Question

我是PHP / SQL和通用服务器端脚本的新手。

我有以下代码：

$username = $_POST['username'];
$password = crypt($_POST['password']);
$emailAddress = $_POST['emailAddress'];

$con =  mysqli_connect($db_host, $db_user, $db_pass, $db_name);

    if (!$con) {
        die('Could not connect: ' . mysql_error());
    };

    if ($stmt = $con->prepare("SELECT * FROM Users WHERE Username=?")) {
        $stmt->bind_param("s", $username);
        if(!$stmt->execute()) {
            echo "This Username isn't taken!";
        } else {
            echo 'This Username is already taken!';
        };
        $stmt->close();
    };

$con->close();

我运行此代码以检查用户是否在我的数据库中，但我遇到的问题是它产生相同的输出，即使用户名不在数据库中。我该如何解决这个问题？

如果可以修复我的代码，如果执行不成功/如果没有找到用户名，我想运行：

if ($stmt = $con->prepare("INSERT INTO Users (Username, EmailAddress, ePassword) VALUES (?,?,?)")) {
    $stmt->bind_param("sss", $username, $emailAddress, $password);
    $stmt->execute();
    $stmt->close();
};

然后将用户添加到我的数据库。

感谢您的帮助。

Answer 1

我认为你应该删除<{1}}周围的<{p}}

if-statement

相反，尝试这样的事情：

if ($stmt = $con->prepare("SELECT * FROM Users WHERE Username=?"))`

Answer 2

我做了一个小项目，从数据库中搜索并给出输出..你可以尝试一下：）

  <?php
        $con = mysql_connect("localhost","root","");
        mysql_select_db("library", $con);
        echo "<table border=1 id='tab' width='100%' cellpadding=0 cellspacing=0>";

        if(isset($_POST['search'])){

        $mysearch=$_POST['search1'];

        $locate = mysql_query("select * from book where b_name = '$mysearch'");

        if (mysql_num_rows($locate) > 0)
        {

        echo "<tr>";
        echo "<th>Book ID</th>";
        echo "<th>Book name</th>";
        echo "<th>Author</th>";
        echo "<th>Publication</th>";
        echo "<th>Price</th>";
        echo "<th>Type</th>";
        echo "</tr>";
        }
        else
        {
        echo "<h1>BOOK NOT FOUND</h1>";
        }

        while($rows=mysql_fetch_array($locate)){
            echo "<tr>";
            echo "<td>".$rows['isbn']."</td>";
            echo "<td>".$rows['b_name']."</td>";
            echo "<td>".$rows['b_author']."</td>";
            echo "<td>".$rows['b_publication']."</td>";
            echo "<td>".$rows['b_price']."</td>";
            echo "<td>".$rows['b_type']."</td>";
            /*echo "<td>".$rows['no_copies']."</td>";
            if($rows['no_copies']==!0){
            echo "<td>".$rows['status']."</td>";
            }
            else{
            echo "<td>Not Available</td>";
            }
            if($rows['no_copies']==!0){
            echo "<td title=Borrow><a href='borrowedform.php?name=".$rows['book_title']."'><center><img src='img/brow.png'></center></a></td>";
            echo "<td title=Reserve><a href='reservedform.php?name=".$rows['book_title']."'><center><img src='img/reserved.png'></center></a></td>";
            }*/
            echo "</tr>";
        }

        }

Answer 3

所以我已经解决了我的问题并改变了我正在做的事情，我现在也检查电子邮件地址是否存在（即使它在数据库中设置为唯一）

$username = $_POST['username'];
$password = crypt($_POST['password']);
$emailAddress = $_POST['emailAddress'];

$con =  mysqli_connect($db_host, $db_user, $db_pass, $db_name);

if (!$con) {
    die('Could not connect: ' . mysql_error());
};

$stmt = $con->prepare("SELECT * FROM Users WHERE EmailAddress=?");
$stmt->bind_param("s", $emailAddress);
$stmt->execute();
$stmt->store_result();

$emailCheck = $stmt->num_rows;

if($emailCheck > 0) {
    echo "Email Address in use";
} else {            
    $stmt = $con->prepare("SELECT * FROM Users WHERE Username=?");
    $stmt->bind_param("s", $username);
    $stmt->execute();
    $stmt->store_result();

    $userCheck = $stmt->num_rows;
    if($userCheck > 0) {
        echo "Taken";
    } else {
        $stmt = $con->prepare("INSERT INTO Users (Username, EmailAddress, ePassword) VALUES (?,?,?)");
        $stmt->bind_param("sss", $username, $emailAddress, $password);
        $stmt->execute();
        $stmt->close();

        echo "Registered";
    };
    $stmt->close();
};
$con->close();

感谢您帮助我理解它

PHP检查SQL条目是否存在产生不同的输出

3 个答案: