我想在html文本框中输入数字,并检查该ID的用户是否存在于数据库Delovni_Cas中,如果它存在echo Success否则失败。我设法让它在没有.$value.
的情况下工作,只是手动输入数字,如1,2,3,…
<html>
<head>
<form method="get">
<INPUT TYPE = "Text" value="1" NAME = "IDU">
<INPUT TYPE = "Submit">
</form>
<?php
$servername = "localhost";
$username = "root";
$password = "perkowich";
$conn = new mysqli($servername, $username, $password);
$value = $_POST['IDU'];
$query = mysqli_query($conn, "SELECT * FROM Delovni_Cas WHERE Zaposleni_ID_Osebe='".$value."'");
if(mysqli_num_rows($query) > 0) {
echo "Sucess";
} else {
echo "Failed";
}
$conn->close();
?>
<html>
<head>
</head>
</html>
您在表单中使用Get方法而在php中使用$_POST,请使用$_GET而不是$_POST来获取表单值