Question

我查看过以前的问题，但无济于事。我有一个PHP脚本，我从请求流中提取信息并尝试进行更新：

<?php
include 'database_conn.php';

$pCDTitle = filter_has_var(INPUT_GET, 'title') ? $_GET['title']: null;
$pPrice = filter_has_var(INPUT_GET, 'price') ? $_GET['price']: null;
$pYear = filter_has_var(INPUT_GET, 'year') ? $_GET['year']: null;
$pcdID = filter_has_var(INPUT_GET, 'cdID') ? $_GET['cdID']: null;

$pCDTitle = isset($_REQUEST['title']) ? $_REQUEST['title'] : null;
$pYear = isset($_REQUEST['year']) ? $_REQUEST['year'] : null;
$pPrice = isset($_REQUEST['price']) ? $_REQUEST['price'] : null;
$pcdID = isset($_REQUEST['cdID']) ? $_REQUEST['cdID'] : null;


$pCDTitle = trim($pCDTitle);
$pYear = trim($pYear);
$pPrice = trim($pPrice);

$pCDTitle = filter_var($pCDTitle, FILTER_SANITIZE_STRING);
$pCDTitle = filter_var($pCDTitle, FILTER_FLAG_NO_ENCODE_QUOTES);
$pCDTitle = filter_var($pCDTitle, FILTER_SANITIZE_FULL_SPECIAL_CHARS);

$pYear = filter_var($pYear, FILTER_SANITIZE_STRING);
$pYear = filter_var($pYear, FILTER_FLAG_NO_ENCODE_QUOTES);
$pYear = filter_var($pYear, FILTER_SANITIZE_FULL_SPECIAL_CHARS);

$pPrice = filter_var($pPrice, FILTER_SANITIZE_STRING);
$pPrice = filter_var($pPrice, FILTER_FLAG_NO_ENCODE_QUOTES);
$pPrice = filter_var($pPrice, FILTER_SANITIZE_FULL_SPECIAL_CHARS);

$sql = "UPDATE `nmc_cd` SET `CDTitle` = '$pCDTitle', `CDYear` = '$pYear',     `CDPrice` = $pPrice WHERE CDID=$pcdID";

mysqli_query($conn, $sql)
or die(mysqli_error($conn));

if(mysqli_affected_rows($conn)>0)
{
    echo "Update succesful. Affected rows:".mysqli_affected_rows($conn);
    echo "<a href='administrator.php'>Return to Admin Area</a>";
}
else

mysqli_close($conn);

?>

我无法查看代码或SQL语法的任何问题，但我一直收到此错误：

＆＃34;您的SQL语法出错了;查看与您的MySQL服务器版本对应的手册，以获得正确的语法，以便在＆＃39; WHERE CDID = 657＆＃39;附近使用。在第1行＆＃34;

任何人都可以协助我出错的地方。我也试过把＆＃39; 1＆＃39;在WHERE之后？

mysql更新错误 - ＆＃34;您的SQL语法错误＆＃34;

0 个答案: