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Message #38545
Re: [Question #295370]: variable in app.open command
Question #295370 on Sikuli changed:
https://answers.launchpad.net/sikuli/+question/295370
Status: Open => Answered
RaiMan proposed the following answer:
import os
dir = r"C:\Users\win7\Desktop|\textfiles"
for e in os.listdir(dir):
fileName = os.path.join(dir, e)
print fileName
print "all files processed"
... and then
App.open ("notepad" + fileName)
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