Question

我有一个简单的PHP代码，只显示是否已建立与数据库的连接。它有一个HTML部分，但似乎没有建立连接。使用的数据库是MySql。这些表不是作为程序的一部分创建的。

需要一些建议......

提前谢谢....

这是代码......

<h1>LOGIN PAGE </h1>
<form action="" method="post">
Username   <input name="usname" type="text" />
Password <input name="paswd" type="password" />
<input name="submit" type="submit" />
</form>

<?php
if (isset($_POST['submit'])) {
    $uname = $_POST['usname'];
    $pswd = $POST['paswd'];
    $conn = mysqli_connect("localhost","root","archana","details");
    if(mysqli_connect_errorno())
    {
        echo "Failed to connect to Mysql" ;
    }
    else
    {
        echo "connection established";
    }
    $data = mysqli_query($conn, "SELECT * FROM user");
    while($a = mysqli_fetch_array($data))
    {
        echo $a['username'];
    }
    mysqli_close($conn);
}
?>

Answer 1

我根据Marcel Balzer评论重写您的代码

<h1>LOGIN PAGE </h1>
<form action="" method="post">
Username   <input name="usname" type="text" />
Password <input name="paswd" type="password" />
<input name="submit" type="submit" />
</form>

<?php
if (isset($_POST['submit'])) {
    $uname = $_POST['usname'];
    $pswd = $POST['paswd'];
    $conn = mysqli_connect("localhost","root","archana","details");
    if(mysqli_connect_errno())
    {
        echo "Failed to connect to Mysql. Error: ".mysqli_connect_errno() ;
    }
    else
    {
        echo "connection established";
    }
    $data = mysqli_query($conn, "SELECT * FROM user");
    while($a = mysqli_fetch_array($data))
    {
        echo $a['username'];
    }
    mysqli_close($conn);
}
?>

使用Mysql无法在PHP中与数据库建立连接

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