[Resolved] Mysql_fetch ARRAY? Problem

[cmpennington]

I have a script that passes a drop-down selection to a results page which grabs the information for the town selected in the drop down menu. I have an error in my fetch array though?

CAN ANYBODY HELP A BROTHER OUT?

Anywhoo. here is the first page:

$result = mysql_query("SELECT * FROM funeralhomes ORDER BY city ASC",$db);

while ($myrow = mysql_fetch_assoc($result))
{
$name = $myrow["name"];
$id = $myrow["id"];
$city = $myrow["city"];
$address = $myrow["address"];
$state = $myrow["state"];
$zip = $myrow["zip"];
$phone = $myrow["phone"]; 
$fax = $myrow["fax"];
$email = $myrow["email"];
echo "<option value=$city >$city&nbsp;,$state&nbsp;&nbsp;($name)";
}


echo "</select>";
echo "<p>";
echo "<center>";
?>

<input type="submit" name="view" value="View Info" />
	</form> 

And here is the proccessing part:

//// display individual record

if ($id) {

   $result = mysql_query("SELECT * FROM funeralhomes WHERE city=$city",$db);

   $myrow = mysql_fetch_array($result);
$name = $myrow["name"];	
$address = $myrow["address"];
$city = $myrow["city"];
$state = $myrow["state"];
$zip = $myrow["zip"];
$phone = $myrow["phone"];
$fax = $myrow["fax"];
$email = $myrow["email"];

?>
<table width="35%" border="0" cellspacing="0" cellpadding="0" align=center>
  <tr> 
    <td width="25%">&nbsp;</td>
    <td width="75%"><font size="4" color="blue"><b><font color="blue"><?php echo $name; ?></font></b></font> 
    </td>
  </tr>
  <tr> 
    <td width="25%"><b><font size="3"></font></b></td>
    <td width="75%"><b><font size="3"><?php echo $address; ?></font></b></td>
  </tr>
  <tr> 
    <td width="25%"><b><font size="3"></font></b></td>
    <td width="75%"><b><font size="3"><?php echo "$city,&nbsp;$state&nbsp;&nbsp;$zip"; ?></font></b></td>
  </tr>
  <tr> 
    <td width="25%"><b><font size="3"></font></b></td>
    <td width="75%"><b><font size="3"></font></b></td>
  </tr>
  <tr> 
    <td width="25%"><b><font size="3"></font></b></td>
    <td width="75%"><b><font size="3"><?php echo $phone; ?></font></b></td>
  </tr>
  <tr> 
    <td width="25%"><b><font size="3"></font></b></td>
    <td width="75%"><b><font size="3"><?php echo $fax; ?></font></b></td>
  </tr>
  <tr> 
    <td width="25%">&nbsp;</td>
    <td width="75%">&nbsp;</td>
  </tr>
  <tr> 
    <td width="25%">
      <div align="right"><a href=\"mailto:$email\"><img src=images/icon_email.gif border=0></a>&nbsp;</div>
    </td>
    <td width="75%"><font color="green"><b><font size="3"><?php echo $email; ?> 
      </font></b></font></td>
  </tr>
  <tr> 
    <td width="25%">&nbsp;</td>
    <td width="75%">&nbsp;</td>
  </tr>
  <tr> 
    <td width="25%" height="25">&nbsp;</td>
    <td width="75%" height="25"><b><font color="blue"><a href=newspapers_view.php>Click Here To Go Back</a></font></b></td>
  </tr>
</table>
<?php




} else {

// show employee list

   $result = mysql_query("SELECT * FROM funeralhomes ORDER BY id DESC",$db);

// begin to show results set
$count = 1 + $s ;


// now you can display the results returned
  while ($row= mysql_fetch_array($result)) {
$title = $row["firm"];

// ****************TEST EDITS***************
$name = $row["name"];	
$address = $row["address"];
$city = $row["city"];
$state = $row["state"];
$zip = $row["zip"];
$phone = $row["phone"];
$fax = $row["fax"];
$email = $row["email"];

echo "<tr><td><i><font size=2>Result $count:</font></i></td></tr>
<tr><td><b><font color=blue size=3>$title</font></b></td></tr>
<tr><td><font size=2>$address</font></td></tr>
<tr><td><font size=2>$city,&nbsp;$state&nbsp;&nbsp;$zip</font></td></tr>
<tr><td><font size=2>$phone</font></td></tr>
<tr><td><font size=2>$fax (f)</font></td></tr>
<tr><td><font size=3 color=green><b>$contact</b> &nbsp;<a href=\"mailto:$email\"><img src=images/icon_email.gif border=0></a></font></td></tr><tr><td>&nbsp;</font></td></tr>
<p>";
  $count++ ;

}    

}


?>
</table>
</html> 

I am getting the error on line 18 saying that

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/scfdaorg/wwwroot/scfda.org/member/funeralhome_view_results.php on line 18

Thanks for trying to help this problem.

CMPENNIGNTON

[bad76]

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/scfdaorg/wwwroot/scfda.org/member/funeralhome_view_results.php on line 18

When you get this kind of error the problem is in the query string.
Try to debug with this:

$query="SELECT ...";
$result=mysql_query($query) or die(mysql_error()." - ".$query);

I hope this helps...

[barand]

Try it with single quotes around $city

$result = mysql_query("SELECT * FROM funeralhomes WHERE city= '$city' ",$db); 

[cmpennington]

I get the error now: Query is empty.

I'm not sure if I'm passing the right variables?

When I changed the method to GET just to see what variables where going through i found this:

funeralhome_view_results.php?city=Abbeville&view=View+Info

So I think that the city variable is being passed? I'm just not sure why I can't call it up in the query?

Angain thanks for considering the help.

cmpennington

[barand]

Have you picked up the POSTed value?

$city = $_POST['city'];

[cmpennington]

It picks up the variable. but I need to find a way to loop through and get each database entry from that town. Currently it only shows the first entry from the town?

Would this be a do While? or no.
I'm not sure but you are close to solving the problem.

Thanks Barand,
cmpennington

[barand]

<table width="35%" border="0" cellspacing="0" cellpadding="0" align=center>

while ($myrow = mysql_fetch_array($result) )
{
$name = $myrow["name"];

$address = $myrow["address"];
$city = $myrow["city"];
$state = $myrow["state"];
$zip = $myrow["zip"];
$phone = $myrow["phone"];
$fax = $myrow["fax"];
$email = $myrow["email"];

table rows code here

}

echo "</table>";

[cmpennington]

Once again barand saves the day. You've gained a new student. I appreciate the help.

cmpennington