Re: Why doesn't this code work?

On 30-1-2015 19:04, Mark Montague wrote:

On 2015-01-30 11:39, Abdul Abdul wrote:

Thanks for your reply. I get a blank screen. On Fri, Jan 30, 2015 at 6:33 AM, Ashley Sheridan <ash@ashleysheridan.co.uk> wrote:

On 30 January 2015 14:26:51 GMT+00:00, Abdul Abdul <abdul.sw84@gmail.com> wrote:

I have the following PHP script, where I'm trying to retrieve the image type. Why doesn't it work? What am I doing wrong? <?php $image="4.2.06" $imagetype = exif_imagetype($image); echo "Image type is: " $imagetype; ?>

What does "doesn't work" mean? What output are you getting? Are there any errors or warnings in your logs?

Ash's advice is correct: check your error log. However, I think that one problem is that you don't have a semicolon at the end of the first assignment statement above, which results in a syntax error occurring. Also, please do not top-post. Put your replies at the bottom instead.

The problem is actually threefold; 1 - Errors aren't being displayed (display_errors in your php.ini is turned OFF; turn it on to see them), and YOU are not looking in the log for the errors (as suggested by Ashley and Mark). 2. You have a parse error online 2 as hinted by Mark: no semicolon to terminate the line. You don't see this error due to #1 3. You have another parse error on line 4 between the double quote and the variable you're trying to concatenate. You're missing the concatenation operator: the 'dot' (.). You don't see this error due to #1 So, fix #1 either by turning on display_errors or read your logfiles. Then you'll be able to identify problems #2 and #3 for yourself, and fix them easily. - Tul --- This email has been checked for viruses by Avast antivirus software. http://www.avast.com