stat work at the command line but not when passed into a variable

Question

I am debugging a bash script (on ubuntu) to obtain an image file block size using stat. The result from stat is correct at the command line but not when passed into a variable (as it is in the script).

If I use the stat command at the command line, I get what I want (the number of blocks, %b):

stat --format=%b image.png

Output, e.g.: 72

But if I pass the same into a variable (at the command line or in a script), like so:

b = $(stat --format=%b image.png); echo $b

I get this output:

 15:16:57 up  3:47,  0 users,  load average: 0.52, 0.58, 0.59
USER     TTY      FROM             LOGIN@   IDLE   JCPU   PCPU WHAT

Yet, when I "x check" my script with /bin/bash -x ../script.sh, the variable b, defined as above, except that the image filename is passed as a variable, is shown to hold this value:

+ b = 328
 15:47:39 up  4:18,  0 users,  load average: 0.52, 0.58, 0.59
USER     TTY      FROM             LOGIN@   IDLE   JCPU   PCPU WHAT

What did I miss?

This is my script:

#!/bin/bash

## Run in a folder containing image files to return the html <img /> links.

touch img_reflinks.html

generate_list ()
{
  ls  .| egrep '\.png|\.svg|.jpg|.jpeg|.tiff'
}

for f in $(generate_list)
do
    str=''
    style=''

    # the filename prints correctly:
    echo "$f"

    # this is the problematic assignment, as in CL:
    b = $(stat --format=%b "$f")

    style="\"width:" + "$b" + "px;\""

    str="<img src=\"" + "$f" + "\", style=" + "$style" + "/>\n"

    echo "$str" >> img_reflinks.html
done

Aren't you getting a pile of "command not found" type errors for this script? — that other guy, Apr 07 '19 at 20:34
Please explain the downvotes as they are totally [unwarranted](https://stackoverflow.com/help/privileges/vote-down). Thanks. — MCC, Apr 11 '19 at 20:39
in bash no spaces are allowed surrounding = sign during assignment so its `b=$(stat --format=%b "$f")` and not `b = $(stat --format=%b "$f")` — Scott Stensland, Aug 23 '21 at 19:31

Gilles Quénot · Accepted Answer · 2019-04-07T21:06:30.097

0

No space allowed in a variable declaration; so instead of

b = $(stat --format=%b image.png); echo $b

do

b=$(stat --format=%b image.png); echo $b

And instead of

ls  .| egrep '\.png|\.svg|.jpg|.jpeg|.tiff'

do :

for i in *.png *.svg *.jpg *.jpeg *.tiff; do
    # ...
done

edited Apr 07 '19 at 21:06

answered Apr 07 '19 at 20:36

Gilles Quénot

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stat work at the command line but not when passed into a variable

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