bash in urxvt - backspacing on alrady typed character while reading variable

84 views Asked by At

The intent is to put the cursor at some position on the screen (around the center of the screen), enter some initial text, and prompt the user to enter more text to be saved in a variable, while leaving him the normal readline line editing capabilities.

My initial attemp was to:

  1. center the cursor vertically echo/printfing some whitespace,
  2. center the cursor horizontally again through echo/printf,
  3. issue read -i "editable pre-text" -e answer.

However I noticed the behavior described below, and crafted the following exemplifying two lines-code to demonstrate it.

When executing the following script

echo -n "______________"
read -e risp

as long as no input is typed, Backspace is ineffective (and this is kind of the behavior I like, as the characters entered by echo before read cannot be deleted during read).

However, typing something (e.g. some text) and then deleting it (completely, up to and including the first typed character) with Backspace, will result in the cursor to jump to the first colon of the terminal as soon as Backspace deletes the first typed character.

In other words, as the script starts, Backspace does nothing, while XBackspace will result in the cursort to jump to the first column of the terminal.

What is the reason for this behvaior, and how can I avoid it?

Given the title of this question I would expect it to be related to the present one, but I can't understand if it really is.

1

There are 1 answers

0
rici On BEST ANSWER

As mentioned in a comment, you should use read's -p option to print the prompt, rather than trying to set it up before the read command.

The -e option asks read to use the readline library to handle the input, allowing a wider range of line-editing characters. However, in order to implement these behaviours, readline needs to be able to redraw the current line, and that's not possible if there is anything on the current line when the read starts. It's not possible because Unix provides no mechanism for an application to look at what's being displayed on the console. So under some circumstances, readline will simply clear the line. Using the -p option allows readline to output the prompt, and it can then know what the line currently looks like.