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])
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:
Why does shell ignore quoting characters in arguments passed to it through variables? [duplicate]
(3 answers)
How can I store a command in a variable in a shell script?
(12 answers)
Closed 2 years ago.
I have an external executable that I need to pass arguments to. With a bash script, I have code that determines these arguments. Some arguments may have escaped spaces.
I need to then execute that string, without expanding each of the arguments.
# ... some code that determines argument string
# the following is an example of the string
ARGSTR='./executable test\ file.txt arg2=true'
exec ${ARGSTR}
I must have the $ARGSTR be expanded so that I can pass arguments to ./executable, but each of the arguments should not be expanded. I have tried quoting "test file.txt", but this still does not pass it as one argument to ./executable.
Is there a way to do something like this?
Us an array instead of a string:
#!/usr/bin/env bash
ARGSTR=('./executable' 'test file.txt' 'arg2=true')
exec "${ARGSTR[#]}"
See:
BashFAQ-50 - I'm trying to put a command in a variable, but the complex cases always fail.
https://stackoverflow.com/a/44055875/7939871
This may achieve what you wanted :
ARGSTR='./executable test\ file.txt arg2=true'
exec bash -c "exec ${ARGSTR}"
This question already has answers here:
How to substitute shell variables in complex text files
(12 answers)
Closed 3 years ago.
Trying to variable replace a templated yaml file.
I'm using eval to take the environment shell variables and replace whats in the file dynamically. I can't figure out how to take the output of this and save to a file.
I just want to take the evaluated output and save to a file.
eval "cat <<EOF
$(<${baseFileName})
EOF"
Exmaple test.yaml
---
value: ${PORT}
Bash environment variable:
PORT=8888
output temp.test.yaml
---
value: 8888
Right now the code will just print the evaluated text to the console.
I've tried.
eval "cat <<EOF
$(<${baseFileName})
EOF" > $newBaseFileName
but no joy. Didn't even create the file.
The reason I'm not using sed is because the file could have unlimited variable decelerations, and I want to replace any value matching a defined bash variable or environment variable. This is part of a template engine. For the life of me I can't remember how I did it before with pure bash.
It didn't work for me but what I did is this
renderTemplate() {
eval "cat <<EOF
$(<${1})
EOF"
}
baseFileName=$(basename $fileName)
templateOutput=`renderTemplate ${baseFileName}`
echo "${templateOutput}"
I'm using this as a temp file anyways so what I'll do is save to variable and then pump that variable in to the command to apply the template as a file. That way it's only ever stored in memory. This is a middleware cli to another cli to add variable replacement to in-memory web hosted files before applying them.
Thanks for your help.
This question already has answers here:
File content into unix variable with newlines
(6 answers)
Closed 4 years ago.
In the bash shell, I'm trying to read the json file and load to a variable
eri#xyz:~/Documents/inbound>e1=$(eval echo $(cat ./deploy/request.json))
Upon fetching the output of that variable, I'm seeing -bash - command not found along with the actual contents of the .json file
eri#xyz:~/Documents/inbound>"$e1"
-bash: { type:Pipeline, category:Software, risk:4, short_description:sample short description text, description:sample detailed description text, assignment_group: Services - Retail Services, cmdb_ci:Retail Service, u_version:1.0, start_date:2017-01-04 18:00:00, end_date:2017-01-04 19:00:00, backout_plan:see department for standard backout plan, implementation_plan:sample implementation plan, test_plan:sample text plan, production_system:false }: command not found
Is there a way to suppress the -bash - command not found in the output?.
No need for eval - just e1=$(< ./deploy/request.json) should do the trick. (Thanks to #shellter for the syntax — you don't even need to use cat!)
To show the variable, you want
echo "$e1"
instead of just "$e1". "$e1" by itself on the command line does not print out the value of $e1, unlike many programming-language REPLs. Instead, it tells bash to try to interpret the entire contents of $e1 as the name of a command. It isn't the name of a command, so bash tells you a command by that name cannot be found.
This question already has an answer here:
Bash - Concatenating Variable on to Path
(1 answer)
Closed 7 years ago.
I'm trying to redirect my output to a file which has its path stored in a variable but I cant get it to work.
LOG_TEST="~/AVDS/logs/${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.