Question

我正在尝试从数据库中进行选择但是我得到了这个错误：

解析错误：第46行的C：\ wamp \ www \ Project \ USSD_Misec \ test_USSD2 \ listener.php中的语法错误，意外的'echo'（T_ECHO）

这是我的代码：

<?php
    include "func/database.php";

    //$db = new Database();
    $db =new  Database('localhost', 'username', 'password', 'database_name');
    $db->connect();


    $data = file_get_contents("php://input"); 
    $line = "\n-------------------------------------------------------------\n";
    file_put_contents("logs/ussdgw.txt", $data."\n", FILE_APPEND);  

    $dataDecoded = json_decode($data);  

    $uri = array_reverse(explode("/", $_SERVER["REQUEST_URI"]));
    $usid = $uri[1];
    $action = $uri[0];



    $defaultmenu = array(
        "shouldClose" => false,
        "ussdMenu" => "Hello Welcome to Misec..., choose :
        1.Enrolment Enquiry 
        2.Enrolment Date 

        3.Exit",
        "responseExitCode" => 200,
        "responseMessage" => ""
        );

    switch($action)
    {
        case 'response':    

                switch($dataDecoded->text) 
                {
                    case 1:
                        $ussdMenu = "1. Enrolment Enquiry";
                        $q = "SELECT * FROM ussd ORDER BY id DESC";
                        $r = $db->query($q);
                        // if we have a result loop over the result
                        if($db->num_rows($r) > 0){
                            while($a = $db->fetch_array_assoc($r)){

                                echo "{$a['id']} wrote $a['name']\n";

                            }

                        }
                    break;
                    case 2:
                        $ussdMenu = "2. Enrolment Date";
                    break;
                }   
            $response = array(
                "shouldClose" => true,
                "ussdMenu" => $ussdMenu,
                "responseExitCode" => 200,
                "responseMessage" => ""
            );
            $r_encoded = json_encode($response);
            echo $r_encoded;
        break;

        case 'status':
            $sessions = file_get_contents("logs/sessions.txt");
                $sessionsActive = explode("\n", $sessions);
                    if(in_array($usid, $sessionsActive))
                        $sessionActive = true;
                    else 
                        $sessionActive = false;

                    $end = array(
                        "sessionActive" => $sessionActive,
                        "responseExitCode" => 200,
                        "responseMessage" => ""
                    );
                $endEncoded = json_encode($end);
                echo $endEncoded;   
        break;

        case 'start':
            echo json_encode($defaultmenu);
            file_put_contents("logs/sessions.txt", $usid."\n", FILE_APPEND);    
        break;

        case 'end':

            $sessions = file_get_contents("logs/sessions.txt");
                $sessionsActive = explode("\n", $sessions);
                    unset($sessionActive[$usid]);
                $sessionsModified = implode("\n", $sessionActive);
                file_put_contents("logs/sessions.txt", $sessionsModified); 


            $end = array(
                "responseExitCode" => 200,
                "responseMessage" => "Connection ended"
            );
            $endEncoded = json_encode($end);
            echo $endEncoded;
        break;
}

Answer 1

您已将其中一个变量括在大括号中，这是正确的，但不是第二个。更改以下行：

echo "{$a['id']} wrote $a['name']\n";

为：

echo "{$a['id']} wrote {$a['name']}\n";

解析错误：语法错误，意外＆＃39;回声＆＃39; （T_ECHO）在第46行的C：\ test_USSD2 \ listener.php中

1 个答案: