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php - 我们向用户提示的详细信息(通过 php 编码)未显示在 MySQL 数据库(本地主机)中

转载 作者：行者123 更新时间：2023-11-29 18:52:15 27 4
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我最近编写了一个 php 代码，提示用户输入他们的姓名、电子邮件、网站、评论和性别(我正在使用 php 编码)。当用户已经输入数据时，数据应该显示在我的本地主机 MySQL 数据库( phpmyadmin )中。但这里的问题是，无论我将 INSERT INTO 函数放在哪里，用户键入的数据都不会插入到 phpmyadmin 中的表中。是的，我已经成功创建了该表(使用php 代码)，但详细信息似乎没有插入到我创建的表中。

这是我的代码:

<?php
    include "header.php";
    include "lib/db.php";
    include "module/functions.php";

    error_reporting(0);
?>

<!DOCTYPE html>
<html>
<body>
    <form action="index.php" method="POST">

    <?php
        $name = "";
        $email = "";
        $gender = "";
        $comment = "";
        $website = "";
    ?>

    <center>
    <table border="13" width="400" height="400">
        <tr>
            <td colspan="5" align="center" bgcolor="blue">guest's information</td>
        </tr>

        <tr>
            <td align="right">Name:</td><td> <input type="text" name="name" value="<?php echo $name;?>"></td>
        </tr>

        <tr>
            <td align="right">E-mail: </td><td><input type="text" name="email" value="<?php echo $email;?>"></td>
        </tr>

        <tr>
            <td align="right">Website: </td>
            <td><input type="text" name="website" value="<?php echo $website;?>"></td>
        </tr>

        <tr>
            <td align="right">Comment: </td>
            <td><textarea name="comment" rows="10" cols="65"><?php echo $comment;?></textarea></td>
        </tr>

        <tr>
            <td align="right">Gender: </td>
        <td>
            <input type="radio" name="gender"<?php if ($gender=="female"); ?> value="female">Female
            <input type="radio" name="gender"<?php if ($gender=="male"); ?> value="male">Male
        <td>
        </tr>

        <tr>
            <td colspan="5" align="center"><input type="submit" name="submit" value="submit"></td>
        </tr>

        <tr>
            <td colspan="5" align="center"><input type="reset" name="reset" value="reset"></td>
        </tr>


<?php
    include "config.inc.php";
?>

<?php
    {
        $name = $_POST['name'];
        $email = $_POST['email'];
        $website = $_POST['website'];
        $comment = $_POST['comment'];
        $gender = $_POST['gender'];

        $query = "INSERT INTO myguests (name, email, website, comment, gender)
            VALUES ('$name', '$email', '$website' , '$comment' , '$gender')";   

    }

    if(isset($_POST['submit'])) {
        if($name !="" && $email !="" && $website !="" && $comment !="" && $gender !="") {
            $query = "INSERT INTO myguests (name, email, website, comment, gender)
                        VALUES ('$name', '$email', '$website' , '$comment' , '$gender')";

        echo "You've entered the data.";


        echo '<script language="javascript">';
        echo 'alert("Yo brother, sudah la brother.")';
        echo '</script>';

        } else {

            echo '<script language="javascript">';
            echo 'alert("Yo brother, fuck off. Isi elok-elok la brother.")';
            echo '</script>';       
        }
    } 


    ?>  
     </form>
   </body>
</html>

这是我的配置代码(连接到本地主机数据库)

<?php
    header("Content-Type: text/html; charset=utf-8");

    //Set connection time limit/memory limit to infinity/unlimited
    set_time_limit(0);
    ini_set('default_charset', 'UTF-8');
    ini_set('memory_limit', '-1');

    $servername = "localhost";
    $username = "root";
    $password = "";
    $dbname = "first_test";

    // Create connection
    $conn = mysqli_connect($servername, $username, $password, $dbname);

    // Check connection
    if (!$conn) {
        die("Connection failed: " . mysqli_connect_error());
    }

    // sql to create table
    $sql = "CREATE TABLE MyGuests (
        id INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
        name VARCHAR(30) NOT NULL,
        email VARCHAR(50),
        website VARCHAR(100),
        comment VARCHAR(420),
        gender VARCHAR(6)
        )";

        $name = "";
        $email = "";
        $website = "";
        $comment = "";
        $gender = "" ;

    if (mysqli_query($conn, $sql)) {
        echo "<br><br>Table MyGuests created successfully";
    }

    $query = "INSERT INTO myguests (name, email, website, comment, gender)
                VALUES ('$name', '$email', '$website' , '$comment' , '$gender')";

    if(mysqli_query($conn, $query)) {   
        echo "<br><br>Guest's data is inserted successfully into the database<br><br>";
    } else {
            echo "<br><br>Fail to insert guest's data into the database<br><br>";
    }

    mysqli_close($conn);
?>

谢谢您，祝您有愉快的一天。

最佳答案

你的问题实际上在这里:

 if(isset($_POST['submit'])) {
        if($name !="" && $email !="" && $website !="" && $comment !="" && $gender !="") {
            $query = "INSERT INTO myguests (name, email, website, comment, gender)
                        VALUES ('$name', '$email', '$website' , '$comment' , '$gender')";

        echo "You've entered the data.";


        echo '<script language="javascript">';
        echo 'alert("Yo brother, sudah la brother.")';
        echo '</script>';

        }

您实际上正在构建字符串，但没有传递到数据库来查询它。

您实际上缺少的是这一行:

$result = mysql_query($query);

修改代码:

        $query = "INSERT INTO myguests (name, email, website, comment, gender)
                        VALUES ('$name', '$email', '$website' , '$comment' , '$gender')";

        $result = mysql_query($query);

        echo "You've entered the data.";

关于php - 我们向用户提示的详细信息(通过 php 编码)未显示在 MySQL 数据库(本地主机)中，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/44320261/

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