curl-users
RE: Apostrophe Causing "Output" To Fail?
Date: Fri, 21 Nov 2003 12:17:58 -0500
-----Original Message-----
From: Art Pena
Sent: Friday, November 21, 2003 10:06 AM
> Thanks for the tip. I did as you suggested, but it does appear to be
> (the start of) a valid image file:
You might also look at the end of the file to see if some extra text got in there.
What are the exact file sizes when it does and doesn't work? That info might yield a clue.
Is it possible that your webcam was in the middle of updating its photo when you tried the URL with the apostrophe in it? (e.g. is this repeatable every single time, or did you just get unlucky a couple times and assume the apostrophe caused it)
> Or, how can I incorporate "-f" into my string to try and debug
Normally you'd just add it in; if a non-successful http_code was returned, instead of outputting your (possibly corrupted) file, curl would instead 'die' with an exit code of 22, and would print an error message (to stderr I assume) explaining why it just quit.
But, I notice you've got a -w (--write-out) statement in use, which already shows you the same thing, so the -f may not be useful.
- Kevin
On Friday, November 21, 2003, at 08:22 AM, Roth, Kevin P. wrote:
>> "curl --silent " + cameraSource + " --max-time 300 -w
>> \"%{http_code}\\n\" --output Resources/Cache/cache.jpg"
>
>> But if "cameraSource" is something like this, the file saved has
>> errors:
>> http://192.168.1.152/Webcam%20Repository/Jim's%20Webcam/001.jpg
>
>> Could not open "cache.jpg" because a JPEG marker segment length is
>> too short (the file may be truncate or incomplete)
>
> Try opening the resulting output file in a text editor (rename it
> to cache.txt if needed) and see what's in there. I wonder if you
> won't find some kind of web server error rather than the image
> data you're expecting.
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Received on 2003-11-21