0

MYSQL テーブルにレコードを挿入するときに少し問題が発生しました。データは正常に処理されますが、コードがクエリを実行してエントリが作成されたかどうかを確認すると、エラーが発生し続けます。

$sql = mysqli_query($connect,"INSERT INTO testtable (fnhusband,fnwife,surname,email,address,suburb,city,postcode,region,number,altnumber)
VALUES ('$fnhusband','$fnwife','$surname','$email','$address','$suburb','$city','$postcode','$region','$number','$altnumber')");

if (!mysqli_query($connect,$sql))
  {
  die('Error:' . mysqli_error());
  }
else { 
    echo "1 record added";
    mysqli_close($connect);
    }
4

1 に答える 1

2

$connect がリンクであると仮定します。

$connect = new mysqli("localhost", "my_user", "my_password", "world");

変更してみてください

$sql = mysqli_query($connect,"INSERT INTO testtable (fnhusband,fnwife,surname,email,address,suburb,city,postcode,region,number,altnumber)
VALUES ('$fnhusband','$fnwife','$surname','$email','$address','$suburb','$city','$postcode','$region','$number','$altnumber')");
$result = mysqli_query($connect,$sql);

に：

$sql = "INSERT INTO testtable (fnhusband,fnwife,surname,email,address,suburb,city,postcode,region,number,altnumber)
VALUES ('$fnhusband','$fnwife','$surname','$email','$address','$suburb','$city','$postcode','$region','$number','$altnumber')";
$result = $connect->query($sql);

または手続き型：

$sql = "INSERT INTO testtable (fnhusband,fnwife,surname,email,address,suburb,city,postcode,region,number,altnumber)
VALUES ('$fnhusband','$fnwife','$surname','$email','$address','$suburb','$city','$postcode','$region','$number','$altnumber')";
$result = mysqli_query($connect, $sql);
于 2013-03-07T00:23:39.847 に答える