Forum Discussion
aniruddhay
16 years agoVisitor
Hello MarkRoddy,
Thank you for posting a reply. What I mean is that when I parse the feed form Picasa, it gives me the url as 'www.example.com/somehting/xxxx.jpg', whereas in the code, when given a value for the variable, it expects, "www.example.com/somehting/xxxx.jpg". As it is getting a value with single quote, it is treating it as a syntax error. Is my understanding of the code correct? If it is, how do I convert the " to ' so the variable will take the value within the double quotes.
Sorry for being unclear on my first post.
Thank you for posting a reply. What I mean is that when I parse the feed form Picasa, it gives me the url as 'www.example.com/somehting/xxxx.jpg', whereas in the code, when given a value for the variable, it expects, "www.example.com/somehting/xxxx.jpg". As it is getting a value with single quote, it is treating it as a syntax error. Is my understanding of the code correct? If it is, how do I convert the " to ' so the variable will take the value within the double quotes.
Sorry for being unclear on my first post.