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I'm trying to get a list of values to display upon a selection from a dropdown menu, but I think my SQL is not quite correct. This is a modification of a W3Schools tutorial http://www.w3schools.com/php/php_ajax_database.asp

I get the error: "Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/virtual/hafdal.dk/public_html/test/getuser.php on line 28" when I try to do a selection.

I can't seem to find out what the error is, this function should be able to return multiple rows as I want it to. It might be because it's almost 2 am or just cause I'm dense ;-)

See an example here: hafdal.dk/public_html/test/getuser.php

Here is my code:

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" 

"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<?php header('Content-Type:text/html; charset=UTF-8');
$q=$_GET["q"];

require_once 'login.php';
$db_server = mysql_connect($db_hostname, $db_username, $db_password);
if (!$db_server) die("Unable to connect to MySQL: " . mysql_error());
mysql_select_db($db_database, $db_server) or die("Unable to select database: " . mysql_error());

$sql="SELECT * FROM view_bæjartal WHERE hrepparid = '".$q."'";

$result = mysql_query($sql);

echo "<table border='1'>
<tr>
<th>Bær/Hús/Þurrabúð</th>
<th>Sýsla</th>
<th>Lat</th>
<th>Long</th>
</tr>";

while($row = mysql_fetch_array($result))
  {
  echo "<tr>";
  echo utf8_encode("<td>" . $row['bæir'] . "</td>");
  echo utf8_encode("<td>" . $row['slysla'] . "</td>");
  echo utf8_encode("<td>" . $row['hreppur'] . "</td>");
  echo utf8_encode("<td>" . $row['lat'] . "</td>");
  echo utf8_encode("<td>" . $row['long'] . "</td>");
  echo "</tr>";
  }
echo "</table>";

mysql_close($db_server);
?>

<body>
</body>
</html>

I hope someone can help :-)

share|improve this question
1  
are you sure that the collation of your db and table is utf-8? because as I see you're using unicode symbols in the name of table. anyway it's a good idea to print out the error if there any. $result = mysql_query($sql) or die(mysql_error()); – haynar Dec 11 '11 at 0:55
1  
Your first error is using w3schools. – PeeHaa Dec 11 '11 at 0:56
1  
PS: escape the input before doing a query. $sql="SELECT * FROM view_bæjartal WHERE hrepparid = '".mysql_real_escape_string($q)."'"; – PeeHaa Dec 11 '11 at 0:57

2 Answers

up vote 1 down vote accepted

The reason might that that w3schools tutorial, as usual, forgot the proper escaping function. This might lead to a syntax error, and due to the absence of any error checking code a failure with the loop.

Right after your mysql_query() add a mysql_error() call:

$result = mysql_query($sql)  or  print(mysql_error());

It could also just be your Unicode table name. (Then add backticks.)

share|improve this answer
Thanks for helping! I went back to the original query and didn't get the same error again. So I'm thinking it is an sql query error. And now I'm not sure if I should post a new question or continue this tread...? – user1088537 Dec 11 '11 at 14:14
Look again what PeeHaa said. Might explain your issue. – mario Dec 11 '11 at 14:32
It didn't help :-( – user1088537 Dec 12 '11 at 17:58
up vote 1 down vote

You can display errors by calling echo mysql_error($db_server); right after your queried the database (with mysql_query($sql)).

Maybe there is a problem with your table/view name, but to know this for sure you need to have a look at the actual error message returned by mysql.

share|improve this answer
I do believe you are right with regards to there being a problem with my view. – user1088537 Dec 11 '11 at 14:16

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