PHP를 사용하여 간단한 채용정보 공개 기능을 개발하는 방법

CREATE TABLE jobs (
    id INT AUTO_INCREMENT PRIMARY KEY,
    title VARCHAR(255) NOT NULL,
    description TEXT NOT NULL,
    company VARCHAR(255) NOT NULL,
    location VARCHAR(255) NOT NULL,
    salary DECIMAL(10, 2) NOT NULL,
    created_at TIMESTAMP DEFAULT CURRENT_TIMESTAMP
);

<form method="POST" action="submit_job.php">
    <input type="text" name="title" placeholder="职位名称" required>
    <textarea name="description" placeholder="职位描述" required></textarea>
    <input type="text" name="company" placeholder="公司名称" required>
    <input type="text" name="location" placeholder="工作地点" required>
    <input type="text" name="salary" placeholder="薪资" required>
    <button type="submit">发布</button>
</form>

<?php
// 连接到数据库
$connection = mysqli_connect("localhost", "root", "password", "database_name");

// 检查是否连接成功
if (!$connection) {
    die("数据库连接失败：" . mysqli_connect_error());
}

// 获取表单中的数据
$title = $_POST['title'];
$description = $_POST['description'];
$company = $_POST['company'];
$location = $_POST['location'];
$salary = $_POST['salary'];

// 将数据插入到数据库中
$sql = "INSERT INTO jobs (title, description, company, location, salary) VALUES ('$title', '$description', '$company', '$location', $salary)";

if (mysqli_query($connection, $sq)) {
    echo "招聘信息发布成功！";
} else {
    echo "发布失败：" . mysqli_error($connection);
}

// 关闭数据库连接
mysqli_close($connection);
?>

<?php
// 连接到数据库
$connection = mysqli_connect("localhost", "root", "password", "database_name");

// 检查是否连接成功
if (!$connection) {
    die("数据库连接失败：" . mysqli_connect_error());
}

// 查询数据库中的所有招聘信息
$sql = "SELECT * FROM jobs";
$result = mysqli_query($connection, $sql);

// 显示查询结果
if (mysqli_num_rows($result) > 0) {
    while ($row = mysqli_fetch_assoc($result)) {
        echo "<h2>{$row['title']}</h2>";
        echo "<p>{$row['description']}</p>";
        echo "<p>{$row['company']}</p>";
        echo "<p>{$row['location']}</p>";
        echo "<p>{$row['salary']}</p>";
        echo "<hr>";
    }
} else {
    echo "暂无招聘信息";
}

// 关闭数据库连接
mysqli_close($connection);
?>