MySQL question

Discussion in 'OT Technology' started by Tucker McElroy, Jan 29, 2005.

  1. Tucker McElroy

    Tucker McElroy New Member

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    In PHP, I have a filename being passed in from the database, into the variable $filename

    I want to display an image based on this filename

    This is what I have right now:

    echo "<td><img src="portfolio<? echo $filename; ?>" width="70" height="70"></td>"

    so if $filename = "/image01.jpg", it would then be parsed to:

    echo "<td><img src="portfolio/image01.jpg" width="70" height="70"></td>"

    but this is giving me a parse error

    I'm new to this, wtf am i doing wrong? :hs: Thanks!
     
  2. P07r0457

    P07r0457 New Member

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    here's the prob:

    echo "<td><img src='portfolio".$filename."' width='70' height='70'></td>";
     
  3. Tucker McElroy

    Tucker McElroy New Member

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    fuckin' beautiful :bowdown:

    thanks for your time
     

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