Question

所以我尝试将新节点发布到json文件。

这是我的js：

        var newDate = new Date;             
        var markerId = newDate.getTime();
        var markerData = { 'id': markerId, 'lat':markerId, 'long':markerId };


        $.ajax({ 
            type: 'POST',
            url: "dataPath.php",
            data: {
                    marker: markerData
                },
            dataType: 'json',
            async: false,
            success: function(result)
                {
                    alert("Added OK");
                }               
        });

这是php文件

$dataPath = 'file_path';
$markerDataFile = 'adauga.json';
$markerText = file_get_contents($markerDataFile);
$markerList = json_decode($markerText,true);
 if( !empty($_POST['marker'])  ){

    $markerData =  $_POST['marker'];
    $markerData['ip'] = $_SERVER['REMOTE_ADDR'];
    $markerData['created'] = time();
    $markerList['markers'][] = $markerData;
    $markerText = json_encode($markerList);
    file_put_contents($markerDataFile, $markerText);
    echo json_encode($markerData);
}else{
    echo "Invalid request";

问题是JSON文件显示：

{"markers":[{"id":"1310499027672","lat":"47.1405","long":"7.243839999999977","ip":"127.0.0.1","created":1310499032},"1object Object]","1object Object]","1object Object]","1object Object]","1object Object]","1object Object]","1object Object]","1object Object]"]}

Answer 1

好像你的$markerData有其他嵌套对象需要显式读取并显式添加到json中。您需要定位这些对象的属性并逐个添加它们。

如果你要回复$_POST['marker']，你会得到[object Object]，但是如果你这样打印出来的话：

echo "<pre>";
print_r($_POST['marker']);
echo "</pre>";

然后您将看到所有标记对象和属性，您可以将它们单独添加到您的json中，或者至少在for循环中引用它们。

JSON添加到文件问题。仅显示对象对象

1 个答案: