Renaming multiple files with a bash loop

I am trying to replace all instances of "heat_1" to "heat_21" in the following file names, but I am having issues with producing a consise regex that covers all four files:

5JUP_N2_UUCGCU_heat_1.out  5JUP_N2_UUCGCU_heat_1.rst  mdcrd_heat_1  mdinfo_heat_1

I did the following, but the file extensions ".out" and ".rst" got removed. How do I write the regex so that they don't get removed?

for i in ls *
do 
mv $i ${i/heat_*/heat_21}
done

With Perl's standalone rename (or sometimes prename) command:

rename -n 's/heat_1/heat_21/' *

Output:
<pre>
rename(5JUP_N2_UUCGCU_heat_1.out, 5JUP_N2_UUCGCU_heat_21.out)
rename(5JUP_N2_UUCGCU_heat_1.rst, 5JUP_N2_UUCGCU_heat_21.rst)
rename(mdcrd_heat_1, mdcrd_heat_21)
rename(mdinfo_heat_1, mdinfo_heat_21)
</pre>

If output looks okay, remove -n.

With bash:

for i in *; do echo mv "$i" "${i/heat_1/heat_21}"; done

Output:
<pre>
mv 5JUP_N2_UUCGCU_heat_1.out 5JUP_N2_UUCGCU_heat_21.out
mv 5JUP_N2_UUCGCU_heat_1.rst 5JUP_N2_UUCGCU_heat_21.rst
mv mdcrd_heat_1 mdcrd_heat_21
mv mdinfo_heat_1 mdinfo_heat_21
</pre>

If output looks okay, remove echo.

In your regular expression, you are changing "heat_*" to "heat_21".

So, you are changing every occurence of string "heat_" and all letters after it to "heat_21". If you will place "heat_1" in a middle of your file name you will see, that after running your initial code, all the rest of the file name (after "heat_") will be gone.

So, to fix your problem, you should use proper regexp:

for i in *
do
mv "$i" "${i/heat_?/heat_21}"
done

I replaced * in your regular expression with ?, which stands for any character. But, since you only want to switch "heat_1" to "heat_21", you can replace '?' with '1'.