Em Quinta 31 Maio 2007 16:25, Jason Pruim escreveu: > Hi Everyone, I am attempting to setup a search field on a database > application I'm dinking around with and running into problems that > I'm hoping someone might be able to shed some light on. > > Here is the code I am using to display the results of the search: > > echo ('<table border="1">'); > echo "<tr><th>First Name</th><th>Author</th><th>Pages</th></tr>"; > > $result_row[] = mysql_query($query) or die(mysql_error()); > echo "Result_row dump: $result_row <BR>"; > $num=mysql_numrows($result); > echo "Num dump: $num <BR>"; > $i= 0; > /* > while($i < $num) { */ $result=$result_row[0]; while($result_row = mysql_fetch_array($result) { > echo "<tr><td>"; > echo $result_row[0] . '</td><td>'; > echo $result_row[1] . '</td><td>'; > echo $result_row[2] . '</td></tr>'; > $i++; > } > > echo ("</table>"); > > > the problem is instead of printing out the actual results it prints > out 6 fields that say: "Resource id #5" > > the query I'm using is: SELECT FName, LName, Add1, Add2 FROM current > WHERE FName like '%jason%' which works in MySQL > > I can't find any errors in my log files to even give me a hint as to > what is going on. > > If someone could take a look I would greatly appreciate it. > HTH -- Davi Vidal davividal@xxxxxxxxxxxxxxxx davividal@xxxxxxxxx -- "Religion, ideology, resources, land, spite, love or "just because"... No matter how pathetic the reason, it's enough to start a war. " -------------------------------------------------------- Por favor não faça top-posting, coloque a sua resposta abaixo desta linha. Please don't do top-posting, put your reply below the following line. --------------------------------------------------------
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