我開發一個Web應用程序,允許用戶通過Web表單條目添加到MySQL數據庫MySQL的條目。 Web表單的帖子在同一個頁面,一些PHP代碼捕獲數據並將其發送到MySQL。無論出於何種原因,任何事情都不會讓它成爲MySQL。這是我的代碼至今不工作:添加從PHP網頁
<?php
include("config.inc.php");
$Name=$_POST['AddName'];
$Group=$_POST['AddGroup'];
$Grade=$_POST['AddGrade'];
$Position=$_POST['AddPosition'];
$Email=$_POST['AddEmail'];
$HomeAddress=$_POST['AddHomeAddress'];
$City=$_POST['AddCity'];
$State="SC";
$Zip=$_POST['AddZIP'];
$CellPhone=$_POST['AddCellNumber'];
$HomePhone=$_POST['AddHomeNumber'];
$FirstPeriod=$_POST['AddFirstPeriod'];
$SecondPeriod=$_POST['AddSecondPeriod'];
$ThirdPeriod=$_POST['AddThirdPeriod'];
$FourthPeriod=$_POST['AddFourthPeriod'];
$FifthPeriod=$_POST['AddFifthPeriod'];
$SixthPeriod=$_POST['AddSixthPeriod'];
$SeventhPeriod=$_POST['AddSeventhPeriod'];
$Homeroom=$_POST['AddHomeroom'];
$dbpassword=$_POST['AddPassword'];
$con = mysql_connect("127.0.0.1","$username","******");
if (!$con)
{
die();
}
mysql_select_db("$database", $con);
$sql="INSERT INTO names (Name,Grp,Grade,Position,Email,HomeAddress,City,State,Zip,CellPhone,HomePhone,FirstPeriod,SecondPeriod,ThirdPeriod,FourthPeriod,FifthPeriod,SixthPeriod,SeventhPeriod,Homeroom)
VALUES('$Name','$Group','$Grade','$Position','$Email','$HomeAddress,'$City','$State','$Zip','$CellPhone','$HomePhone','$FirstPeriod','$SecondPeriod','$ThirdPeriod','$FourthPeriod','$FifthPeriod','$SixthPeriod','$SeventhPeriod','$Homeroom')";
if (!mysql_query($sql,$con))
{
die();
}
echo "1 record added";
mysql_close($con)
?>
</body>
</html>
建議:羅斯閱讀有關SQL注入 –
二忠告:永遠不會離開密碼,當您發佈的代碼! –
用你的'die()'迴應一些錯誤信息,你可能會得到一個線索是什麼問題...... – weltraumpirat