How to escape a percentage sign

Question

I have a simple one line MySQL command that I want to run from AppleScript.
The command itself is
UPDATE exif SET datevalue = tagvalue WHERE tagname LIKE "% DATE %" ;
Unfortunately, every time I run it I get some error or improper result.
I've tried escaping out the quotes and percentage sign with a backslash, using "quoted form of" statements, parsing out individual words and using "quote" operators, using curly quotes, and even substituting "ASCII character 37" or "character id 37" for my percent symbol. Most often I get backslashes before my quotes. Here is a recent example.
Code:
set part1 to "UPDATE exif SET datevalue = tagvalue WHERE tagname LIKE "
set part2 to " DATE "
set part3 to " ;"
set convertDate to part1 & quote & (ASCII character 37) & part2 & (ASCII character 37) & quote & part3
return convertDate
Result:
"UPDATE exif SET datevalue = tagvalue WHERE tagname LIKE \"% DATE %\" ;"
I know I can always call a 1 line shell script, but I hope there is a simple solution.

Answer 1

Quote-Mania
mabye this example helps:
do shell script "echo UPDATE exif SET datevalue = tagvalue WHERE tagname LIKE \"% DATE %\" >~/Desktop/testoutput.txt"
do shell script "echo UPDATE exif SET datevalue = tagvalue WHERE tagname LIKE \\\"% DATE %\\\" >>~/Desktop/testoutput.txt"
the result in Skript-Editor looks like
"UPDATE exif SET datevalue = tagvalue WHERE tagname LIKE % DATE %
UPDATE exif SET datevalue = tagvalue WHERE tagname LIKE \"% DATE %\""
when you open that text file now you will see
UPDATE exif SET datevalue = tagvalue WHERE tagname LIKE % DATE %
UPDATE exif SET datevalue = tagvalue WHERE tagname LIKE "% DATE %"
Just keep in mind, that the result of a script could be directly inserted into another script, so the output will be quoted if needed to make this possible.