如何在PHP中将这部分代码转换为准备好的语句？

Question

陈述如何将部分代码转换为准备好的语句？edit.phpif (isset($_GET['id']))      {          $id = $_GET['id'];          $update = true;          $sql = "SELECT * FROM crimenews WHERE crimenews_id=$id";          $query = mysqli_query($conn,$sql);        if(mysqli_num_rows($query) == 1)          {              $row = mysqli_fetch_array($query);              $category = $row['crimenews_cat'];              $url = $row['crimenews_url'];              $datetime = $row['crimenews_datetime'];              $lat = $row['crimenews_locationLat'];              $lng = $row['crimenews_locationLong'];          }      }上面的给定链接中已经提供了更多代码。

Answer 1

// while testing add

ini_set('display_errors', 1); 

ini_set('log_errors',1); 

error_reporting(E_ALL); 

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

if (isset($_GET['id']))

{

    $update = true;

    $sql = "SELECT * FROM crimenews WHERE crimenews_id=?";

    $stmt = $conn->prepare($sql);

    $stmt->bind_param('i', $_GET['id']);

    $stmt->execute();

    $res = $stmt->get_result();

    if($res->num_rows == 1)

    {

        // fetch_assoc() returns just an assoc array and not the numeric array as well

        $row = $res->fetch_assoc();

        $category = $row['crimenews_cat'];

        $url = $row['crimenews_url'];

        $datetime = $row['crimenews_datetime'];

        $lat = $row['crimenews_locationLat'];

        $lng = $row['crimenews_locationLong'];

    }

}