这是我的代码至今如何在一个php if语句中运行2个sql查询?
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')";
}
这两个$ sql语句很好地工作在那里自己,但是当我在if语句有两似乎绕过第一个,只要运行最后$ SQL声明。
我怎样才能让它运行尽可能多的$ sql语句,因为我需要它....将有大约15 - 20在那里。
在此先感谢。
根据要求提供更多代码。
$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>" . $conn->error;
}
那么因为你发布的代码实际上并没有运行任何SQL语句,所以也许你应该显示不起作用的真实代码 – 2015-03-31 09:05:34
当你设置“插入”查询而不做任何事情时,你正在覆盖$ sql值$ sql =“DELETE FROM UpcomingRota”; – Olvathar 2015-03-31 09:05:37
您需要执行查询。在这里,你只需要在查询中设置'$ sql'变量。 – D4V1D 2015-03-31 09:07:04