我试图理解这种错误插入SQL值,但我不如何解决呢 这是我的JS代码:无法通过AJAX的PHP
$.ajax({
url: 'C:\\inetpub\\wwwroot\\VisionwareHelp\\Php/CriaUserEempresa.php',
type: "POST",
data: ({Pname: Pname, Uname: Uname, email: email, Ename: Ename, Sigla: Sigla}),
complete:function(data)
{
resposta = data;
console.log(resposta);
}
});
这是我的PHP代码:
$serverName = $server;
$uid = $uid;
$pwd = $pass;
$connectionInfo = array("UID" => $uid, "PWD" => $pwd,"Database"=>"Portal");
//$connectionInfo = array("Database"=>"Portal");
$conn = sqlsrv_connect($serverName, $connectionInfo);
$Pname = $_POST["Pname"];
$Uname = $_POST["Uname"];
$email = $_POST["email"];
$Ename = $_POST["Ename"];
$Sigla = $_POST["Sigla"];
if($conn)
{
$sqlCliente = "INSERT INTO Portal.dbo.Empresa VALUES ($Ename, $Sigla)";
if (mysqli_query($conn, $sqlCliente)) {
echo "New record created successfully";
} else {
echo "Error: " . $sqlCliente . "<br>" . mysqli_error($conn);
}
}
什么是错我的代码?
,没有u得到任何错误? – user3040610
与此相关的HTML表单呢? –
您的查询中也有语法错误。 'mysqli_error($ conn);'应该把东西扔给你;它是什么?我知道它是什么,我们需要“你”来告诉我们。哦,你可以在这里看到一些主要的SQL注入。 –