Python throw error open image

Question

img_data=open('assets\pblogo.png','rb').read()
img_title=MIMEImage(img_data,'png')
img_title.add_header('Content-Id','<pblogo>')
img_title["Content-Disposition"]="inline"

img_data=open('data\report.jpg','rb').read()
img_report=MIMEImage(img_data,'jpeg')
img_report.add_header('Content-Id','<reportimg>')
img_report["Content-Disposition"]="inline"

I am using IDLE on win7, and the first png captured well

as to the second:

img_data=open('data\report.jpg','rb').read()

IOError: [Errno 22] invalid mode ('rb') or filename: 'data\report.jpg'

why can't I open the second pic in the same way?

falsetru · Accepted Answer · 2013-07-26 05:24:30Z

1

\r is carriage return.

Escape the \:

'data\\report.jpg'

or use raw string:

r'data\report.jpg'

answered Jul 26, 2013 at 5:24

falsetru

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1 Comment

Ethan Over a year ago

thanks~There are same Q&A and I do as you said and still not work~Guess what, the file name is "reprot" not "report"...Anyway thanks LOL

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