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 mysqli_stmt_bind_result'
What is my error??
FYI - MR comes from here --
<center><strong>Do you Wish to Enter Visit Data?</center>
<center>If Yes:</center>
<form name="form2" method="post">
<center>Choose the Medical Record that You Wish to Use</strong></center>
<form method="post">
<center><input type="text" name="MR"></input></center>
<center><input type="submit" value="Enter Medical Record"/></center>
<input type="hidden" name="next_step" value="step10" />
</form>
Advice and help, please.
Ethan
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