Question

我试图回应PHP中的HTML表单，但我无法完成它。它只是回显预先格式化的HTML。我没有收到表格。

这是我的PHP脚本
do-reset.php

<?php

require_once 'connect.php';

session_start();

if($_SERVER['REQUEST_METHOD'] === 'GET') {


if(isset($_GET['email']) && !empty($_GET['email']) && isset($_GET['hash']) && !empty($_GET['hash'])) {

    $email = htmlentities(mysqli_real_escape_string($connection, trim($_GET['email'])));
    $hash = htmlentities(mysqli_real_escape_string($connection, trim($_GET['hash'])));

    $search_query = "SELECT email, hash, status FROM users WHERE email = '{$email}' AND forgot_password_hash = '{$hash}' AND
               status = '1'";

    $do_search_query = mysqli_query($connection, $search_query);

    if($do_search_query) {

        $count_rows = mysqli_num_rows($do_search_query);

        if($count_rows > 0) {

            $_SESSION['email'] = $email;
            $_SESSION['hash'] = $hash;


            echo "<form method='post' action='do-reset.php'><input type='password' name='password'><br><input type='submit' value='Reset My Password'></form>";


        }
        else {
            $data = array("result" => -3, "message" => "Invalid URL");
        }


    }
    else {
        $data = array("result" => -2, "message" => "Something Went Wrong! Try Again Later.");
    }
}
else
{
    $data = array("result" => -1, "message" => "Certain Request Parameters Are Missing!");
}

}
else {
 $data = array("result" => 0, "message" => "Incorrect Request Method!");   
}



mysqli_close($connection);
/* JSON Response */
header('Content-type: application/json');
echo json_encode($data, JSON_PRETTY_PRINT);



?>

Answer 1

1。：删除header('Content-type: application/json'); 这基本上会告诉浏览器将输出显示为文本。

<强> 2：要保留格式，您可以使用<pre> - 标记：

echo "<pre>";
echo json_encode($data, JSON_PRETTY_PRINT);`
echo "</pre>";

不同方法： 仅在填充$data - 数组时将内容类型设置为application / json

if(!empty($data)){
    header('Content-type: application/json');
    echo json_encode($data, JSON_PRETTY_PRINT);
}

Answer 2

我找到了出路！ <?php require_once 'connect.php'; session_start(); $type_json = true; if($_SERVER['REQUEST_METHOD'] === 'GET') { if(isset($_GET['email']) && !empty($_GET['email']) && isset($_GET['hash']) && !empty($_GET['hash'])) { $email = htmlentities(mysqli_real_escape_string($connection, trim($_GET['email']))); $hash = htmlentities(mysqli_real_escape_string($connection, trim($_GET['hash']))); $search_query = "SELECT email, hash, status FROM users WHERE email = '{$email}' AND forgot_password_hash = '{$hash}' AND status = '1'"; $do_search_query = mysqli_query($connection, $search_query); if($do_search_query) { $count_rows = mysqli_num_rows($do_search_query); if($count_rows > 0) { $_SESSION['email'] = $email; $_SESSION['hash'] = $hash; $type_json = false; echo "<form method='post' action='do-reset.php'><input type='password' name='password'><br><input type='submit' value='Reset My Password'></form>"; } else { $data = array("result" => -3, "message" => "Invalid URL"); } } else { $data = array("result" => -2, "message" => "Something Went Wrong! Try Again Later."); } } else { $data = array("result" => -1, "message" => "Certain Request Parameters Are Missing!"); } } else { $data = array("result" => 0, "message" => "Incorrect Request Method!"); } mysqli_close($connection); /* JSON Response */ if($type_json) { header('Content-type: application/json'); echo json_encode($data, JSON_PRETTY_PRINT); } ?>为我做了诀窍。

{{1}}

Answer 3

PHP在服务器端执行，被视为脚本而不是标记语言，这意味着请求页面上的HTML对服务器无关紧要，只关心PHP。所以，如果你做了

<?php
if(true) {
    ?>
    <form>Hello</form>
    <?php
}
?>

如果语句为true，则html表单将仅显示文本hello，其中true始终为true。您可以将其替换为任何声明，例如检查某人是否从他们提交的表单中输入某个字段中的内容。

希望这有帮助！

无法在PHP中回显HTML <form> </form>

3 个答案: