Question

json发送到服务器php文件作为来自ibm worklight Http适配器的发布请求即使解码后也无法访问：
以下是代码：

Http适配器：

function storeRegistrationData(emailVal, passVal, fname1, gender, phone, userType, vehType) {
    var credentials = JSON.stringify({jemailVal: emailVal, jpassVal: passVal, jfname1: fname1, jgender: gender, jphone: phone, juserType : userType, jvehType : vehType});

    var input = {
        method : 'post',
        returnedContentType : 'json',
        path : "/carbikepooling/index.php",
        parameters: {credentials: credentials}
    };
    return WL.Server.invokeHttp(input);
}

服务器php代码：

$jsonObj = (isset($_POST['credentials']) ? $_POST['credentials'] : null);
        $credentials = json_decode($jsonObj);
        $email = $credentials->jemailVal;
        $pass  = $credentials->jpassVal;
        $uname = $credentials->jfname1;
        $gender = $credentials->jgender;
        $phone = $credentials->jphone;
        $usertype = $credentials->juserType;
        $vehtype = $credentials->jvehType;
        $boolean = false;
        $userId; $userTypeId;

        // sanitation, database calls, etc

        $connection = mysqli_connect("localhost","root","","carbikepooling");


                     if (mysqli_connect_errno($connection))
                     {
                      echo "Failed to connect to MySQL: " . mysqli_connect_error();
                     }

                    mysqli_query($connection,"insert into userInfo values(0,".$email.",'".$pass."','".$uname."','".$gender."','".$phone."')");

                        $result1 = mysqli_query($connection, "select u.userId from userInfo u order by u.userId desc limit 1");

                        if($result1){
                            $row1 = mysqli_fetch_row($result1);
                            $userId = $row1[0];
                            mysqli_query($connection,"insert into userType values(0,".$userId.",'".$usertype."')");
                            $result2 = mysqli_query($connection, "select t.userTypeId from userType t order by t.userTypeId desc limit 1");

                            if($result2){
                                    $row2 = mysqli_fetch_row($result2);
                                    $userTypeId = $row2[0];
                                    mysqli_query($connection, "insert into vehicleType values(0,".$userTypeId.",'".$vehtype."')");
                                    $boolean = true;
                                }
                        }

                    mysqli_close($connection);  

        $returnData[] = array(
            'boolean' => "$boolean"
        );

        echo json_encode($returnData);
?>

显示来自服务器php的返回结果的Javascript代码：

function storeFormData(emailVal, passVal, fname1, gender, phone, userType, vehType){

    var invocationData = {
            adapter : 'registerUser',
            procedure : 'storeRegistrationData',
            parameters : [emailVal, passVal, fname1, gender, phone, userType, vehType]
    };

    WL.Client.invokeProcedure(invocationData, {
        onSuccess : storeFormDataSuccess,
        onFailure : storeFormDataFailure,
    });
}

function storeFormDataSuccess(result){
    WL.Logger.debug('Data stored Successfully');
    alert(JSON.stringify(result.invocationResult));
}

function storeFormDataFailure(result){
    WL.Logger.error('Error in storing Data');
}

当我在insert语句中使用服务器php文件中的解码json数据时，没有任何反应。我尝试了电子邮件，性别等的回显值，但没有打印出来，好像它们不包含任何值。但是，如果我在returnData数组中发送了电子邮件，性别等的已解码json值，因为它们在insert语句中使用，则由发出请求的应用程序成功接收的值，即：在js代码中的警报中显示这些值。
请解决这个问题？

Answer 1

嗯，这很奇怪。我复制并粘贴了您的确切代码，我能够回显从Worklight适配器传递的变量。也许您尝试将变量用于sql语句的方式是错误的？

你能改变你的mysql查询看起来更像这样吗？

mysqli_query($connection, "insert into vehicleType values(0, '$userTypeId','$vehtype')");

请注意我是如何删除连接的'。'从查询和我现在我只是用单引号包围变量。

echo在服务器php中发布了json数据

1 个答案: