這是我的代碼至今如何在一個php if語句中運行2個sql查詢?
if (isset($_POST['button1']))
{
$sql = "DELETE FROM UpcomingRota";
$E1 = $_POST["MondayAMFirstEmployee"]; $E2 = $_POST["MondayAMSecondEmployee"]; $E3 = $_POST["MondayAMThirdEmployee"];
$sql = "INSERT INTO UpcomingRota (DayAndTime, FirstEmployee, SecondEmployee, ThirdEmployee) VALUES ('MondayAM', '$E1', '$E2', '$E3')";
}
這兩個$ sql語句很好地工作在那裏自己,但是當我在if語句有兩似乎繞過第一個,只要運行最後$ SQL聲明。
我怎樣才能讓它運行儘可能多的$ sql語句,因爲我需要它....將有大約15 - 20在那裏。
在此先感謝。
根據要求提供更多代碼。
$servername = "db568845851.db.1and1.com";
$username = "dbo568845851";
$password = "";
$dbname = "db568845851";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
if (isset($_POST['button1']))
{
$sql = "DELETE FROM UpcomingRota";
$E1 = $_POST["MondayAMFirstEmployee"]; $E2 = $_POST["MondayAMSecondEmployee"]; $E3 = $_POST["MondayAMThirdEmployee"];
$sql = "INSERT INTO UpcomingRota (DayAndTime, FirstEmployee, SecondEmployee, ThirdEmployee) VALUES ('MondayAM', '$E1', '$E2', '$E3')";
}
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
那麼因爲你發佈的代碼實際上並沒有運行任何SQL語句,所以也許你應該顯示不起作用的真實代碼 – 2015-03-31 09:05:34
當你設置「插入」查詢而不做任何事情時,你正在覆蓋$ sql值$ sql =「DELETE FROM UpcomingRota」; – Olvathar 2015-03-31 09:05:37
您需要執行查詢。在這裏,你只需要在查詢中設置'$ sql'變量。 – D4V1D 2015-03-31 09:07:04