MySQL Result Resource Error

[egenart]

Hi Wizards,
I'm working to make a little application where visitors can search for a therapist in our country (Sweden). The options are "part of country" and "female or male therapist".

Everything has gone well until I want the visitor to be able to click a submit button that is there after each therapist name in an array (a little form in every row, with the <input name=\"$terapeutid\">), to send a variable to the presentation page of the therapist.
The presentation page is supposed to make a mysql_query "WHERE terapeutid = $teraputid" and line up all the fields of information for this one therapist (including an image) in a table.

I'm stuck here. I get a Warning that this is not a legal MySQL result resource.
Can anyone see what I have done wrong (or if anything is right ;-)?
Warmly, Lars

Here is the code for the file "teraput.php":

<TABLE CELLSPACING=5 width="500" BORDER="0">
<?
// Create database connection
$link = @mysql_connect ("host", "user", "password")
or die ("Anslutningen funkade inte! Ledsen.");
$databas = mysql_select_db("a0088600_db1");

// Define what'll be selected
$sql = "SELECT * FROM teraputer WHERE terapeutid = $terapeutid";
$result = mysql_query("$sql");
while($rad = mysql_fetch_array($result)) {

echo "<TR>";

echo "<TD VALIGN=\"top\"> &nbsp;</TD>";

echo "<TD VALIGN=\"top\" ALIGN=\"left\">";

echo "<H2><B>";

echo $rad[fornamn]." ".$rad[efternamn];

echo "</B></H2><BR>";

echo "</TD></TR>";

echo "<TR><TD WIDTH=\"150\" VALIGN=\"top\">";

echo "<IMG SRC=\"../images/";

echo $rad[bild];

echo "\" WIDTH=\"150\">";

echo "</TD><TD WIDTH=\"350\" VALIGN=\"top\">";

echo "<TABLE WIDTH=\"350\" BORDER=\"0\"  CELLSPACING=\"5\" >";

echo "<TR><TD VALIGN=\"top\"><B>Inriktning:</B></TD>";

echo "<TD VALIGN=\"top\">";

echo $rad[inriktning];

echo "</TD>";

echo "</TR><TR>";

echo "<TD VALIGN=\"top\"><B>Bakgrund:</B></TD>";

echo "<TD VALIGN=\"top\">";

echo $rad[bakgrund];

echo "</TD>";

echo "</TR><TR>";

echo "<TD VALIGN=\"top\"><B>Adress:</B></TD>";

echo "<TD VALIGN=\"top\">";

echo $rad[adress].", ".$rad[ort];

echo "<BR>";

echo "Tel: ";

echo $rad[telefon];

echo ", ";

echo $rad[mobil];

echo ". Fax: ";

echo $rad[fax];

echo "<BR>";

echo "<I>e-post ";

echo $rad[email];

echo "</I><BR>";

echo "Hemsida: ";

echo $rad[web];

echo "</TD></TR></TABLE>";

echo "</TD></TR>";

} 
mysql_close($link);
?>

</TABLE>

[kbrij]

You will get such warning message when you are fetching array result from the query result without 1st checking how many rows have been selected in your query.

what if the value of $terapeutid is empty or not existing in the table.

so check 1st

if ($result && mysql_num_rows($result) > 0) {
while($rad = mysql_fetch_array($result)) {
....
}
}

// Define what'll be selected
$sql = "SELECT * FROM teraputer WHERE terapeutid = $terapeutid";
$result = mysql_query("$sql");
while($rad = mysql_fetch_array($result)) {

---

hope that is helpful to you.