Question

任何能够帮助解决这个令人讨厌的错误的人都试着等了30分钟。

<?php
if(isset($_POST['statusBtn']))
{
    $username = htmlspecialchars($_POST['address']);
    strtolower($username);
    if(empty($username))
    {
        echo '<center><div class="alert alert-icon alert-danger alert-dismissible fade in col-md-7" role="alert"><button type="button" class="close" data-dismiss="alert" aria-label="Close">
<span aria-hidden="true">&times;</span></button><i class="mdi mdi-check-all"></i>Fill in all fields</div></center>';
    } else if($username == "" || $username == ""){
        echo '<center><div class="alert alert-icon alert-danger alert-dismissible fade in col-md-7" role="alert"><button type="button" class="close" data-dismiss="alert" aria-label="Close"><span aria-hidden="true">&times;</span></button><i class="mdi mdi-check-all"></i>Domain Whitelisted</div></center>';
    } else {
        $geo = @file_get_contents("http://ip-api.com/json/{$username}");
        $json_a = json_decode($geo, true);
        $proxyid = $json_a['status'];
        if($proxyid == success)
        {
            echo '<center><div class="alert alert-icon alert-success alert-dismissible fade in col-md-7" role="alert"><button type="button" class="close" data-dismiss="alert" aria-label="Close"><span aria-hidden="true">&times;</span></button><i class="mdi mdi-check-all"></i>'.htmlspecialchars($username).' is Online!</div></center>';
        } else {
            echo '<center><div class="alert alert-icon alert-danger alert-dismissible fade in col-md-7" role="alert"><button type="button" class="close" data-dismiss="alert" aria-label="Close"><span aria-hidden="true">&times;</span></button><i class="mdi mdi-check-all"></i>'.htmlspecialchars($username).' is Offline!</div></center>';
        }
    }
}
?>

当我以表格形式提交google.com时，它会显示结果以及错误

注意：使用未定义的常数成功 - 假设成功＆＃39;

Answer 1

将if($proxyid == success)更改为if($proxyid == 'success')

错误告诉您success没有' '表示这是一个常量，而不是字符串。

注意：使用未定义的常量成功 - 假设成功＆＃39;

1 个答案: