Ahem. It's not very often that someone gets to tell Jeff that he's wrong, AND can prove it...

for %%a in ("E:\Source\APP_30\SQL\Structure\*.sql") do SQLCMD -S ServerName -d DatabaseName -E -i %%a -o "E:\output.txt"

for %%a in ("E:\Source\APP_30\SQL\SP\*.sql") do SQLCMD -S ServerName -d DatabaseName -E -i %%a -o "E:\output.txt"

for %%a in ("E:\Source\APP_30\SQL\Triggers\*.sql") do SQLCMD -S ServerName -d DatabaseName -E -i %%a -o "E:\output.txt"

Jeff's suggestion is to change it to:

for %%a in ("E:\Source\APP_30\SQL\Structure\*.sql") do SQLCMD -S ServerName -d DatabaseName -E -i %%a -o "E:\output.txt"

for %%a in ("E:\Source\APP_30\SQL\SP\*.sql") do SQLCMD -S ServerName -d DatabaseName -E -i %%a >> "E:\output.txt"

for %%a in ("E:\Source\APP_30\SQL\Triggers\*.sql") do SQLCMD -S ServerName -d DatabaseName -E -i %%a >> "E:\output.txt"

The first line is processing each .sql file in the "Structure" directory. For each file, it runs the SQLCMD, which will have the -o parameter, overwriting the output file if it already exists. So if there are 3 .sql files in this directory, the output will only have the output from the third file.

It's not until you get to the second "for" line that Jeff's suggestion will start appending the results to the file.

Edit: added this suggestion:

if you were to start off with:

echo %date% %time% > E:\output.txt

you could then replace ALL of the -o with >>