This question already has an answer here:
Bash - Concatenating Variable on to Path
(1 answer)
Closed 7 years ago.
I'm trying to redirect my output to a file which has its path stored in a variable but I cant get it to work.
LOG_TEST="~/AVDS/logs/${HOSTNAME}_testlog.log"
echo "foo" >> ~/AVDS/logs/${HOSTNAME}_testlog.log
echo "bar" >> $LOG_TEST
The "foo" line will work fine but the "bar" line returns the error:
./testarea.sh: line 9: ~/AVDS/logs/tvpc-office_testlog.log: No such file or directory
What am I doing wrong here?
Tilde expansion only happens when unquoted.
Get in the habit of using $HOME, not ~, in scripts.
Related
This question already has answers here:
Command not found error in Bash variable assignment
(5 answers)
Closed 2 years ago.
i am trying to give an file as input to me my shell script:
#!/bin/bash
file ="$1"
externalprogram "$file"
echo 'unixcommand file '
i am trying to give the path to my file but it says always
cannot open `=/home/username/documents/file' (No such file or directory)
my path is this /home/username/Documents/file
i do this in terminal : ./myscript.sh /home/username/Documents/file
can someone help me please?
When you say
file ="$1"
with a space after "file", you're running something called file with =$1 as an argument. There probably actually is a utility called file. If you want to assign $1 to a variable called file, you don't need the space:
file="$1"
there shouldn't be a space before = in the second line.
file=$1 should be good enough.
Check what shellcheck says about your code:
^-- SC1068: Don't put spaces around the = in assignments (or
quote to make it literal).
You can read more about SC1068 case on its Github
page.
#!/bin/bash
file=$1
code $file
echo "aberto o arquivo ${file} no vscode"
I made this code snippet to demonstrate, I pass a path and it opens the file in vscode
This question already has answers here:
I just assigned a variable, but echo $variable shows something else
(7 answers)
Closed 2 years ago.
Execute the following command in bash shell:
export sz1='"authorities" : ["uaa.resource"]'
Now, try echo $sz1
I expect to see the following output:
"authorities" : ["uaa.resource"]
But instead I get this:
"authorities" : c
The interesting thing is that I have dozens of servers where I can execute this type of variable assignment and it works except on this server. This server has exactly the same OS version, profile, bash version etc. What could be the reason for this behavior?
Always quote your variables. Use
echo "$sz1"
When you don't quote the variable, word splitting and wildcard expansion is done on the variable expansion. On ["uaa.resource"] is a wildcard that will match any of the following filenames:
"
u
a
.
r
e
s
o
u
c
On that one machine you have a file named c, so the wildcard matches and gets replaced with that filename.
This question already has answers here:
Replace a string in shell script using a variable
(12 answers)
sed substitution with Bash variables
(6 answers)
Closed 4 years ago.
I need to replace current value in configuration file with new value which is assigned to variable ,
like
file_name=abc.txt
needs to be replaced like
file_name=xyz.txt
where $file=xyz.txt
I tried
sed -i 's/file_name=.*/file_name=$file/g' conf_file.conf
however the variable is not getting expanded,
it comes like file_name=$file.
any pointers?
This should work,assuming that variable file has value:xyz.txt assigned to it:
sed "s/file_name=.*/file_name=${file}/g" file_name
Output:
file_name=xyz.txt
This question already has an answer here:
Execute an shell program with node.js and pass node variable to the shell command [closed]
(1 answer)
Closed 4 years ago.
I have the following code:
exec('sh cert-check-script-delete.sh', req.body.deletedCert);
console.log(req.body.deletedCert);
The console log correctly shows the req.body.deletedCert is non-empty.
And in cert-check-script-delete.sh I have:
#!/bin/sh
certs.json="./" # Name of JSON file and directory location
echo -e $1 >> certs.json
But it's just writing an empty line to certs.json
I've also tried:
exec('sh cert-check-script-delete.sh' + req.body.deletedCert)
But neither formats work
Use execFile(), and pass your arguments out-of-band:
child_process.execFile('./cert-check-script-delete.sh', [req.body.deletedCert])
That way your string (from req.body.deletedCert) is passed as a literal argument, not parsed as code. Note that this requires that your script be successfully marked executable (chmod +x check-cert-script-delete.sh), and that it start with a valid shebang.
If you can't fix your file permissions to make your executable, at least pass the arguments out-of-band:
child_process.execFile('/bin/sh', ['./check-cert-script-delete.sh', req.body.deletedCert])
This question already has answers here:
Command not found error in Bash variable assignment
(5 answers)
Variable variable assignment error -"command not found"
(1 answer)
Why does a space in a variable assignment give an error in Bash? [duplicate]
(3 answers)
Closed 4 years ago.
So I've got a shell script to do some lazy stuff for if the directory isn't changing for a user. It's below. Essentially, it should be an if statement that if the user enters "default" for the directory, it'll pull them to the default directory for the files. However, I'm getting a command not found on line 16, which is the reassignment statement.
The entire if statement below:
if [ $directory = "default" ];
then
echo Enter your ldap:
read $ldap
$directory = "/usr/local/home/google/${ldap}/Downloads"
fi
I've tried doing it without the dollar sign too...nothing. What's going on here? New to shell, couldn't find this question asked before either.
This is how you should assign a value to a variable in shell:
directory="/usr/local/home/google/${ldap}/Downloads"
No dollar ($) sign.
No space around equal (=) sign.
Also, you should wrap your variables inside double quotes ("). This way, you avoid errors with undefined variables, arguments with spaces, etc.
That gives us:
if [ "$directory" = "default" ]
then
echo "Enter your ldap:"
read $ldap
directory="/usr/local/home/google/${ldap}/Downloads"
fi