大家好我是從一個URL獲取值並將其傳遞到更新語句,當我把WHERE ID = 1時,它工作正常,但是當我把ID = $ ID,代碼工作,但沒有及時更新,記錄保持不變,可有的幫我解決這個問題,請更新記錄,通過從網址獲取一個ID
<?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=".$id);
header ("Location: index.php");
}
else
echo "failed";
}
else
echo "Image file size max 1 MB";
}
else
echo "Invalid file format..";
}
else
echo "Please select image..!";
exit;
}
}
}
echo"error";
}
你檢查了什麼是executin查詢 –
你很容易受到[SQL注入攻擊](http://bobby-tables.com)。你有沒有確認你的'$ _GET ['id']'是你認爲的?就php而言,'example.com?ID = foo'和'example.com?id = foo'是兩個完全不同的查詢字符串。 –
「WHERE ID = 1有效」OK,那麼爲什麼如果字段是'ID',爲什麼要使用'WHERE MovieID ='?這是什麼? – MrCode