Question

我的程序没有插入到数据库中，尝试了各种方法，我是php的新手并试图用这个来测试自己，但我发现很难得到。我认为问题在于与数据库的连接，但我没有得到它

<!DOCTYPE html>
<html>
    <head>
        <title>LIST</title>
    </head>
    <body>
        <h1> TODO LIST </h1>
        <?php
            if(isset($_POST["submit"])){
                $servername = "localhost";
                $username = "root";
                $password = "";

                try {
                    //create a database conneection 
                    $conn = mysqli_connect("localhost", "root", "");
                    if(!$conn){
                        die("Database connection failed: ". mysql_error());
                    }
                    $sql = "INSERT INTO tasks (task, date, time) VALUES (:task, :date, :time)";
                    $query = $conn->prepare($sql);
                    $query->execute(array(':task'=>$task, 
                        ':date'=>$date,':time'=>$time));
                }
                catch(PDOException $e)
                {
                    echo "Connection failed: " . $e->getMessage();
                }
            }
        ?>

        <form action = "todolist.php" method = "post">
          Task: <input type="text" name="task" id="task"><br/>
          Date: <input type="date" name="date" id="date"><br/>
          Time: <input type="time" name="time" id="time"><br/>
        <input type = "submit" value = "submit" name="submit">
        </form>
    </body>
</html>

Answer 1

您不是选择任何数据库。尝试这样的事情。

$conn = mysqli_connect("localhost", "root", "", "Yourdatabasename");

Answer 2

您是否已连接到数据库？你没有选择任何bdd，日志是正确的吗？

也许试试：

ax.set_yticks([])

首先，查看您是否已连接到数据库

Answer 3

我在我的一个项目中使用这种方式。创建 connect.php 文件并创建如下连接：

<?php
$conn = mysqli_connect("localhost", "root", "", "Yourdatabasename");
?>

现在在index.php文件中检查是否按下了提交按钮：

<?php

if(isset($_POST["submit"]))

{

// include connection file
include("connect.php");

// get values
$subject = mysqli_real_escape_string($con, $_POST["subject"]);
$comment = mysqli_real_escape_string($con, $_POST["comment"]);

// Insert in database

$query = "INSERT INTO comments(comment_subject, comment_text)VALUES ('$subject', '$comment')";

mysqli_query($con, $query);

}

?>

程序未插入数据库

3 个答案: