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sjonni

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  1. Thanx :) I got it to work after adding the error code .. newbie mistake, misspelled the table name! ::)
  2. thank you for your replys, but neither of this seems to work :( Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /var/www/netmynd.is/left.php on line 35 maybe there is a better solution to do this than this one?
  3. Hi I am totally new to php and MySQL. Let my try and explain my problem. I know this is REALLY simple, i´m just not there yet ;) I am working on a site with multiple pages. I want the same content to be on the left and right side of the pages, so I figured the easiest way was to create a MySQL table to handle the contents. So it would look something like this |                |                                |                | | same for all |        main content      | same for all | |    pages    |                                |    pages    | The table consists of 2 fields, ID and content. the problem is, how do I make a command to show a content from a desired ID. This is the code in which I thought would do it, but no :( <?php $sql = "SELECT content from sidebar WHERE id = 1"; $result = mysql_query($sql); echo $result;?> can anyone help me with this?
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