通过从网址获取ID更新记录

Question

大家好，我通过get从URL中获取值并将其传递到更新语句中，当我输入WHERE ID = 1时，它可以正常工作，但是当我输入ID = $ id时，代码可以工作，但是没有更新，记录保持不变，请帮我解决这个问题

   <?php
   require 'db2.php';
    $id = null;
   if ( !empty($_GET['id'])) {
    $id = $_REQUEST['id'];

 $dbc = mysqli_connect (DB_HOST, DB_USER, DB_PASSWORD, DB_NAME) OR die ('Could not connect to MySQL: ' . mysqli_connect_error() );
    $q = mysqli_query($dbc,"SELECT * FROM movie WHERE MovieID = '$id' ");
   while($r=mysqli_fetch_array($q))
   {   
    $title = $r["Title"];
    $tag = $r["Tag"];
    $year = $r["YEAR"];
    $cast = $r["Cast"];
    $comment = $r["Comment"];
    $IDBM = $r["IMDB"];
  }


  }


 if (!empty($_POST) ) {
 if ( !empty($_GET['id'])) {
    $id = $_REQUEST['id'];

    // keep track post values
    $cast = $_POST['cast'];
    $title = $_POST['title'];
    $comment =$_POST['comment'];
     $year = $_POST['year'];
      $tag = $_POST['tags'];
       $IDBM = $_POST['idbm'];
    $cast = htmlspecialchars($cast);
    $title = htmlspecialchars($title);
    $comment = htmlspecialchars($comment);

    // validate input
    $valid = true;
    if (empty($cast)) {
        $castError = 'Please enter Cast';
        $valid = false;
    }

    if (empty($title)) {
        $titleError = 'Please enter Title';
        $valid = false;
    }
      if (empty($comment)) {
        $commentError = 'Please enter Comment';
        $valid = false;
    }


    if ($valid) {

    $path = "uploads/";



$valid_formats = array("jpg", "png", "gif", "bmp");
if(isset($_POST) and $_SERVER['REQUEST_METHOD'] == "POST")
    {
        $name = $_FILES['photoimg']['name'];
        $size = $_FILES['photoimg']['size'];

        if(strlen($name))
            {
                list($txt, $ext) = explode(".", $name);
                if(in_array($ext,$valid_formats))
                {
                if($size<(1024*1024))
                    {
                        $actual_image_name = time().substr(str_replace(" ", "_", $txt), 5).".".$ext;
                        $tmp = $_FILES['photoimg']['tmp_name'];
                        if(move_uploaded_file($tmp, $path.$actual_image_name))
                            {

                                mysqli_query($dbc,"UPDATE movie SET Title='$title',Year = '$year',Cast='$cast',Cover='$actual_image_name',Tag='$tag',Comment='$comment',IMDB ='$IDBM' WHERE MovieID=".$id);
                                header ("Location: index.php");
                            }
                        else
                            echo "failed";
                    }
                    else
                    echo "Image file size max 1 MB";                    
                    }
                    else
                    echo "Invalid file format..";   
            }

        else
            echo "Please select image..!";

        exit;
    }


    }
    }
    echo"error";
    }

Answer 1

听起来您的MovieID可能没有定义为整数，但我们无法确定，因为您没有告诉我们mysqli_query抛出的错误消息。

您需要检查由mysqli_query创建的错误消息才能知道。 见http://www.php.net/manual/en/mysqli.error.php

Answer 2

尝试这个

$id = $_GET['id']; // taking the value from URL 

mysqli_query($dbc,"UPDATE movie SET Title='$title',Year = '$year',Cast='$cast',Cover='$actual_image_name',Tag='$tag',Comment='$comment',IMDB ='$IDBM' WHERE MovieID=".$id); // the sql statement of the query

最好通过使用intval()防止注入来保护get

 $id = intval($_GET['id']); // taking the value from URL 

Answer 3

这个怎么样：

$ id = strip_tags（intval（$ _ GET ['id']））;

mysqli_query（$ dbc，“ UPDATE`movie` SET`Title` ='{$ title}'，`Year` ='{$ year}'，`Cast` ='{$ cast}'，`Cover` ='{ $ actual_image_name}'，`Tag` ='{$ tag}'，`Comment` ='{$ comment}'，`IMDB` ='{$ IDBM}'在哪里`MovieID` ='{$ id}';“ ）;

要验证$ id是否具有相同的值：

echo $ id;