Default displaying problem

[zang8027]

So i have a page with 3 drop down menus. Each menu's values are recieved from a database. When you select the first value, it reloads the page and displays the proper values in the dropdown below it. Then that happens again. The problem I am having is that because my drop downs are not getting values until the one above it is selected, I am getting:

 

Warning: mysql_fetch_array(): supplied argument is not a valid MYSQL result.

 

 

Is there like a if(null) statement I can use to set the default value to something like "Please Select a  State"?

 

 

<FORM action="findrestaurants.php" method="POST">
<SELECT NAME="stateList" onChange="this.form.submit()">
<OPTION>Please Select a State
<?php
//Make a connection to the server... connect("servername","username","password")
$link=mysql_connect("localhost","root") or die("No server connection".mysql_error());
//connect to our database
$db=mysql_select_db("dineonline_db") or die("No Database Connection ".mysql_error());
//construct a SQL query that shows all hotels in the city
$query="SELECT * FROM states";
//run the query
$result=mysql_query($query);	

//for each row returned by our query, do what is in curly braces
while($row = mysql_fetch_array($result)){

//store into a variable the ID # for the current row 
$stateID=$row['state_ID'];
//store into a variable the Name for the current row
$stateName=$row['state_name'];
  
//Print out each "<OPTION> with the proper values
print "<OPTION VALUE='$stateID'>$stateName";
}
//close the connection
mysql_close($link);

?>
</SELECT>
</FORM>

<!-- Below is the cityBig List drop down code -->

<FORM action="findrestaurants.php" method="POST">
<SELECT NAME="cityLargeList" onChange="this.form.submit()">
<OPTION>Please Select an Area
<?php
//Get the value from City Large
$userState = $_POST['stateList'];
//Make a connection to the server... connect("servername","username","password")
$link=mysql_connect("localhost","root") or die("No server connection".mysql_error());
//connect to our database
$db=mysql_select_db("dineonline_db") or die("No Database Connection ".mysql_error());
//construct a SQL query that shows all hotels in the city
$query="SELECT * FROM cityLarge WHERE state_ID = $userState";
//run the query
$result=mysql_query($query);	

//for each row returned by our query, do what is in curly braces
while($row = mysql_fetch_array($result)){

//store into a variable the ID # for the current row 
$cityLargeID=$row['cityLarge_ID'];
//store into a variable the Name for the current row
$cityLargeName=$row['cityLarge_name'];
  
//Print out each "<OPTION> with the proper values
print "<OPTION VALUE='$cityLargeID'>$cityLargeName";

}
//close the connection
mysql_close($link);

?>
</SELECT>
</FORM>

<!-- Below is the citySmall List drop down code -->
<FORM action="displayRest.php" method="get">
<SELECT NAME="citySmallList">
<OPTION>Please Select a City
<?php
//Get the value from City Large
$userCityLarge = $_POST['cityLargeList'];
//Make a connection to the server... connect("servername","username","password")
$link=mysql_connect("localhost","root") or die("No server connection".mysql_error());
//connect to our database
$db=mysql_select_db("dineonline_db") or die("No Database Connection ".mysql_error());
//construct a SQL query that shows all hotels in the city
$query="SELECT * FROM citySmall WHERE cityLarge_ID = $userCityLarge";
//run the query
$result=mysql_query($query);	

//for each row returned by our query, do what is in curly braces
while($row = mysql_fetch_array($result)){

//store into a variable the ID # for the current row 
$citySmallID=$row['citySmall_ID'];
//store into a variable the Name for the current row
$citySmallName=$row['citySmall_name'];
  
//Print out each "<OPTION> with the proper values
print "<OPTION VALUE='$citySmallID'>$citySmallName";

}
//close the connection
mysql_close($link);

?>
</SELECT>
<br/>
<br/>
<input type="submit" value="Find restaurants" />
</FORM>

[GingerRobot]

The correct approach would be to check to see if those values actually exist in the $_POST array (using isset) and only execute the query if they do.

[zang8027]

didn't even think of that. Would i do like..

 

if(isset($_POST['myvar']))

{

PHP

}

 

[zang8027]

Now i added this:

 

<?php
//Get the value from City Large

$userState = $_POST['stateList'];
if(isset($userState))
	{
//Make a connection to the server... connect("servername","username","password")
$link=mysql_connect("localhost","root") or die("No server connection".mysql_error());
//connect to our database
$db=mysql_select_db("dineonline_db") or die("No Database Connection ".mysql_error());
//construct a SQL query that shows all hotels in the city
$query="SELECT * FROM cityLarge WHERE state_ID = $userState";
//run the query
$result=mysql_query($query);	

//for each row returned by our query, do what is in curly braces
while($row = mysql_fetch_array($result)){

//store into a variable the ID # for the current row 
$cityLargeID=$row['cityLarge_ID'];
//store into a variable the Name for the current row
$cityLargeName=$row['cityLarge_name'];
  
//Print out each "<OPTION> with the proper values
print "<OPTION VALUE='$cityLargeID'>$cityLargeName";
	}else{

	}

}
//close the connection
mysql_close($link);

?>

 

And i get

 

Unexpected T_ELSE on line 76.

 

If i delete the else part, i get the drop down boxes again but i get:

mysql_close():2 is not a valid MYSQL-Link

 

[GingerRobot]

You're while statement is closed by the brace on this line:

 

}else{

 

Evidently you intended to close the while loop first. If you indented your code properly, it'd make things like that a whole stack easier to spot.