How to inject a variable as `VARIABLE` in the following?

```bash
text='```python
VARIABLE
```'

curl 'URL' \
-F 'mode=comment' \
-F "text=$text"
```
May 19, 11:33pm

To inject a variable as VARIABLE in your Bash script, you need to ensure that the text is properly escaped and that the variable is expanded correctly. Here’s how you can do it:

Solution

bash
VARIABLE="your_value_here"
text='```python
'"\$VARIABLE"'
```'

curl 'URL' \
  -F 'mode=comment' \
  -F "text=$text"

Explanation

  1. Define the Variable: Set the value of VARIABLE as needed.

  2. Construct the Text: Use double quotes for the parts where you want variable expansion. Here, "'"$VARIABLE"'" ensures that the value of VARIABLE is inserted between the triple backticks.

  3. Use in cURL: Pass the constructed text variable to cURL as before.

Example

If VARIABLE is set to "print('Hello, World!')", the resulting text will be:

python
print('Hello, World!')

This ensures the variable is correctly injected and expanded.

May 19, 11:33pm

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