0

Heres my PHP code:

<?php
    $manName = $_POST['manName'];
    $equipType = $_POST['equipType'];
    $serNum = $_POST['serNum'];
    $locId = $_POST['locId'];
    $version = $_POST['version'];
    $altNum = $_POST['altNum'];
    $rack= $_POST['rack'];


    $conn = new mysqli('localhost','root','','inventory');
    if($conn->connect_error){
        die('Connection Failed : '.$conn->connect_error);
    } else {
        $stmt = $conn->prepare("insert into items(manName, equipType, serNum, locId, version, altNum, rack)values(?, ?, ?, ?, ?, ?, ?)");
        $stmt->bind_param('sssssss', $equipType, $equipType, $serNum, $locId, $version, $altNum, $rack);
        $stmt->execute();
        echo "Regitration added successfully";
        $stmt->close();
        $conn->close();
    }

?>

I get the "Regitration added successfully" message but nothing appears in the db table.

ADyson
  • 57,178
  • 14
  • 51
  • 63
craftee
  • 1
  • 1
  • 1
    Turn on error reporting so you can see what's going wrong. Read [What to do with mysqli problems? Errors like mysqli\_fetch\_array(): Argument #1 must be of type mysqli\_result and such](https://stackoverflow.com/questions/22662488/what-to-do-with-mysqli-problems-errors-like-mysqli-fetch-array-argument-1-m) for details – ADyson Sep 28 '22 at 20:00
  • You should bind variables correctly in your prepared statement. Hence, change `$stmt->bind_param('sssssss', $equipType, $equipType, $serNum, $locId, $version, $altNum, $rack);` to `$stmt->bind_param('sssssss', $manName, $equipType, $serNum, $locId, $version, $altNum, $rack);` please – Ken Lee Sep 28 '22 at 20:12
  • That was it Ken, thx. had $equipType twice. – craftee Sep 28 '22 at 20:39

0 Answers0