Programming Forums - View Single Post - can't seem to get my app working on the web

glevine · Aug 30th, 2005, 12:24 PM

I have 'somewhat' successfully written an app using html, javascript, and xmlhttprequest that grabs data from a js file residing on another website and displays the data found in that file. I say 'somewhat' because I can run the app locally, but it fails when I attempt to run it from the web. After storing the file in a directory on my webserver I can't seem to get the page to work. I can input data, but I can't get any output. Using Firefox, I get absolutely nothing. Using IE, I get an error. That error claims to be for the line...

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http.open("GET", url + escape(file), true);

'http' is the value that is returned by 'new XMLHttpRequest();' and 'file' is the string representing the js file that needs to be opened. 'url' is the string that represents the URL of the file location without the filename concatenated. I was hoping someone might know a reason why the open function wouldn't work in my case. If you need more info please ask. Thanks.

Aug 30th, 2005, 12:24 PM	#1
glevine Newbie Join Date: Jun 2005 Posts: 22 Rep Power: 0	can't seem to get my app working on the web I have 'somewhat' successfully written an app using html, javascript, and xmlhttprequest that grabs data from a js file residing on another website and displays the data found in that file. I say 'somewhat' because I can run the app locally, but it fails when I attempt to run it from the web. After storing the file in a directory on my webserver I can't seem to get the page to work. I can input data, but I can't get any output. Using Firefox, I get absolutely nothing. Using IE, I get an error. That error claims to be for the line... (Toggle Plain Text) http.open("GET", url + escape(file), true); http.open("GET", url + escape(file), true); 'http' is the value that is returned by 'new XMLHttpRequest();' and 'file' is the string representing the js file that needs to be opened. 'url' is the string that represents the URL of the file location without the filename concatenated. I was hoping someone might know a reason why the open function wouldn't work in my case. If you need more info please ask. Thanks.