Question

我知道这个问题早已被问过很多次，但其他答案都没有对我有用。我在这方面遇到了麻烦：

$row = $conn->query("SELECT * FROM urls WHERE id = '$id'");

我遵循了一个教程，所以我不知道是否还有其他信息我应该提供

编辑：

见下全文：

<?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'];
    }
?>

初始代码

    <?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";
?>

行上的错误：7,18

Answer 1

你忘记了数据库名称：

$conn = new mysqli($servername, $username, $password);

// Create connection like this :
$conn = new mysqli($servername, $username, $password, $dbname);

更改

$row = $conn->query("SELECT * FROM urls WHERE id = '$id'");

到

$row = $conn->query("SELECT * FROM urls WHERE id = ".$id);

也改变了：

if($row -> num_rows > 0)

到

if($row->num_rows > 0)

PHP错误：＆＃34;试图获取非对象的属性＆＃34;

1 个答案: