problem with passing value through a pull down menu in php

[Anonymous]

Hello Every one!!

I am having problem with passing value through a pull down menu and then using that value in a PHP while to execute a query using mysql. So it would be great if you take a look of my code and give me some suggestion. Thank you. :?

Here is my code of HTML pull down menu:

<?php 
require(""); 
$link = mysql_connect($host,$user,$pass); 
if (!$link) 
die("Couldn't connect to MySQL"); 

mysql_select_db($db) 
or die("Couldn't open $db: ".mysql_error()); 


$result = mysql_query("SELECT Name FROM Domain"); 
$num_rows = mysql_num_rows($result); 
if ($num_rows) 
{ 
print "There are currently $num_rows Domain in the Database <p>"; 
print "<table border = 1>\n"; 
while($a_row = mysql_fetch_row($result)) 
{ 
print "<tr>\n"; 
foreach ($a_row as $field) 
print "\t<td>$field</td>\n"; 
$Domain_Name[] = $field; 
print "</tr>\n"; 
} 
print "</table>\n"; 
?> 


<form action = "ShowDomainDescription1.php" Method = "POST"> 

Select any of the domain for more information<br> 

<select name = "DomainMenu"> 
<?php 
foreach ($Domain_Name as $Existing_Domain) 
print " <option value=\"$Existing_Domain \">$Existing_Domain</option>"; 
?> 
</select> 
<input type = "submit" value = "Show" > 

</form> 

<?php } 
else 
print "There is no data available in the database"; 
mysql_close($link); ?>

I am getting following worning
Warning: Supplied argument is not a valid MySQL result resource in /home/public_html/ShowDomainDescription1.php on line 12
There is no Domain called , Please check the Domain list again.


Here is my code for ShowDomainDescription1.php

<? 
require(""); 
$link = mysql_connect($host,$user,$pass); 
if (!$link) 
die("Couldn't connect to MySQL"); 

mysql_select_db($db) 
or die("Couldn't open $db: ".mysql_error()); 

if($_POST['DomainMenu'] == $Existing_Domain) 
$result = mysql_query("SELECT Description FROM Domain where Name = '$Existing_Domain'"); 
$num_rows = mysql_num_rows($result); 
if ($num_rows > 0) 
{ 
$Domain_Description = mysql_result($result, 0, 'Description'); 
print "<h1> $Existing_Domain Description </h1>\n"; 
print "$Domain_Description"; 
} 
else 
print "There is no Domain called $Existing_Domain, Please check the Domain list again."; 
mysql_close($link); 

?>

 

[Anonymous]

Hi!

In your code

if($_POST['DomainMenu'] == $Existing_Domain) 
$result = mysql_query("SELECT Description FROM Domain where Name = '$Existing_Domain'"); 
$num_rows = mysql_num_rows($result);

If the condition $_POST['DomainMenu'] == $Existing_Domain is true, then you will store in $result a resource identifier for the results of the SELECT request...
But if it is false, then $result will not be set up and mysql_num_rows($result) will return the error you got
As a conclusion, look after the missing {} after your condition.

/Flood