I'm writing a shell script for an assignment. The script is supposed to read regex argument and then use that regex. The problem is, if the user doesn't use single/double quotations then the regex will be evaluated before I need it to be evaluated.
Here's an example; let's say that the user executed the script and passed *f* as an argument, and the code has the following lines:
arg=$1 #*f*
echo "$arg"
In the above example echo returns a file name in the current directory that contains the regex *f*. I need echo to print *f*. It works if the user passes '*f*', but is there no other way?
Related
This is my script code
#!/bin/bash
timestamp=$(date +%F-%T)
clinet_id="123"
STRING=s3://<bucketname>/folder/$client_id/$client_id_gdpr_access_report_$timestamp.csv
echo "$STRING"
$SHELL
If i run this code am getting timestamp value.csv file
how can i concatenate variable with string.
am expecting out put like below
s3://<bucketname>/folder/123/123_report_2022-01-25-14:55:47.csv
i can able to concatenateaccess_report_$timestamp.csv
if i add $client_id_ in the beginning, it will print
2022-01-25-14:55:47.csv
Expecting a better advice
You need to look better at the names of your variables; it's 'client_id' not 'clinet_id' ...
And you should take care of double quoting your string, and put braces around variables when in doubt:
STRING="s3://<bucketname>/folder/${client_id}/${client_id}_gdpr_access_report_${timestamp}.csv"
I'm trying to reassign the value of the positional command line argument $1 to another variable to use elsewhere in the script, but it never works. I have put the following code in a new script and found that the value of the new variable is always blank.
dirname = $1
echo "This is \$1:$1"
echo "This is \$dirname:$dirname"
The output of this after running ./test.sh someval is:
This is $1:someval
This is $dirname:
As you can see, 'dirname' is blank, even after being assigned the value of $1. I'm new to bash so likely missing something. Any help is appreciated.
Assignments cannot have whitespace around the =. An assignment is a single word containing a =: the name precedes the (first) =, and the value follows it.
dirname=$1
I would like to achieve this in Bash: echo $(a=1)and print the value of variable a
I test eval, $$a,{}, $() but none of them work as most of the times either I got literally a=1 or in one case (I don't remember which) it tried to execute the value.
I known that I can do: a=1;echo $a but because I'm little fun one command per line (even if sometimes is getting little complicated) I was wondering if is possible to do this either with echo or with printf
If you know that $a is previously unset, you can do this using the following syntax:
echo ${a:=1}
This, and other types of parameter expansion, are defined in the POSIX shell command language specification.
If you want to assign a numeric value, another option, which doesn't depend on the value previously being unset, would be to use an arithmetic expansion:
echo $(( a = 1 ))
This assigns the value and echoes the number that has been assigned.
It's worth mentioning that what you're trying to do cannot be done in a subshell by design because a child process cannot modify the environment of its parent.
I'm very new to Unix scripting (.ksh). I have to implement a functionality to check whether my argument says "welcome" present in an string array e.g.
{"welcome","test","exit"}
The logic is similar to String.contains in Java.
Any help will be appreciated.
You can do something like this. Following is in bash, you need to change it accordingly to ksh.
script
array=(welcome test exit)
string='welcome';
for item in ${array[*]}
do
if [[ $string =~ .*$item.* ]]
then
echo "It's present!"
fi
done
Output
It's present!
To iterate over arguments passed to a shell script, use for with empty in, that default the iteration over arguments, or in '$#'.
How do I write a shell script (bash on HPUX) that receives a string as an argument containing an asterisk?
e.g. myscript my_db_name "SELECT * FROM table;"
The asterisk gets expanded to all the file names in the current directory, also if I assign a variable like this.
DB_QUERY="$2"
echo $DB_QUERY
The asterisk "*" is not the only character you have to watch out for, there's lots of other shell meta-charaters that can cause problems, like < > $ | ; &
The simple answer is always to put your arguments in quotes (that's the double-quote, " ) when you don't know what they might contain.
For your example, you should write:
DB_QUERY="$2"
echo "$DB_QUERY"
It starts getting awkward when you want your argument to be used as multiple parameters or you start using eval, but you can ask about that separately.
You always need to put double quotes around a variable reference if you want to prevent it from triggering filename expansion. So, in your example, use:
DB_QUERY="$2"
echo "$DB_QUERY"
In the first example, use single quotes:
myscript my_db_name 'SELECT * FROM table;'
In the second example, use double quotes:
echo "$DB_QUERY"