Question

我的代码中有什么问题..我尝试单击下一步按钮显示数据库中的下一条记录..但是当我点击时没有任何反应。

这是我获取记录的功能

public function getBooks()
{
     $limit = 1;
     $start=0;

    //SELECT loginUser.username, Library.nameOfBook FROM loginUser JOIN userBook JOIN Library ON userBook.user_id = loginUser.id AND userBook.book_id = Library.id WHERE loginUser.username="loay";
    $query = "SELECT Library.nameOfBook FROM loginUser JOIN userBook JOIN Library ON userBook.user_id = loginUser.id AND userBook.book_id = Library.id WHERE loginUser.username=:username LIMIT $start, $limit";
    $statment = $this->db->prepare($query);
    $statment->execute([
        ':username' => $this->username
        //,':start' => $start, ':limit' => $limit
    ]);
    $result = $statment->fetchAll();

   echo "<table border='1'>
    <tr>
    <th>Books</th>
    </tr>";
    foreach($result as $row){
        echo "<tr>";
        echo "<td>" . $row['nameOfBook'] . "</td>";
        echo "</tr>";
    }
    echo "</table>";

}

这是我在index.php中的代码

  $object->getBooks();

  if( isset($_POST['next'])){
    $start +=1;
}

这是我的表单代码

＆＃13;

<form method="post" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>">
  Name: <input type="text" name="user" value="<?php echo $user;?>">
  <span class="error">* <?php echo $nameErr;?></span>
  <br><br>
  password: <input type="password" name="pass" value="<?php echo $password;?>" >
  <span class="error">* <?php echo $passwordErr;?></span>
  <br><br>
  <input type="submit" name="submit" value="Login">
  <br><br>

 
  <button type="button" name="button" onclick="window.location.href='/ooRigester.php'">SignUp</button>
  <br><br>
  <input type='submit' name='next' value='next' method="post"><br>
</form>

＆＃13;

表单后的代码

<?php
include_once('User.php');

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

$username = $_POST["user"];
$password = $_POST["pass"];

$object = new User();
$object->username= $username;
$object->Password=$password;
if( $object->isAuthenticated() ){

  echo "User Verified . <br><br>";
  $object->getBooks();

  if( isset($_POST['next'])){
    $start +=1;
    $object->getBooks($start); 
}

}

else{

  echo "Wrong User Name Or Password . <br>";
}}


?>

和之前的形式

＆＃13;

<!DOCTYPE HTML>
<html>
<head>
<style>
.error {color: #FF0000;}
</style>
<title>Login Form</title>
</head>
<body>
<?php
/*$str = "Hello";
echo password_hash($str,1);*/
// define variables and set to empty values
$nameErr = "";
$passwordErr = "";
$user = "";
$password = "";

if ($_SERVER["REQUEST_METHOD"] == "POST") {
  if (empty($_POST["user"])) {
    $nameErr = "Name is required";
  } else {
    $user = test_input($_POST["user"]);
    // check if name only contains letters and whitespace
    if (!preg_match("/^[a-zA-Z ]*$/",$user)) {
      $nameErr = "Only letters and white space allowed";
    }
  }
  if (empty($_POST["pass"])) {
    $passwordErr = "Password is required";
  } else {
    $password = test_input($_POST["pass"]);
  }
  
}

function test_input($data) {
  $data = trim($data);
  $data = stripslashes($data);
  $data = htmlspecialchars($data);
  return $data;
}
?>

＆＃13;

Answer 1

控制器中的
代码：

function getBooks($start = 0)
{
    $limit = 1;

//SELECT loginUser.username, Library.nameOfBook FROM loginUser JOIN userBook JOIN Library ON userBook.user_id = loginUser.id AND userBook.book_id = Library.id WHERE loginUser.username="loay";
    $query = "SELECT Library.nameOfBook FROM loginUser JOIN userBook JOIN Library ON userBook.user_id = loginUser.id AND userBook.book_id = Library.id WHERE loginUser.username=:username LIMIT $start, $limit";
    $statment = $this->db->prepare($query);
    $statment->execute([
        ':username' => $this->username,
//,':start' => $start, ':limit' => $limit
    ]);
    $result = $statment->fetchAll();

    echo "<table border='1'>
            <tr>
                <th>Books</th>
            </tr>";

    foreach ($result as $row) {
        echo "<tr>";
        echo "<td>" . $row['nameOfBook'] . "</td>";
        echo "</tr>";
    }
    echo "</table>";
}

index.php中的代码。

 $object->getBooks();

if( isset($_POST['next']))
{
    $start +=1;
    $object->getBooks($start);  
}

Answer 2

在Index.php中 - 您没有正确设置$ start值。以下代码设置$start的{{1}}值，如果不存在$_POST['start']默认为0。

而且你也没有正确接受表单提交。我通过对$ _POST变量进行isset（）检查来修改代码以接受提交类型（下一个和登录）。

使用：

$start

在表单代码中，您必须添加一个隐藏的输入，该输入保存下一个数据库记录的偏移值。

使用此代码：

<?php include_once('User.php'); if(isset($_POST['submit']) OR isset($_POST['next'])){ $username = $_POST["user"]; $password = $_POST["pass"]; $object = new User(); $object->username= $username; $object->Password=$password; if( $object->isAuthenticated() ){ echo "User Verified . <br><br>"; $start = 0; if(isset($_POST['start'])){ $start = $_POST['start']; } $object->getBooks($start); } } else{ echo "Wrong User Name Or Password . <br>"; } ?>

您的类方法<form method="post" action=""> Name: <input type="text" name="user" value="<?php echo $user;?>"> <span class="error">* <?php echo $nameErr;?></span> <br><br> password: <input type="password" name="pass" value="<?php echo $password;?>" > <span class="error">* <?php echo $passwordErr;?></span> <br><br> <input type="submit" name="submit" value="Login"> <br><br> <input type="hidden" value="<?php echo (intval(isset($_POST['start'])?$_POST['start']:0)+1);?>" name="start"/> <button type="button" name="button" onclick="window.location.href='/ooRigester.php'">SignUp</button> <br><br> <input type='submit' name='next' value='next' method="post"><br> </form>必须具有getBooks()参数，以便它知道下一条要显示的记录。

<强>尝试：

$start

php使用按钮

2 个答案: