Peter,
after querying the database, you need to fetch the returned values from
the result before you can use them:
> $result = mysql_query("SELECT company FROM resellers WHERE Company LIKE
> '%$query%'");
$row = mysql_fetch_row($result);
$company = $row[0];
OR (shorter):
list($company)=mysql_fetch_row($result);
> printf("<tr><td> $company </td></tr>\n");
otherwise, $company would not be set.
Ben
Peter Houchin wrote:
>
> Hiya, i was wondering if some one could help me with this ... i have a
> little form where you can enter a value say company then on submit its ment
> to check the db for value or one like it my sql line works fine
> (mysql_error() shows nothing and it works from command line) ..
>
> .
> .
> .
> <form name=search method=POST action="<? $PHP_SELF ?>">
> <input type="text" name="query">
> <input type="submit" name="submit" value="submit">
> </form>
> <? if ($submit){
> echo "<TABLE BORDER=\"0\" CELLPADDING=\"3\" CELLSPACING=\"1\"
> BGCOLOR=\"#CCCCCC\" ALIGN=\"CENTER\">\n<tr BGCOLOR=\"#CCCCCC\">\n<td
> bgcolor=\"#006699\"><div align=\"center\"><font face=\"Helvetica,
> sans-serif\" size=\"3\" color=\"#FFFFFF\"><b>Company</b> </td></tr>";
> $result = mysql_query("SELECT company FROM resellers WHERE Company LIKE
> '%$query%'");
> printf("<tr><td> $company </td></tr>\n");
> echo "</table>";
> echo mysql_error();
> }
> ?>
>
> Thanks in advance
>
> Peter
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