Undefined variable and parameters

[ITNerd1002]

I am having trouble with my php code and end up getting the following two problems. Below are the two problems I keep getting which are marked with asterisks.

 

**Warning: mysqli_query() expects parameter 1 to be mysqli, null given in

 

**Notice: Undefined variable: con in

 

This is my php code:

 

<?php
/**
 * WARNING!
 * This EXAMPLE code contains severe SECURITY ISSUES
 * which should be addressed before real-life usage!!
 *
 * These issues have been ignored intentionally to
 * simplify the code so we can focus on the topic
 * in discussion.
 */
 
// Lets test if the form has been submitted
if(isset($_POST['SubmitCheck'])) {
    // The form has been submited
    // Check the values!
  //  if($_POST['Username'] == "John" and $_POST['Password'] == "Doe") {
        // User validated!
        
//Variable declaration

        //$username = $_POST["Username"];
        //$first = $_POST["First"];
        
        $dev_id = $_POST["dev_id"];
        $dev_name = $_POST["dev_name"];
        //$term_id = $_POST["term_id"];
        $cust_id = $_POST["cust_id"];
        $cust_fname = $_POST["cust_fname"];
        $cust_lname = $_POST["cust_lname"];
        //$date_of_loan = $_POST["date_of_loan"];
        //$date_returned = $_POST["date_returned"];
        
        
        

        
        
        $result = mysqli_query($con, "SELECT '".$dev_id."', ta.TERM_ID, '".$cust_id."', '".$dev_name."', '".$cust_fname."', '".$cust_lname."', DATE_OF_LOAN, DATE_RETURNED
                from device d
                join abc_audit au on d.DEV_ID = '".$dev_id."'
                join term_agreement ta on au.TERM_ID = ta.TERM_ID
                join customer cu on ta.CUST_ID = '".$cust_id."'
                where au.abc_status = 'O'");

              echo "<table border='1'>
                <tr>
                    <th>DEV_ID</th>
                    <th>DEV_NAME</th>
                    <th>TERM_ID</th>
                    <th>CUST_ID</th>
                    <th>CUST_FNAME</th>
                    <th>CUST_LNAME</th>
                    <th>DATE_OF_LOAN</th>
                    <th>DATE_RETURNED</th>
                </tr>";

                while($row = mysqli_fetch_array($result))
                    {
                        echo "<tr>";
                            echo "<td>" . $row['DEV_ID'] . "</td>";
                            echo "<td>" . $row['DEV_NAME'] . "</td>";
                            echo "<td>" . $row['TERM_ID'] . "</td>";
                            echo "<td>" . $row['CUST_ID'] . "</td>";
                            echo "<td>" . $row['CUST_FNAME'] . "</td>";
                            echo "<td>" . $row['CUST_LNAME'] . "</td>";
                            echo "<td>" . $row['DATE_OF_LOAN'] . "</td>";
                            echo "<td>" . $row['DATE_RETURNED'] . "</td>";
                        echo "</tr>";
                    }
                        

                mysqli_close($con);
    
        
        //$con=mysqli_connect("localhost", "root", "", "abc");
         //mysqli_query($con,"INSERT INTO Customer (CUST_ID, CUST_FNAME) VALUES ('".$username."', '".$first."' ) ");
        //mysqli_close($con);
  //  }
   // else {
        // User info invalid!
   //     echo "Sorry mate, try again!";
   // }
}
else {
    // The form has not been posted
    // Show the form
?>

<form id="Form1" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
    Device_ID: <input type="integer" name="dev_id"><br>
    Device_Name: <input type="text" name="dev_name"><br>
    Customer_ID: <input type="integer" name="cust_id"><br>
    Customer_Firstname: <input type="text" name="cust_fname"><br>
    Customer_Lastname: <input type="text" name="cust_lname"><br>
    <br>
     <input type="hidden" name="SubmitCheck" value="sent"><input type="Submit" name="Form1_Submit" value="Submit">
</form>
<?php
}
?>

Edited March 18, 2014 by requinix
please use [code] tags when posting code

[Psycho]

Both errors appear to be due to the same problem, I don't see where $con is defined before this line is executed:

 

$result = mysqli_query($con, "SELECT '".$dev_id."',  . . .

[Ch0cu3r]

This line is responsible for connecting to mysql. 

//$con=mysqli_connect("localhost", "root", "", "abc");

However it is commented out (meaning it will be ignored) and secondly you need to be connected to mysql before performing any database operations, such as a query.

 

To solve the error,  you need to first uncomment it (by deleting the // infront of it), then make sure you give it the correct database credentials (hostname, username, password and database name) and then move it so it is a line before performing a query on the database.

Edited March 19, 2014 by Ch0cu3r

[ITNerd1002]

Alright I uncommented out the line and I cannot believe that I missed that. Even though that is working now it stil doesnt pull the data back but just shows the column titles of the table.

[mac_gyver]

most of the things you have selected in your query are literal string values. each item in the select list that are between single-quotes will be the literal strings those variables contain, not the values from the corresponding columns.

 

i recommend that you form the sql query statement in a php variable, then echo out that variable so that you can see what it actually is and so that you can copy/paste it to run it directly against your database using your favorite database management tool (phpmyadmin or similar.)

[ITNerd1002]

Yeah dude I do not really understand what that means I am pretty new to php and I have been teaching myself. I appreciate the help but I really have no idea what you mean sorry.

[mac_gyver]

this is real basic stuff and means you skipped over the first few php chapters where it talks about variables, assigning values to them, and echoing them.

// build the sql query in a php variable
$query = "SELECT '".$dev_id."', ta.TERM_ID, '".$cust_id."', '".$dev_name."', '".$cust_fname."', '".$cust_lname."', DATE_OF_LOAN, DATE_RETURNED
                from device d
                join abc_audit au on d.DEV_ID = '".$dev_id."'
                join term_agreement ta on au.TERM_ID = ta.TERM_ID
                join customer cu on ta.CUST_ID = '".$cust_id."'
                where au.abc_status = 'O'";

// echo the sql query for debugging purposes
echo $query;

// run the query
 $result = mysqli_query($con,$query);

Edited March 19, 2014 by mac_gyver