
read -p Enter_the_DEVICE_Bssid "device3"
That command works, but when I set it as a variable ie:
com="read -p Enter_the_DEVICE_Bssid "device1" ;
read -p Enter_the_DEVICE_Bssid "device2" ;
read -p Enter_the_DEVICE_Bssid "device3""
and execute it as: $com it does not work. Probably because the read command is trying to set my input to the variables device1 and ; .
Any ideas on how to fix it?
You're running into problems with the order in which things are expanded by the shell.
A simpler example:
$ command='echo one ; echo two'
$ $command
one ; echo two
The semicolon in the value of $command is taken as part of the argument to echo, not as a delimiter between two echo commands.
There might be a way to resolve this so it works the way you want, but why bother? Just define a shell function. Using my simple example:
$ command() { echo one ; echo two ; }
$ command
one
two
$
Or using yours:
com() {
read -p "Enter_the_DEVICE_Bssid: " device1
read -p "Enter_the_DEVICE_Bssid: " device2
read -p "Enter_the_DEVICE_Bssid: " device3
}
Note that I've added ": " at the end of the prompts. I've also removed the unnecessary semicolons and the quotation marks around the variable names (since the argument has to be a valid variable name, it doesn't need to be quoted).
You are not completing the quotes.
com="read -p Enter_the_DEVICE_Bssid "device1"
Quotes always look for a pair and you are missing that.
> com="read -p Enter_the_DEVICE_Bssid: device1"
> $com
Enter_the_DEVICE_Bssid:abc123
> echo $device1
abc123
Here I am using bash shell.
I'm trying to write a shell script that calls another script that then executes a rsync command.
The second script should run in its own terminal, so I use a gnome-terminal -e "..." command. One of the parameters of this script is a string containing the parameters that should be given to rsync. I put those into single quotes.
Up until here, everything worked fine until one of the rsync parameters was a directory path that contained a space. I tried numerous combinations of ',",\",\' but the script either doesn't run at all or only the first part of the path is taken.
Here's a slightly modified version of the code I'm using
gnome-terminal -t 'Rsync scheduled backup' -e "nice -10 /Scripts/BackupScript/Backup.sh 0 0 '/Scripts/BackupScript/Stamp' '/Scripts/BackupScript/test' '--dry-run -g -o -p -t -R -u --inplace --delete -r -l '\''/media/MyAndroid/Internal storage'\''' "
Within Backup.sh this command is run
rsync $5 "$path"
where the destination $path is calculated from text in Stamp.
How can I achieve these three levels of nested quotations?
These are some question I looked at just now (I've tried other sources earlier as well)
https://unix.stackexchange.com/questions/23347/wrapping-a-command-that-includes-single-and-double-quotes-for-another-command
how to make nested double quotes survive the bash interpreter?
Using multiple layers of quotes in bash
Nested quotes bash
I was unsuccessful in applying the solutions to my problem.
Here is an example. caller.sh uses gnome-terminal to execute foo.sh, which in turn prints all the arguments and then calls rsync with the first argument.
caller.sh:
#!/bin/bash
gnome-terminal -t "TEST" -e "./foo.sh 'long path' arg2 arg3"
foo.sh:
#!/bin/bash
echo $# arguments
for i; do # same as: for i in "$#"; do
echo "$i"
done
rsync "$1" "some other path"
Edit: If $1 contains several parameters to rsync, some of which are long paths, the above won't work, since bash either passes "$1" as one parameter, or $1 as multiple parameters, splitting it without regard to contained quotes.
There is (at least) one workaround, you can trick bash as follows:
caller2.sh:
#!/bin/bash
gnome-terminal -t "TEST" -e "./foo.sh '--option1 --option2 \"long path\"' arg2 arg3"
foo2.sh:
#!/bin/bash
rsync_command="rsync $1"
eval "$rsync_command"
This will do the equivalent of typing rsync --option1 --option2 "long path" on the command line.
WARNING: This hack introduces a security vulnerability, $1 can be crafted to execute multiple commands if the user has any influence whatsoever over the string content (e.g. '--option1 --option2 \"long path\"; echo YOU HAVE BEEN OWNED' will run rsync and then execute the echo command).
Did you try escaping the space in the path with "\ " (no quotes)?
gnome-terminal -t 'Rsync scheduled backup' -e "nice -10 /Scripts/BackupScript/Backup.sh 0 0 '/Scripts/BackupScript/Stamp' '/Scripts/BackupScript/test' '--dry-run -g -o -p -t -R -u --inplace --delete -r -l ''/media/MyAndroid/Internal\ storage''' "