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#1
June 1st, 2001, 03:33 AM
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php amateur...
Guys, help us out here please...
----------------------------------- the script... ----------------------------------- <html> <body> <?php require("mydb.inc"); mysql_connect("***.**.***.**",$user,$password); $database="ellel"; @mysql_select_db("$database") or die( "Unable to select database"); $sql = "SELECT company_name,link_ref,region FROM advertisers WHERE region='"; //switch loop switch ($select){ //scotland search case 1: $sql .= "Scotland'"; break; //north england search case 2: $sql .= "North England'"; break; //central england search case 3: $sql .= "Central England'"; break; //wales search case 4: $sql .= "Wales'"; break; //south east england search case 5: $sql .= "South East England'"; break; //south west england search case 6: $sql .= "South West England'"; break; //channel islands search case 7: $sql .= "Channel Islands'"; break; //northern ireland search case 8: $sql .= "North Ireland'"; break; //ireland search case 9: $sql .= "Ireland'"; break; //overseas search case 10: $sql .= "Overseas'"; break; //all search //case 11 //$sql .= "%'"; //break; } $query=mysql_query($sql); $rslt=mysql_fetch_array($query); //display results echo "<table>\n<tr>\n"; echo "<td>Company Name</td>\n"; echo "<td>URL</td>\n"; echo "<td>Region</td>\n"; echo "</tr>\n"; while ($row = mysql_fetch_array($query)) { // or list($all, $the, $retrieved, $columns) echo "<tr>\n"; echo "<td>$row[company_name]</td>\n"; echo "<td>$row[link_ref]</td>\n"; echo "<td>$row[region]</td>\n"; echo "</tr>\n"; } echo "</table>\n"; ?> </body> </html> ----------------------------------- returns... ----------------------------------- Warning: Supplied argument is not a valid MySQL result resource in /home/christianh/public_html/out3.php on line 76 Warning: Supplied argument is not a valid MySQL result resource in /home/christianh/public_html/out3.php on line 86 Company Name URL Region ----------------------------------- HELP?? i cant pick anything wrong with this current syntax... much appreciated JD
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#2
June 1st, 2001, 04:45 PM
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Keep in mind I am fairly new to php as well. But I will give it a shot.
The first thing is it is not your php but your MYSQL queries which are failing. $sql = "SELECT company_name,link_ref,region FROM advertisers WHERE region='"; There are no () around your query plus in the section "WHERE region = '"; it looks like you only have one ' instead of ' ' $sql = ("SELECT company_name,link_ref,region FROM advertisers WHERE region=''"  ;I hope that helps E.
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#3
June 1st, 2001, 05:14 PM
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I'm new to php also, but i think this will work:
echo "<td>" . $row["company_name"] . "</td>\n"; echo "<td>" . $row["link_ref"] . "</td>\n"; echo "<td>" . $row["region"] . "</td>\n"; I hope this helps. Last edited by undodigital : June 2nd, 2001 at 01:23 PM.
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#4
June 10th, 2001, 03:38 PM
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theeggman should be right - almost...
don't need brackets around your query, but supplied argument is missing (or misspeled), thats the key. try: //switch loop switch ($select){ //scotland search case 1: $selection="Scotland"; break; and so on... $sql = "SELECT company_name,link_ref,region FROM advertisers WHERE region='$selection'"; should be easier with the variable rather than adding something to the querystring bombel problem solved already? just post a "worked" or something so no one bothers with your questions anymore Last edited by bombel : June 10th, 2001 at 03:43 PM.
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