Question

我有一个错误：

意外的令牌D

当我尝试将JSON从JavaScript发送到PHP时。我知道错误是由服务器端引起的，我试图修复它，但后来我又收到了Unexpected input，Unexpected end等类似错误。

我试图在1周内修复它并且没有结果。我的目标是将数据发送到服务器并使服务器显示它没有错误，我该怎么办？

我的客户代码：

$("#sendRoute").live('click', function () {
    trackCoords_str = JSON.stringify(trackCoords);
    final_time_m_str = JSON.stringify(final_time_m);
    final_time_s_rounded_str = JSON.stringify(final_time_s_rounded);
    aver_speed_km_h_rounded_str = JSON.stringify(aver_speed_km_h_rounded);
    total_km_rounded_str = JSON.stringify(total_km_rounded);
    $.ajax({
        url: "http://test.whirlware.biz/server/index.php",
        type: "POST",
        data: {
            route: trackCoords_str,
            timeInMinutes: final_time_m_str,
            timeInSeconds: final_time_s_rounded_str,
            averageSpeed: aver_speed_km_h_rounded_str,
            distance: total_km_rounded_str,
        },
        dataType: "json",
        success: function (){alert("success!");},
        error: function (xhr, ajaxOptions, thrownError) {
            alert(xhr.responseText);
            alert(thrownError);
        }
    });
});

我的服务器代码：

 $route = $_POST['route'];
 $timeInMinutes = $_POST['timeInMinutes'];
 $timeInSeconds = $_POST['timeInSeconds'];
 $averageSpeed = $_POST['averageSpeed'];
 $distance = $_POST['distance'];

 $trackCoords1 = json_encode($route);
 $final_time_m1 = json_encode($timeInMinutes);
 $final_time_s_rounded1 = json_encode($timeInSeconds);
 $aver_speed_km_h_rounded1 = json_encode($averageSpeed);
 $total_km_rounded1 = json_encode($distance);

 echo "Distance: </br>"; echo $distance;
 echo "Time in minutes: </br>"; echo $timeInMinutes;
 echo "Time in seconds: </br>"; echo $timeInSeconds;
 echo "Average speed: </br>"; echo $averageSpeed;

Answer 1

您的错误是：

意外的令牌D

然后转到your site’s URL,输出就是：

Distance: </br>Time in minutes: </br>Time in seconds: </br>Average speed: </br>

D的“意外令牌”是指输出的第一个字母，即距离。

我不清楚PHP代码的目标是什么。您是否期望当它连接到该PHP代码时，它会获取数据并将其添加到数据库中？

现在意味着您的路径：

JavaScript Ajax通过$ _POST - ＆gt; PHP解析$ _POST - ＆gt;那个PHP 脚本是在回应内容吗？

echo与此过程有什么关系？当你只注释掉echo s？

时会发生什么？

让我们来看看你的代码：

$route = $_POST['route'];
$timeInMinutes = $_POST['timeInMinutes'];
$timeInSeconds = $_POST['timeInSeconds'];
$averageSpeed = $_POST['averageSpeed'];
$distance = $_POST['distance'];

$trackCoords1 = json_encode($route);
$final_time_m1 = json_encode($timeInMinutes);
$final_time_s_rounded1 = json_encode($timeInSeconds);
$aver_speed_km_h_rounded1 = json_encode($averageSpeed);
$total_km_rounded1 = json_encode($distance);

echo "Distance: </br>"; echo $distance;
echo "Time in minutes: </br>"; echo $timeInMinutes;
echo "Time in seconds: </br>"; echo $timeInSeconds;
echo "Average speed: </br>"; echo $averageSpeed;

这到底是什么意思？所以在第一个中你要分配变量：

$route = $_POST['route'];
$timeInMinutes = $_POST['timeInMinutes'];
$timeInSeconds = $_POST['timeInSeconds'];
$averageSpeed = $_POST['averageSpeed'];
$distance = $_POST['distance'];

好的，然后在下一个变量中使用json_encode：

$trackCoords1 = json_encode($route);
$final_time_m1 = json_encode($timeInMinutes);
$final_time_s_rounded1 = json_encode($timeInSeconds);
$aver_speed_km_h_rounded1 = json_encode($averageSpeed);
$total_km_rounded1 = json_encode($distance);

那么这些是什么？

echo "Distance: </br>"; echo $distance;
echo "Time in minutes: </br>"; echo $timeInMinutes;
echo "Time in seconds: </br>"; echo $timeInSeconds;
echo "Average speed: </br>"; echo $averageSpeed;

你现在用那些json_encode变量做什么？你的脚本的目标是什么？这简直就是一团糟。

也许您的PHP文件中的代码就是您的服务器代码：

$json_output = array();
$json_output['route'] = $_POST['route'];
$json_output['timeInMinutes'] = $_POST['timeInMinutes'];
$json_output['timeInSeconds'] = $_POST['timeInSeconds'];
$json_output['averageSpeed'] = $_POST['averageSpeed'];
$json_output['distance'] = $_POST['distance'];

echo json_encode($json_output);

编辑此外，查看您的JavaScript代码，您的JSON中有一个尾随逗号。并且JSON规范中不允许使用as explained on this question & answer thread。所以我会改变这个：

data: {
    route: trackCoords_str,
    timeInMinutes: final_time_m_str,
    timeInSeconds: final_time_s_rounded_str,
    averageSpeed: aver_speed_km_h_rounded_str,
    distance: total_km_rounded_str,
},

成为这样;请注意删除total_km_rounded_str,后的逗号，使其仅为total_km_rounded_str：

data: {
    route: trackCoords_str,
    timeInMinutes: final_time_m_str,
    timeInSeconds: final_time_s_rounded_str,
    averageSpeed: aver_speed_km_h_rounded_str,
    distance: total_km_rounded_str
},

Answer 2

客户端希望获得JSON，因此您需要发回有效的JSON ...和仅有效的JSON。

这意味着你真的应该只打一次json_encode。你的代码应该更像是：

echo json_encode(array(
    "distance" => $distance,
    "time in minutes" => $timeInMinutes
    //etc.
));

或者只是使用HTML而不是JSON并删除客户端上的dataType: json行。

Answer 3

删除此处的逗号

distance: total_km_rounded_str
                              ^ HERE

PHP + Javascript错误：意外的令牌D

3 个答案: