porta325 Posted August 31, 2007 Share Posted August 31, 2007 $judet = $_GET["judet"]; $result = mysql_query ("SELECT oras, id_oras FROM orase WHERE id_oras = '{$judet}' ORDER BY oras"); This is a piece of my code. The variable $judet is working perfect but i have a problem with another variable in the same table with judet witch i can't call.Everything from table orase works i just don't know how to simply display on my page another insert from a different table. Quote Link to comment https://forums.phpfreaks.com/topic/67445-solved-variable-problem/ Share on other sites More sharing options...
soycharliente Posted August 31, 2007 Share Posted August 31, 2007 Why are you using {} around your variable? You don't need to. Have you tried taking them out? Quote Link to comment https://forums.phpfreaks.com/topic/67445-solved-variable-problem/#findComment-338575 Share on other sites More sharing options...
pocobueno1388 Posted August 31, 2007 Share Posted August 31, 2007 Your going to have to give a lot more detail than that for anyone to be able to help you. i just don't know how to simply display on my page another insert from a different table. By that do you mean you are wanting to select a row from a different table in the same query? If thats the case, the two tables have to relate in some way...give more information on the table your trying to get the row from. Quote Link to comment https://forums.phpfreaks.com/topic/67445-solved-variable-problem/#findComment-338578 Share on other sites More sharing options...
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