add background image

[learnPHP]

Hi,

I have 2 background images.
Ineed to add the first image if the value of variable test is 1 otherwise i need to add the second background image.

I have written the following code:

<?PHP
			   if ($test ==1)
			   echo "<td width='50%' valign='top'  STYLE='background-image: url('img/table_faoback.jpg');'>";
			   else
			   echo "<td width='50%' valign='top'  STYLE='background-image: url('img/table_back.jpg');'>";
			  
			   ?>

However it does not display any image. Please let me know how to do this.

Thanks.

[Markus]

Hi,

I have 2 background images.
Ineed to add the first image if the value of variable test is 1 otherwise i need to add the second background image.

I have written the following code:

<?PHP
			   if ($test ==1)
			   echo "<td width='50%' valign='top'  STYLE='background-image: url('img/table_faoback.jpg');'>";
			   else
			   echo "<td width='50%' valign='top'  STYLE='background-image: url('img/table_back.jpg');'>";
			  
			   ?>

However it does not display any image. Please let me know how to do this.

Thanks.

Check your source code: is there <td width='... printed out?

If so: this is not a php problem: it is echoing through the conditional statement and your problem lies in the html!

regards

[nomad]

Hi,

I have 2 background images.
Ineed to add the first image if the value of variable test is 1 otherwise i need to add the second background image.

I have written the following code:

<?PHP
			 if ($test ==1)
			 echo "<td width='50%' valign='top' STYLE='background-image: url('img/table_faoback.jpg');'>";
			 else
			 echo "<td width='50%' valign='top' STYLE='background-image: url('img/table_back.jpg');'>";
 
			 ?>

However it does not display any image. Please let me know how to do this.

Thanks.

You might try this...
background-image: url(img/table_back.jpg) ;

Also you need to look at your html coding as well as markusn00b suggested. Something is not right with the ;

nomad

[Markus]

You might try this...
background-image: url(img/table_back.jpg) ;

Also you need to look at your html coding as well as markusn00b suggested. Something is not right with the ;

nomad

It may be the single quotes within single quotes that are causing the problem.

[ronverdonk]

You should at least escape the single quotes within the single quotes in this statement[php]STYLE='backgrou nd-image: url('img/table_faoback.j pg');'>";[/php]Ronald

[ronverdonk]

Markus you beat me again!

Ronald

[Markus]

Markus you beat me again!

Ronald

That's just how i get down

8]

You've been slacking ron!
Where you been?

[ronverdonk]

That's just how i get down

8]

You've been slacking ron!
Where you been?

Very busy with making some websites.

Ronald