0
我知道這個問題早已被問過很多次,但沒有其他答案爲我工作。我在這條線的麻煩:PHP錯誤:「試圖獲取非對象的屬性」
$row = $conn->query("SELECT * FROM urls WHERE id = '$id'");
我跟着一個教程,所以我不知道是否有任何其他信息,我應該提供
編輯:
繼承人整個文本文檔:
<?php
function idExists($id){
include $_SERVER['DOCUMENT_ROOT'] . '/short/includes/init.php';
$row = $conn->query("SELECT * FROM urls WHERE id = '.$id'");
if($row -> num_rows > 0){
return true;
} else {
return false;
}
}
function urlHasBeenShortened($url){
include $_SERVER['DOCUMENT_ROOT'] . '/short/includes/init.php';
$row = $conn->query("SELECT * FROM urls WHERE link_to_page = '$url'");
if($row->num_rows > 0){
return true;
} else {
return false;
}
}
function getURLID($url){
include $_SERVER['DOCUMENT_ROOT'] . '/short/includes/init.php';
$row = $conn->query("SELECT id FROM urls WHERE link_to_page = '$url'");
return $row->fetch_assoc()['id'];
}
function insertID($id, $url){
include $_SERVER['DOCUMENT_ROOT'] . '/short/includes/init.php';
$conn->query("INSERT INTO urls (id, link_to_page) VALUES ('$id', '$url')");
if(strlen($conn->error) == 0){
return true;
}
}
function getUrlLocation($id){
include $_SERVER['DOCUMENT_ROOT'] . '/short/includes/init.php';
$row = $conn->query("SELECT link_to_page FROM urls WHERE id = '$id'");
return $row->fetch_assoc()['link_to_page'];
}
?>
初始化代碼
<?php
$servername = "localhost";
$username = "root";
$password = "";
// Create connection
$conn = new mysqli($servername, $username, $password);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
?>
錯誤上線7條,18
'ID = '$ id''應該是'ID =' $ id.'' –
你忘記添加數據庫名稱:/你的連接 – yoeunes
您應該檢查query()的返回值是否調用「=== FALSE」。如果是這樣的話,你的查詢中可能會出現語法錯誤(例如傳入的用戶數據)。您可以嘗試打印整個查詢,以便查看正在查詢的內容。 – PhilMasterG