Question

所以这是我的PHP代码，我不明白为什么我的查询总是失败！我从phpMyAdmin网站复制了$sql。但它一直都是假的。

$servername = "localhost";
        $dbname = "gumdb";

        $conn = mysqli_connect($servername, $dbname);
        if (!$conn) {
            die("Connection failed: " . mysqli_connect_error());
        } else {
            echo "Connected! <br />";
        }

        $sql = 'SELECT * FROM `test` WHERE 1;';
        $result = mysqli_query($conn, $sql);

        if ($result === FALSE) {
            echo "result failed!";
            die(mysql_error());
        }

Result of the test

Answer 1

我认为您遇到了这个问题，因为您没有声明MySQL连接的用户名和密码，例如：

    $servername = "localhost";
    $user = "root"; // Your MySQL username here
    $password = "root"; // Your MySQL password here
    $dbname = "gumdb";

    $conn = mysqli_connect($servername,$user,$password, $dbname);

Answer 2

您的查询存在问题

$sql = 'SELECT * FROM `test` WHERE 1;';

您是否注意到了2个半克隆。请删除第一个并尝试它将工作。使用如下

$servername = "localhost";
$dbname = "gumdb";
$pass = "";
//$conn = mysqli_connect(servername, "dbuser", "dbpass", "dbname");
$conn = mysqli_connect($servername, "root", "", "gumdb");
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
} else {
    echo "Connected! <br />";
}

$sql = 'SELECT * FROM `test` WHERE 1';
$result = mysqli_query($conn, $sql);

if ($result === FALSE) {
    echo "result failed!";
    die(mysql_error());
}
else {
    var_dump($result);
}

我已编辑完整代码。数据库连接也存在问题。

SQL Query无故失败

2 个答案: