Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
975 views
in Technique[技术] by (71.8m points)

php - mysqli bind_param() fatal error

I Have an Error at my Code could someone help me?

<?php
  $db = new mysqli("localhost","root","","karmintalender");

  $owner_ID = 1;

  $sql = "SELECT name, kalender_ID FROM kalender WHERE ersteller_ID = ?";
  $stmt = $db->prepare($sql);
  $stmt->bind_param("i", $owner_ID);
  $stmt->execute();
  $stmt->bind_results($name, $kalender_ID);

  while ($stmt->fetch()) {
    echo $name . " " . $kalender_ID;
  }
?>

When I open it this error appears "Fatal error: Call to a member function bind_param() on a non-object in G:xampphtdocsKarmintalenderest.php on line 8"

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

One of your fields on this line doesn't exist,check them.

$sql = "SELECT name, kalender_ID FROM kalender WHERE ersteller_ID = ?";

Also, you should be checking for $stmt.

$db = new mysqli("localhost","root","","karmintalender");

 $owner_ID = 1;

 $sql = "SELECT name, kalender_ID FROM kalender WHERE ersteller_ID = ?";
 $stmt = $db->prepare($sql);
 if($stmt){
     $stmt->bind_param("i", $owner_ID);
     $stmt->execute();
     $stmt->bind_results($name, $kalender_ID);

     while ($stmt->fetch()) {
       echo $name . " " . $kalender_ID;
     }
 }

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...