php - 总是 MySQL 什么也不输出。告诉我为什么？

Question

<?php
    $uname = $_POST["username"]$uname = 
    stripslashes(trim($uname));
    $pward = $_POST["password"];
    $pward = stripcslashes(trim($pward));

    $servername = "localhost";
    $username = "root";
    $password = "";
    $dbname = "index_data";

    $conn = new mysqli($servername, $username, 
    $password, $dbname);
    if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
    }

    $sql = "SELECT * FROM 00_user_details WHERE 
    username = '$uname'  ";
    $result = mysqli_query($conn, $sql);
    $rows = mysqli_fetch_row($result);
    echo "_".$rows["first_name"]."_";
?>

我是一个更大的。这是我的代码，但它什么也没输出。如果您尊重先生/女士，请帮助我...

Answer 1

请在 $_POST["username"] 之后加上分号，并使用 mysqli_fetch_assoc 代替 mysqli_fetch_row。

mysqli_fetch_assoc 返回一行作为关联数组，其中列名将是存储相应值的键。

请找到以下实际代码：

$uname = $_POST["username"];
$uname = stripslashes(trim($uname));

$pward = $_POST["password"];
$pward = stripcslashes(trim($pward));

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "index_data";

$conn = new mysqli($servername, $username,$password, $dbname);
if ($conn->connect_error) 
{
    die("Connection failed: " . $conn->connect_error);
}

$sql = "SELECT * FROM 00_user_details WHERE username = '$uname'";
$result = mysqli_query($conn, $sql);
$rows = mysqli_fetch_assoc($result);
echo "_".$rows["first_name"]."_";