我的代码出了什么问题..我尝试单击下一个按钮以显示数据库中的下一条记录..但是当我单击时什么也没有发生。
这是我获取记录的函数
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'])){
$start +=1;
}
这是我的表单代码
<form method="post" action="<?php echo htmlspecialchars($_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 $passwordErr;?></span>
<br><br>
<input type="submit" name="submit" value="Login">
<br><br>
<button type="button" name="button" onclick="window.location.href='/ooRigester.php'">SignUp</button>
<br><br>
<input type='submit' name='next' value='next' method="post"><br>
</form>此代码在表单之后
<?php
include_once('User.php');
if(isset($_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>";
}}
?>
这在表格之前
<!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;
}
?>最佳答案
在 Index.php 中 - 您没有正确设置 $start 值。以下代码从 $_POST['start'] 设置 $start 值，如果 $start 不存在，则默认为 0。
而且您也没有正确接受表单提交。我修改了代码，通过对 $_POST 变量执行 isset() 检查来接受两种提交类型(下一步和登录)。
用途:
<?php
include_once('User.php');
if(isset($_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>";
}
?>
在表单代码中，您必须添加一个隐藏输入，用于保存下一条数据库记录的偏移值。
使用此代码:
<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>
您的类方法 getBooks() 必须具有 $start 参数，以便它知道接下来要显示哪条记录。
尝试:
public function getBooks($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>";
}
关于php使用next按钮限制查询pdo，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/50227074/