Dear List -

Here is some code:

            $sql13 = "SELECT * FROM Intake3 WHERE MedRec = ?";
            $stmt = mysqli_stmt_init($cxn);
            mysqli_stmt_prepare( $stmt, $sql13 );
            $_SESSION['stmt'] = $stmt;
            $args = array();

            $args[0] = &$_POST['MR'];
            $types = 'i';
            if( !$stmt )
                throw new Exception( 'Error preparing statement' );

// Put the statement and types variables at the front of the params to pass to mysqli_stmt_bind_param() array_unshift($args, $stmt, $types ); // Note that I've moved this call. Apparently it doesn't pass back the result.
        //    I guess sometimes I just forget these things.

            // mysqli_stmt_bind_param()
            if( !call_user_func_array( 'mysqli_stmt_bind_param', $args ) )
            throw new Exception( 'Failed calling mysqli_stmt_bind_param' );

            if( !mysqli_stmt_execute( $stmt ) )
                throw new Exception( 'Error while executing statement' );

mysqli_stmt_bind_result( $stmt, $Site, $MedRec, $Fname, $Lname, $Phone, $Height, $Sex, $Hx, $Bday, $Age); throw new Exception( 'Failed calling mysqli_stmt_bind_result' );

Fatal error: Uncaught exception 'Exception' with message 'Failed calling 

What is my error??

FYI - MR comes from here --

                        <center><strong>Do you Wish to Enter Visit 
                        <center>If Yes:</center>
                        <form name="form2" method="post">
                        <center>Choose the Medical Record that You Wish to 
                        <form method="post">
                        <center><input type="text" name="MR"></input></center>
                        <center><input type="submit" value="Enter Medical 
                        <input type="hidden" name="next_step" value="step10" />

Advice and help, please.


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