I had a simple python code working with URLs which caused an error of the form: "unsupported format character 'p' (0x70) at index 72".
The code looks like:
num = 1
abc = "http://<site>?value=[abc%20def],value2=%d"
URL = abc % (num)
This is caused because of using the % sign in the string. We need to escape the % sign with another % sign.
So the new string looks like:
abc = "http://<site>?value=[abc%%20def],value2=%d"
References:
1. http://yuji.wordpress.com/2009/01/09/python-valueerror-unsupported-format-character-percent-sign-python-format-string/
The code looks like:
num = 1
abc = "http://<site>?value=[abc%20def],value2=%d"
URL = abc % (num)
This is caused because of using the % sign in the string. We need to escape the % sign with another % sign.
So the new string looks like:
abc = "http://<site>?value=[abc%%20def],value2=%d"
References:
1. http://yuji.wordpress.com/2009/01/09/python-valueerror-unsupported-format-character-percent-sign-python-format-string/
No comments:
Post a Comment