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  1. #1
    New to the CF scene
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    Jan 2013
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    Unhappy Problem with displaying images using variable in php

    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

  • #2
    Senior Coder
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    Feb 2011
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    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.
    I can't really think of anything to write here now...


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