本文介紹了mysqli bind_param() 致命錯誤的處理方法，對大家解決問題具有一定的參考價值，需要的朋友們下面隨著小編來一起學習吧！
問題描述
我的代碼有錯誤有人可以幫助我嗎?
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;
}
?>
當我打開它時，這個錯誤出現致命錯誤:在第 8 行的 G:xampphtdocsKarmintalender est.php 中的非對象上調用成員函數 bind_param()"
When I open it this error appears "Fatal error: Call to a member function bind_param() on a non-object in G:xampphtdocsKarmintalender est.php on line 8"
推薦答案
您在此行中的一個字段不存在，請檢查它們.
One of your fields on this line doesn't exist,check them.
$sql = "SELECT name, kalender_ID FROM kalender WHERE ersteller_ID = ?";
此外，您應該檢查 $stmt.
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;
}
}
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