set UserName to do shell script "whoami"
set AD to do shell script ("dscl "/Active Directory/YC/All Domains/" read /Users/" & UserName)
My error: Expected “,” but found identifier
So Im trying to find a users SMBHome in active directory because we map our drives based on that. But the issue with that command is the double strings. "/Active Directory/YC/All Domains/" HAS to a string otherwise it wont properly queue.
But to do shell script ("COMMAND HERE") has to be a string as well..
How do I combat this?
And is there a easier way todo this?
Full script:
set UserName to do shell script "whoami"
set AD to do shell script ("dscl "/Active Directory/YC/All Domains/" read /Users/" & UserName)
if AD contains "StaffShare" then
set SMBHome to "smb://domain/StaffShare"
else
set SMBHome to "smb://domain/EmployeeShare"
end if
set mounted_disk to list disk
mount volume SMBHome as user name UserName
Nevermind, I actually figured it out. Similar to Lua
"dscl \"/Active Directory/YC/All Domains/\" read /Users/" & UserName
Although Im still up to see if anyone has any better solutions for this. :)
You can use single quotes:
set AD to do shell script "dscl '/Active Directory/YC/All Domains/' read $HOME"
But the way you've done it, by escaping the double quotes, it perfectly legit.
Related
I'm trying to obtain a secret out of my KeyVault.
The variable name is secretVar.
Obtaining the secret like this: $(secretVar) works fine however I would like to retrieve it from a variable like this:
I keep getting command not found and I've no idea why this shouldn't be working.
So the name of the secret I want to extract is inside a bash variable. For this question I've simplified the problem but in my real use case I have a bash for loop which loops through secret names and inside the for loop I want to extract the appropriate value from the KeyVault with the corresponding secret name like this:
for secretname in secrets; do
echo $($secretname) # This should contain the value of the secret but gives command not found
done
If anyone has an idea what could be happening, any help is very appreciated.
Thanks in Advance!
Look at the syntax you're using.
variable=secretVar
You are creating an environment variable with the literal value secretVar
Then you try to execute the value of the variable $variable with $($variable). So it tries to run the command secretVar, which obviously doesn't exist, and you get an error message.
The syntax you're looking for is
variable=$(secretVar)
just like you used in the first echo command in the script.
If you don't want to run the variable value as a command, the syntax would be $variable, not $($variable)
$variable is the syntax for a Bash environment variable.
$(variable) is the syntax for referencing Azure DevOps variables.
First of all, the script keyword is a shortcut for the command-line task. The task runs a script using cmd.exe on Windows and Bash on other platforms. You need to pay attention to the agent you are using.
If you want to set variables in scripts, you can use task.setvariable logging command. For example:
- script: |
echo $(secretvar)
echo "##vso[task.setvariable variable=variable]$(secretvar)"
- script: |
echo $(variable)
You can find more detailed information in this document.
I am using DevOps pipeline to build and deploy to different environments
For one environment I am encountering this issue where i am using a Pipeline Variable with $$ in the value
For Example:
Password pipeline variable with value = $omeCla$$Password
When i deploy it fails and when i check the logs the password is displayed as $omeCla$Password. So basically when $$ are together it drops one $
For all variable i am using regex __VaraibleValue__ and its working fine
I have tried:
$omeCla$\$Password to try and escape and it displays as $omeCla$\$Password . So basically \ doesn't work.
I tried '$omeCla$$Password' to try and escape and it displays as '$omeCla$Password'
I want to keep this value as a normal pipeline variable before review
So basically how can I escape this?
Or should I add a Secret Token here in the replace token task (see screenshot below)? and then make the pipeline variable secret? If so, what should I set for Secret Token? Also, in app.config in my repo what should I use instead of the regex __VariableName__ that I use for normal variables?
The solution was to use 4 $. So if you have $$ together you need to add $$$$
Example: $someCla$$$$Password
#JaneMa-MSFT as requested
https://developercommunity.visualstudio.com/content/problem/1296808/azure-pipeline-how-to-escape-special-characters-in.html
I have file1.sh file and which internally needs to execute one sql query against two Oracle DBs at a same time and needs to export date to csv fiiles, below is the sample shellscript which executes the query against two dbs.
....
#!bin/bash
set -X
sqlplus -S ${user1}#${DBCONNECTIONNAME_1}/${Pwd} Datesquery.sql & >> ${Targetdirectory}/csvfile1.csv
sqlplus -S ${user1}#${DBCONNECTIONNAME_2}/${Pwd} Datesquery.sql & >> ${Targetdirectory}/csvfile2.csv
sed 1d csvfile2.csv > file2noheader.csv
cat csvfile1.csv file2noheader.csv > ${Targetdirectory}/Expod.csv
....
But it does not connect to DB and does not execute any query and simply displays sqlplus manual as how to use the connection string, please let me know how to call one query against two dbs and execute them in parrallay and binds output to separate csv files.
A given sqlplus session can only connect to one db at a time, so your requirement 'at the same time' is essentially a non-starter. If 'at the same time' really means 'sequentially, in the same script, then you are back to fixing your connect string. And at that you 'have more errors than an early Mets game' (with apologies to any NY Mets fans).
First, your script indicates that your sqlplus command is the very first actual command following specification of your shell processor and 'set -x'. Yet you make heavy use of environment variables as substitutions for username, password, and connection name - without ever setting those variables.
Second, your use of an '&' in the command line is totally confusing to both me and the parser.
Third, you need to preceed your reference to the sql script with '#'.
Fourth, your order of elements in the command line is all wrong.
Try this
#!/bin/bash
orauser1=<supply user name here>
orapw2=<supply password here>
oradb_1=<supply connection name of first database>
#
orauser1=<supply user name here>
orapw2=<supply password here>
oradb_1=<supply connection name of first database>
#
Targetdirectory=<supply value here>
#
sqlplus -S ${orauser1}/${orapw1}#${oradb_1} #Datesquery.sql >> ${Targetdirectory}/csvfile1.csv
sqlplus -S ${orauser2}/${orapw2}#${oradb_1} #Datesquery.sql >> ${Targetdirectory}/csvfile2.csv
Or create a database link form one DB to other and then run both sqls in one db, one over DB link.
select * from tab1
union
select * from tab1#db_link
I'm pretty new to shell scripts and am only doing them because its about time I learnt and I need to for work.
I have been looking around and have tried multiple methods to get this working but can't seem to figure it out.
I have a script in which I want to access an SQLite database and store the result of a select statement in a variable.
What I've Tried So Far
This one just echoes whats inside the apostrophe. If I remove the dollar sign before the apostrophe I get the same outcome.
track_name=$'sqlite3 "$database_name" << EOF
select name from track where id = "$required_track";
exit;
EOF'
Here I get a syntax error near "track_name"
sqlite3 "$database_name" << EOF
track_name='select name from track where id = "$required_track";'
exit;
EOF
I have successfully executed the select statement without trying to store it in a variable but its not much use to me without being able to store it...
Any help would be much appreciated
To store the output of a command into a BASH variable you should use:
VAR_NAME=$(command);
For example, if you want to store your system current time into a variable or
the results of a list directory command ejecution:
DATE_EXAMPLE_VAR=$(date); #Stores 'date' command output into DATE_EXAMPLE_VAR
echo $DATE_EXAMPLE_VAR; #Shows DATE_EXAMPLE_VAR contents
DIRCONTENTS=$(ls); #Stores a list of your current directory contents.
Similarly, this should work for sqlite3:
track_name=$(sqlite3 "$database_name" "select name from track where id = $required_track")
I have a script that needs to know what username it is run from.
When I run it from shell, I can easily use $ENV{"USER"}, which is provided by bash.
But apparently - then the same script is run from cron, also via bash - $ENV{"USER"} is not defined.
Of course, I can:
my $username = getpwuid( $< );
But it doesn't look nice - is there any better/nicer way? It doesn't have to be system-independent, as the script is for my personal use, and will be only run on Linux.
Try getting your answer from several places, first one wins:
my $username = $ENV{LOGNAME} || $ENV{USER} || getpwuid($<);
crontab sets $LOGNAME so you can use $ENV{"LOGNAME"}. $LOGNAME is also set in my environment by default (haven't looked where it gets set though) so you might be able to use only $LOGNAME instead of $USER.
Although I agree with hacker, don't know what's wrong about getpwuid.
Does this look prettier?
use English qw( −no_match_vars );
my $username = getpwuid $UID;
Sorry, why doesn't that "look nice"? That's the appropriate system call to use. If you're wanting an external program to invoke (e.g. something you could use from a bash script too), there are the tools /usr/bin/id and /usr/bin/whoami for use.
Apparently much has changed in Perl in recent years, because some of the answers given here do not work for fetching a clean version of "current username" in modern Perl.
For example, getpwuid($<) prints a whole bunch of stuff (as a list in list context, or pasted-together as a string in scalar context), not just the username, so you have to use (getpwuid($<))[0] instead if you want a clean version of the username.
Also, I'm surprised no one mentioned getlogin(), though that doesn't always work. For best chance of actually getting the username, I suggest:
my $username = getlogin() || (getpwuid($<))[0] || $ENV{LOGNAME} || $ENV{USER};