0

I haven't been able to figure this out. I'm using qpython3 in android 4.4.2 with a metro pcs Samsung galaxy lite. I do very very little developing in android, so I'm unfamiliar with the differences between android and windows and how I could go about figuring it out.

All I could come up with was what i found on stackoverflow:

import android
url = "http://www.Google.com"
android.Android().startActivity('android.intent.action.VIEW'‌​, url)

Which didnt do anything. As soon as you try to run the program, This error pops up and thats all.:

/data/data/com.hipipal.qpy3/files/bin/qpython.sh"/storage/emulated/0/com.hipipal.qpyplus/.last_tmp.py" && exit 
s/.last_tmp.py" && exit  < 
    File "/storage/emulated/0/com.hipipal.qpyplus/.last_tmp.py", line 3 
    android.Android().startActivity('android.intent.action.VIEW'‌​‌​, url) ^     

SyntaxError: invalid character in identifier 1|u0_a194@gardalteMetroPCS:/ $

Any ideas?

  • Please post the full error traceback! – Klaus D. Aug 26 '16 at 22:00
  • Please it to your quextion formatted as code. It's not readable as a comment. – Klaus D. Aug 26 '16 at 22:08
  • I'm using the mobile version of this site, ill have to format it once I get to a computer. Thanks. –  Aug 26 '16 at 22:27
  • the error code has been formatted the best that i can format it. im not totally sure what its saying and how its being said, so i did my best to replicate it the same way its displayed on my phone. –  Aug 27 '16 at 19:42
  • I don't see any error. The exception relates to a disallowed character inside a valiable but there is non I can see. – Klaus D. Aug 28 '16 at 04:42

1 Answers1

1

When I tried the code myself, I got the same error. I fixed by changing 's into "s and import android to import androidhelper so code ended up like this and it worked:

import androidhelper 

url = "http://www.Google.com"
androidhelper.Android().startActivity("android.intent.action.VIEW", url)
Majid Hajibaba
  • 3,105
  • 6
  • 23
  • 55
Fuat Yılmaz
  • 11
  • 2