使用php连接mysql完成以下功能，要求先用workbench的model创建mysql表：user，child user: id,name,sex,age,address,password child: id, parent_id, name, birthdate, sex, time,password 1，管理员登录，添加用户user信息(管理员不需要创建表，验证用户名和密码就可以)2，用户登录， (1) 输入姓，例如王，为child取一个名字，可以多次取名字 (2) 把上一步的名字连同自己的id，保存到孩子信息表中。3，child登陆（假设长大以后），查看个人信息用php实现一个人口管理系统，可以在原来表的基础上，添加医生表，疫苗表，疫苗接种表，child登录可以查自己疫苗接种信息

以下是用PHP连接MySQL实现人口管理系统的代码：

1. 管理员登录，添加用户user信息

<?php

$servername = "localhost";
$username = "admin";
$password = "admin123";
$dbname = "population";

// 创建连接
$conn = mysqli_connect($servername, $username, $password, $dbname);

// 检查连接是否成功
if (!$conn) {
  die("Connection failed: " . mysqli_connect_error());
}

// 获取管理员输入的用户名和密码
$admin_username = $_POST["admin_username"];
$admin_password = $_POST["admin_password"];

// 验证管理员用户名和密码是否正确
$sql = "SELECT * FROM admin WHERE username='$admin_username' AND password='$admin_password'";
$result = mysqli_query($conn, $sql);

if (mysqli_num_rows($result) == 1) {
  // 如果验证成功，处理添加用户信息的请求
  $username = $_POST["username"];
  $sex = $_POST["sex"];
  $age = $_POST["age"];
  $address = $_POST["address"];
  $password = $_POST["password"];

  // 将用户信息插入到user表中
  $sql = "INSERT INTO user (name, sex, age, address, password) VALUES ('$username', '$sex', '$age', '$address', '$password')";
  if (mysqli_query($conn, $sql)) {
    echo "User added successfully!";
  } else {
    echo "Error adding user: " . mysqli_error($conn);
  }

} else {
  // 如果验证失败，提示管理员用户名或密码错误
  echo "Invalid username or password!";
}

mysqli_close($conn);

?>

2. 用户登录，输入姓并为child取名字，保存到孩子信息表中

<?php

$servername = "localhost";
$username = "admin";
$password = "admin123";
$dbname = "population";

// 创建连接
$conn = mysqli_connect($servername, $username, $password, $dbname);

// 检查连接是否成功
if (!$conn) {
  die("Connection failed: " . mysqli_connect_error());
}

// 获取用户输入的用户名和密码
$username = $_POST["username"];
$password = $_POST["password"];

// 验证用户用户名和密码是否正确
$sql = "SELECT * FROM user WHERE name='$username' AND password='$password'";
$result = mysqli_query($conn, $sql);

if (mysqli_num_rows($result) == 1) {
  // 如果验证成功，处理输入姓并为child取名字的请求
  $surname = $_POST["surname"];

  // 生成child的名字
  $sql = "SELECT COUNT(*) AS count FROM child WHERE name LIKE '$surname%'";
  $result = mysqli_query($conn, $sql);
  $row = mysqli_fetch_assoc($result);
  $count = $row["count"] + 1;
  $child_name = $surname . $count;

  // 将child信息插入到child表中
  $parent_id = mysqli_fetch_assoc($result)["id"];
  $birthdate = $_POST["birthdate"];
  $sex = $_POST["sex"];
  $time = $_POST["time"];
  $password = $_POST["password"];
  $sql = "INSERT INTO child (parent_id, name, birthdate, sex, time, password) VALUES ('$parent_id', '$child_name', '$birthdate', '$sex', '$time', '$password')";
  if (mysqli_query($conn, $sql)) {
    echo "Child added successfully!";
  } else {
    echo "Error adding child: " . mysqli_error($conn);
  }

} else {
  // 如果验证失败，提示用户用户名或密码错误
  echo "Invalid username or password!";
}

mysqli_close($conn);

?>

3. child登陆，查看个人信息和疫苗接种信息

<?php

$servername = "localhost";
$username = "admin";
$password = "admin123";
$dbname = "population";

// 创建连接
$conn = mysqli_connect($servername, $username, $password, $dbname);

// 检查连接是否成功
if (!$conn) {
  die("Connection failed: " . mysqli_connect_error());
}

// 获取child输入的用户名和密码
$child_name = $_POST["child_name"];
$child_password = $_POST["child_password"];

// 验证child用户名和密码是否正确
$sql = "SELECT * FROM child WHERE name='$child_name' AND password='$child_password'";
$result = mysqli_query($conn, $sql);

if (mysqli_num_rows($result) == 1) {
  // 如果验证成功，输出child的个人信息和疫苗接种信息
  $row = mysqli_fetch_assoc($result);
  $id = $row["id"];
  $parent_id = $row["parent_id"];
  $birthdate = $row["birthdate"];
  $sex = $row["sex"];
  $time = $row["time"];
  echo "ID: " . $id . "<br>";
  echo "Parent ID: " . $parent_id . "<br>";
  echo "Birthdate: " . $birthdate . "<br>";
  echo "Sex: " . $sex . "<br>";
  echo "Time: " . $time . "<br>";

  // 查询child的疫苗接种信息
  $sql = "SELECT vaccine.name, vaccine.date FROM vaccine JOIN vaccination ON vaccine.id=vaccination.vaccine_id WHERE vaccination.child_id='$id'";
  $result = mysqli_query($conn, $sql);
  if (mysqli_num_rows($result) > 0) {
    echo "<h2>Vaccination Records:</h2>";
    while ($row = mysqli_fetch_assoc($result)) {
      $name = $row["name"];
      $date = $row["date"];
      echo $name . " on " . $date . "<br>";
    }
  } else {
    echo "No vaccination record found!";
  }

} else {
  // 如果验证失败，提示child用户名或密码错误
  echo "Invalid username or password!";
}

mysqli_close($conn);

?>

注意：以上代码仅为示例，实际使用时需要根据具体需求进行修改和完善。同时，为了安全起见，应该对用户输入进行严格的验证和过滤，防止SQL注入等攻击。