我想用php來檢查我的數據庫,看看是否存在一個值。我的主要目標就是利用這個值檢查SQL數據庫的值
$_GET['UDID']
,如果它等於在數據庫中的任何值將返回
echo 'FOUND';
我使用這個代碼:
<?php
$servername = "*****";
$username = "*****";
$password = "*****";
$dbname = "*****";
$connect = new mysqli($servername, $username, $password, $dbname);
if ($connect->connect_error) {
die("CONNECTION FAILED: " . $connect->connect_error);
}
$udid = $_GET['UDID'];
$id = mysqli_real_escape_string($connect, $udid);
$result = mysqli_query($connect, "SELECT udid FROM data WHERE udid = '$id'");
if($result === FALSE) {
die("ERROR: " . mysqli_error($result));
}
else {
while ($row = mysqli_fetch_array($result)) {
if($row['udid'] == $udid) {
$results = 'Your device is already registered on our servers.';
$results2 = 'Please click the install button below.';
$button = 'Install';
$buttonlink = 'https://**link here**';
}
else {
$results = 'Your device is not registered on our servers';
$results2 = 'Please click the request access button below.';
$button = 'Request Access';
$buttonlink = 'https://**link here**';
}
}
}
?>
但由於某種原因,它不能正常工作,我相信我正在尋找一些東西。非常感謝您的幫助。
$ udid = mysqli_real_escape_string($ connect,$ _ GET ['UDID']); –
@Jixone OK ....上面的代碼現在正在工作,但現在我需要它來檢查它是否存在,以及它是否做了某些事情。我更新了問題代碼。 –
@WillMays我的代碼是否顯示正確的結果? – Jixone