How to I use bash to alphabetize entries in a config file, preserving sequential numbering?

Question

I have an .ini file with this pattern:

[Preset0]
Data=78367571EE5E
Data_1=3483290482
Data_2=299230-293
Data_3=2390920391
Name=Mono Lead
Len=3994
[Preset1]
Data=78367571EE5E
Data_1=3483290482
Data_2=299230-293
Data_3=2390920391
Name=Funky Bass
Len=4001
[Preset2]
Data=0200002E3434
Data_1=342EFD340433
Data_2=000000043425
Data_3=EDFE00000000
Name=Airy Pad
Len=3713

Each set of data from [PresetX] to Len=X represents one preset. I want to alphabetize this list by Name=X. However, I need the first preset in the file to be [Preset0] and I need the presets to remain in numeric order. So, my desired output would be:

[Preset0]
Data=0200002E3434
Data_1=342EFD340433
Data_2=000000043425
Data_3=EDFE00000000
Name=Airy Pad
Len=3713
[Preset1]
Data=78367571EE5E
Data_1=3483290482
Data_2=299230-293
Data_3=2390920391
Name=Funky Bass
Len=4001
[Preset2]
Data=78367571EE5E
Data_1=3483290482
Data_2=299230-293
Data_3=2390920391
Name=Mono Lead
Len=3994

How would I do this with bash?

Answer 1

I don't know how to sort and increment with sed, so this solution mixes sed, bash and sort.

sed -r ':a;N;s/\n/|/;/Len=/{s/(.*Name=)(.*)(Len=.*)/\2\1\3/p;d};ba' presets-file |
sort |
while read line; do
    sed -r 's/(.*\|)(\[Preset)[0-9]+(\].*)(Len=.*)/\2'$((i++))'\3\1\4/;s/\|/\n/g' <<<"$line"
done