Question

我正在尝试使用php来检查我的数据库以查看是否存在值。我的主要目标是使用此值

$_GET['UDID']

如果它等于数据库中的任何值，它将返回

echo 'FOUND';

我正在使用此代码：

<?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**';
          }
    }
}

?>

但由于某种原因，它无法正常工作，我相信我一直在寻找一些东西。非常感谢您的帮助。

Answer 1

试试这个：

$sql = mysqli_query($connect, "SELECT udid FROM data WHERE udid = '" .$udid. "'");

另外，请务必设置＆＃39; GET＆＃39; $ udid。应该是这样的：

$udid = $_GET['UDID'];

我们可以使用mysqli_fetch_array（）来获取结果行。我还包括错误处理。现在你的代码必须如下所示：

$udid = $_GET['UDID'];
$id = mysqli_real_escape_string($connect, $udid);

$result = mysqli_query($connect, "SELECT `udid` FROM `wmaystec_WMT-SS`.`data` = '$id'");

if($result === FALSE) {
   die(mysqli_error("error message for the user")); //error handling
}
else {
    while ($row = mysqli_fetch_array($result)) { 
          echo "FOUND :" .$row['thefieldnameofUDIDfromyourDB'];
    }
}

Answer 2

我建议你首先使用mysqli_real_escape_string函数转义字符串，然后调用SQL查询。

RowSpan="Auto"

检查SQL数据库的值

2 个答案: