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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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