Bash Use Origin File as Destination File - bash

Question

I'm working with the ImageMagick bash tool, which uses commands of the form:
<command> <input filename> <stuff> <output filename>
I'm trying to do the following command:
<command> x.png <stuff> x.png
but for every file in a directory. I tried:
<command> *.png <stuff> *.png
But that didn't work. What's the correct way to perform such a command on every file in a directory?

Answer 1

Many Unix command-line tools, such as awk, sed, grep are stream-based or line-oriented, which means they process files one line at a time. When using these, it is necessary to write output to an intermediate file and then rename that (with mv) back over the original input file. The reason for that is that you may be writing over the input file before you have read it all, so you will clobber your inputs. In that case, #sjsam's answer is absolutely correct - especially since he is careful to use the && in between so that the mv is not done if the command is not successful.
In the specific case of ImageMagick however, and its convert, compare, compose commands, this is NOT the case. These programs are file-oriented, not line-oriented, so they read the entire input file into memory before writing any outputs. The reason is that they often need to know if there are any transparent pixels in an image before they can start processing, or how many colours there are in an image and they cannot know this till they have read the entire file. As such, it is perfectly safe, and in fact idiomatic to use the same input filename as output filename, like this:
convert image.jpg ... -crop 100x100 -colors 16 ... image.jpg
In answer to your question, when you have multiple files to process, the preferred method is to use the mogrify tool (also part of the ImageMagick suite) which is specifically intended for processing multiple files. So, you would do this:
mogrify -crop 100x100 -colors 16 *.jpg
which would overwrite your input files with the results - which is what you asked for. If you did not want it to overwrite your input files, you would add a -path telling ImageMagick the path to an output directory like this:
mogrify -path /path/to/thumbnails -thumbnail -colors 16 *.jpg
If you had too many files for the Windows COMMAND.EXE/CMD.EXE to expand and pass to mogrify, you would let ImageMagick expand the glob (the asterisk) internally like this and be able to deal with unlimited numbers of files:
mogrify -crop 100x100 '*.jpg'
The point is that not only is the mogrify syntax concise, and portable across Windows/OSX/Linux, it also means you only need to create a single process to do all your images, rather than having to write a FOR loop and create and execute a convert process and a mv process for each and every one of potentially thousands of files - so it is much more efficient and less error-prone.

Answer 2

For a single file do like this :
<command> x.png <stuff> temp.png && mv temp.png x.png
For a set of files do like this :
#!/bin/bash
find `pwd` -name \*.png | while read line
do
<command> "$line" <stuff> temp.png && mv temp.png "$line"
done
Save the script in the folder containing the png files as processpng. Make it an executable and run it.
./processpng
A general version of the above script which takes the path to folder containing the above files as argument is below :
#!/bin/bash
find $1 -name \*.png | while read line
do
<command> "$line" <stuff> temp.png && mv temp.png "$line"
done
Save the script as processpng anywhere in the computer. Make it an executable and run it like :
./processpng /path/to/your/png/folder
Edit:
Incorporating #anishsane's solution, a compact way of achieving the same
results would be
#!/bin/bash
find $1 -name \*.png -exec bash -c "<command> {} <stuff> temp.png && mv temp.png {}"
In the context of find .... -exec:
{} indicates (contains) the result(s) from the find expression. Note
that empty curly braces {} have no special meaning to shell so we can
get away without escaping {}
Note: Emphasis mine.
Reference: This AskUbuntu Question.