Re: Output Errors 2

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In article <200210221805.MAA08355@www2.ebmstpa.com>, biorn@ebmstpa.com 
says...
> Unless it is a typo, the line that reads:
> 
> if (image == 'NULL') { echo '<p></p></td>'; }
> 
> should read:
> 
> if ($image == 'NULL') { echo '<p></p></td>'; }
>     ^
> 
> You forgot the $ in front of image.
> 
> HTH
> 
> MB
> Bzdpltd@aol.com said:
> 
> > Hi sorry back again.
> > 
> > Thanks for helping me get the code right. Sorry being stupid again, I am 
> having problems with an if / else statement.
> > 
> > I am trying the following but it does not seem to work, what the problem 
> is, that I am saying if the image name field in the database is empty to 
> print nothing, and if it contains data to print that field name, as the a 
> href is in the code which pulls the image, if the field is empty it should 
> show nothing, but what happens is the image is shown in the browser and its 
> not blank.
> > 
> > Here is the revised code:
> > 
> >         echo ("<tr> \n");
> >         echo ("<td width=\"72%\" valign=\"top\"><h3>$title</h3> \n");
> >         echo ("<p>$description</p></td> \n");
> >         echo ("<td width=\"28%\" valign=\"top\">  \n");        
> >             if (image == 'NULL') { echo '<p></p></td>'; }
> >             else { echo '<p><img src="' . $image . '" width="201" 
> height="160"></p></td>'; }
> >         echo ("</tr> \n");        
> >         }}
> > 
> > 
> > If someone can see the problem it would be most helpful.

In addition to Maureen's point, you are currently checking whether the 
variable $image contains the string NULL; you can test for NULL (without 
the quotes) or better, use the function empty() to test whether $image has 
a value.


-- 
David Robley
Temporary Kiwi!

Quod subigo farinam

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