Some more info about this issue that I've discovered is that when using keystroke with numbers, the System Events always sends them as ANSI_Keypad# characters (code 82-92) instead of what you might expect as ANSI_# characters (code 18-29).
For most Mac applications, this does not matter as OS X itself does not care about (use, support, etc) the NUMLOCK function on a Keypad, therefore the Keypad numbers are seen the same as keyboard numbers. However, it will pass the NUMLOCK keypress to the application if you have a keyboard/keypad that has this key.
There are a few applications that do monitor the NUMLOCK key (for example, VMware Fusion application when running a Windows VM) and will change the behavior based on the NUMLOCK state.
So for example, if the AppleScript sends keystroke "456" to a NUMLOCK aware application.
- If the
NUMLOCK state is ON, the numbers "456" will appear.
- If the
NUMLOCK state is OFF, the equivalent keys received are Left Arrow 5 Right Arrow
In the original question, the AppleScript sent abc 123 but most likely his application (which was not mentioned) was aware of the NUMLOCK state, which was currently OFF, and therefore executed the keys as abc [space] End Down Arrow Page Down
I put together a little AppleScript function that loops through the given string sending key code commands for any numbers and keystroke commands for any other characters.
on numberAsKeycode(theString)
tell application "System Events"
repeat with currentChar in (every character of theString)
set cID to id of currentChar
if ((cID ≥ 48) and (cID ≤ 57)) then
key code {item (cID - 47) of {29, 18, 19, 20, 21, 23, 22, 26, 28, 25}}
else
keystroke currentChar
end if
end repeat
end tell
end numberAsKeycode
set myString to "abc 123"
numberAsKeycode(myString)
Which executes the following
tell application "System Events"
keystroke "a"
keystroke "b"
keystroke "c"
keystroke " "
key code {18}
key code {19}
key code {20}
end tell
Hope this helps :)