试图计算数据库中的条目

Question

I'm trying to count entries in a database based on 2 basic criteria. It is returning a blank result, even though there are results to be found. Anyone have any idea what I am doing wrong here? I have tried it so many different ways and they all return no result. (If I enter the query directly in phpmyadmin it returns a result.)

$sql = "SELECT count(*) as total_count from orderOption3Detail WHERE orderDate='$orderDate' AND studentID='$studentID'"; 
$numericalResult = mysql_query($sql, $con);
$row = mysql_fetch_object($numericalResult);
$totalOrders1 =  $row->total_count;
echo "My orders:" . $totalOrders1; 

Answer 1

As others stated, make sure you sanitize variables before they go into query.

$sql = "SELECT * FROM orderOption3Detail WHERE orderDate = '" . $orderDate . "' AND studentID = '" . $studentID . "'";

$sql_request_data = mysql_query($sql) or die(mysql_error());
$sql_request_data_count = mysql_num_rows($sql_request_data);

echo "Number of rows found: " . $sql_request_data_count;

That's all you need.

Answer 2

Edited: providing full code corrected:

$con=mysqli_connect($db_host,$db_user,$db_pass,$db_name); // Check connection 
if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } //global option 1 
$sql = "SELECT count(*) as total_count from orderOption3Detail WHERE orderDate='$orderDate' AND studentID='$studentID'"; 
//echo $sql; 
$numericalResult = $con->query($sql); 
$row = mysqli_fetch_object($numericalResult); 
echo $row->total_count; //echo (int) $row->total_count;

Please test this and let me know. Good luck! ----- End Editing ----

Have you tested assigning values directly as a test in your SQL string, like:

$sql = "SELECT count(*) as total_count from orderOption3Detail WHERE orderDate='05/23/2012' AND studentID='17'";

Also, did you check if the date's format is correct, reading that $orderdate variable and testing it in PHPMyAdmin?

Did you read the $sql with values inserted and test in PHPMyAdmin and worked?

Also, check the connection to assure there is no problem there.

One more thing, sorry. You seem to be using the wrong syntax in your mysql_query statement. That way works for mysqli_query, and the parameters would be inverted. Try only:

$numericalResult = mysql_query($sql);

Provided you made the connection and database selection previously, like in:

$connection=mysql_connect($db_host, $db_username, $db_password);
if (!$connection)
          {
            $result=FALSE;
            die('Error connecting to database: ' . mysql_error());
          }

     // Selects database
     mysql_select_db($db_database, $connection);

Best wishes,