Problem with displaying images using variable in php

[chriswilby]

I am using this syntax in php to display a picture.

$pict="photoids/1368.jpg";

$picture='<img src="$pict" >';

then
echo '<td>' . $picture . '</td>';

but it does not work yet if I use

$pict="photoids/1368.jpg";

$picture='<img src="photoids/1368.jpg" >';

then
echo '<td>' . $picture . '</td>';

the picture is displayed.
But I want to use different values of the image number to display different pictures in the table.

Can you help me sort out the above code used in php.
thankyon

[tangoforce]

It's our use of quotes and variables inside them.

To put it simply, you can't use a $Variable inside 'single quotes' and expect PHP to replace it. You can only use a $Variable inside "double quotes". If you need to use " inside double quotes then you need to escape it like this:

$picture = "<img src=\"$pict\" >";

The \ tells PHP that the character following should be ignored and treated the same as the rest.

See the link in my signatures about quotes for a more indepth explanation.