Bash escape special characters

Question

I am rsyncing directories. using the following.

source_location='/pi/data/2018/image_*.png'
echo $source_location;

Prints /pi/data/2018/image_*.png

mydate=$(date +'%Y');
source_location='/pi/data/$mydate/image_*.png'

Prints /pi/data/2018/image_1.png /pi/data/2018/image_2.png /pi/data/2018/image_3.png etc.

How do I escape the * or is there a better way?

1. what is your desired output? 2. echo $source_location on the second option will not print out what you said it does. — jeremysprofile, Dec 17 '18 at 17:04
I was assuming the second line would print out `/pi/data/2018/image_*.png` just as the first line did. — user3525290, Dec 17 '18 at 17:05
@user3525290: What you're trying to do? What's your expected output? Please make your query more clearer. — User123, Dec 17 '18 at 17:32
My expected outcome would both print lines match. when I use `source_location="/pi/data/$mydate/image_*.png"` printing out source_location prints all of my files in the directory `/pi/data/2018/image_1.png /pi/data/2018/image_2.png /pi/data/2018/image_3.png` instead of just printing the text `/pi/data/2018/image_*.png` — user3525290, Dec 17 '18 at 17:36
Please edit your question and check this if it helps: `source_location=$(ls /pi/data/${mydate}/image_*.png) && echo $source_location` — User123, Dec 17 '18 at 17:40
Unquoted parameter expansions are subject to pathname expansion. I don't see how your *first* example could produce the output you claim. — chepner, Dec 17 '18 at 18:07
Either way, `declare -p source_location` or `printf 'source_location=%q\n' "$source_location"` will be more informative as to the variable's contents. Even the [POSIX spec for `echo`](http://pubs.opengroup.org/onlinepubs/9699919799/utilities/echo.html) advises against its use (see the APPLICATION USAGE and RATIONALE sections). — Charles Duffy, Dec 17 '18 at 18:09

score 0 · Answer 1 · answered Dec 17 '18 at 17:57

0

If you want to store a series of filenames from a glob expansion, use an array:

mydate=$(date +'%Y');
source_location=( "/pi/data/$mydate/image_"*.png )
echo "The first file is ${source_location[0]}"
echo "All files are: " "${source_location[@]}"

answered Dec 17 '18 at 17:57

that other guy

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Paul Hodges · Answer 2 · 2018-12-18T14:28:49.503

-2

Single quotes do not allow interpolation. Double quotes do.

$: touch /pi/data/2018/image_foo.png           # create a file
$: touch /pi/data/2018/image_bar.png           # create a file
$: source_location='/pi/data/2018/image_*.png' # a single-quoted literal
$: echo '$source_location'                     # a single-quoted literal
$source_location
$: echo "$source_location"                     # a double-quoted variable
/pi/data/2018/image_*.png
$: echo $source_location                       # unquoted variable, parsed before passed to echo
/pi/data/2018/image_foo.png /pi/data/2018/image_bar.png

$: mydate=$(date +'%Y');
$: source_location='/pi/data/$mydate/image_*.png' # a single-quoted literal
$: echo '$source_location'                        # a single-quoted literal
$source_location
$: echo "$source_location"                        # a double-quoted variable containing a single-quoted literal
/pi/data/$mydate/image_*.png
$: source_location="/pi/data/$mydate/image_*.png" # a double-quoted variable
$: echo "$source_location"                        # a double-quoted variable containing a double-quoted variable
/pi/data/2018/image_*.png
$: echo $source_location                       # unquoted variable, parsed before passed to echo
/pi/data/2018/image_foo.png /pi/data/2018/image_bar.png

Maybe you should post a more complete question regarding what you are trying to accomplish with what you've tried. Also, please search the site before you do - there are likely a lot of questions out there that are relevant.

edited Dec 18 '18 at 14:28

answered Dec 17 '18 at 17:17

Paul Hodges

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I tried this `source_location='/pi/data/$mydate/image_*.png'` prints `/pi/data/$mydate/image_*.png`. – user3525290 Dec 17 '18 at 17:29
@user3525290 you should not use single quotes if you want some substitution in the text; use double ones. Or, perhaps, try to construct your text step by step. – linuxfan says Reinstate Monica Dec 17 '18 at 17:43
I don't understand bash as well and maybe that is the problem. To me if i use `source_location='/pi/data/2018/image_*.png'` and print that to the screen it prints `/pi/data/2018/image_*.png` but when i introduce a variable `/pi/data/$mydate/image_*.png` it expands and prints all directories `/pi/data/2018/image_21*.png /pi/data/2018/image_2*.png etc` – user3525290 Dec 17 '18 at 17:49
2

The behavior of `echo $source_location` without quotes depends on whether that glob would actually match any files on the system where that test took place -- behavior it's rather critical to expose and discuss in any discussion of the matter. – Charles Duffy Dec 17 '18 at 18:04
Was actually just copy/pasting that line as boilerplate without really looking at it. You're absolutely right. Quoted them. – Paul Hodges Dec 18 '18 at 14:20

Bash escape special characters

2 Answers2