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我有如下代码。我正在阅读一个 JSON URL 并用一些 if 条件回显一些项目。我需要以 JSON 格式重新回显所选项目。

<?php
// Array of trains to list
//Arrival train list
$trainNumbers = array(
    9021,11077
);

$json      = file_get_contents('myURL.json');
$trainData = json_decode($json, true);
foreach ($trainData[0] as $train) {

    $trainNumber = $train[0][0];
    if (in_array($trainNumber, $trainNumbers)) {
        $fields = array(
            'train_no',
            'train_name',
            'dep_date',
            'dep_station',
            'dep_log',
            'dep_lat',
            'arr_station',
            'delay_time',
            'new_lat',
            'new_long',
            'new_station',
            'new_station_name',
            'time_delay',
            'station_left'
        );
        foreach ($train[0] as $i => $dataField) {
            echo $fields[$i] . " - {$dataField}\n";
            $trains[$trainNumber][$fields[$i]] = $datafield;
        }
        echo "\n";

    }
}
?>

上面的代码显示数据如

train_no - 09021
train_name - MUMBAI BANDRA T - JAMMU TAWI Exp (SPL)
dep_date - 2013-05-06
dep_station - BRSQ
dep_log - 28.613196
dep_lat - 77.14046
arr_station - BRAR SQUARE
delay_time - 150
new_lat - 28.659977
new_long - 77.156425
new_station - UMB
new_station_name - AMBALA CANT JN
time_delay - 48
station_left - 67

train_no - 11077
train_name - PUNE - JAMMU TAWI Jhelum Express
dep_date - 2013-05-06
dep_station - HET
dep_log - 26.611628
dep_lat - 77.943449
arr_station - HETAMPUR
delay_time - 56
new_lat - 26.697312
new_long - 77.905769
new_station - DHO
new_station_name - DHAULPUR
time_delay - 44
station_left - 93

如何以 JSON 格式再次回显输出？

按照以下建议进行编辑

$data[$fields[$i]]= " - {$dataField}\n";
      echo json_encode($data);

我收到此错误

在此处输入图像描述

这是 JSON 输出

http://pastebin.com/7EeVg8X9

实际上，在这个脚本中，我们有 100 列火车的列表，所以我从中调用了一些火车。

4

2 回答 2

1

目前，您的问题是您将每个映射的“火车”作为单独的项目输出。为了使输出表示有效的 JSON，至少应将单独的元素收集到一个数组中。这是您的程序的略微修改版本：

<?php
// Fields to be translated
$fields = array(
    'train_no',
    'train_name',
    'dep_date',
    'dep_station',
    'dep_log',
    'dep_lat',
    'arr_station',
    'delay_time',
    'new_lat',
    'new_long',
    'new_station',
    'new_station_name',
    'time_delay',
    'station_left'
);

// Array of trains to list
//Arrival train list
$trainNumbers = array(
    9021,11077
);

$json      = file_get_contents('myurl.json');
$trainData = json_decode($json, true);

foreach ($trainData[0] as $train) {

    $trainNumber = $train[0][0];
    if (in_array($trainNumber, $trainNumbers)) {

        foreach ($train[0] as $i => $dataField) {
            $data[$fields[$i]]= $dataField;
        }

        $translated[] = $data;
    }
}

echo json_encode($translated);
?>

哪些输出（使用jsonlint格式化后）：

[
    {
        "train_no": "09021",
        "train_name": "MUMBAI BANDRA T - JAMMU TAWI Exp (SPL)",
        "dep_date": "2013-05-06",
        "dep_station": "HUK",
        "dep_log": "28.801247",
        "dep_lat": "77.102394",
        "arr_station": "HOLAMBI KALAN",
        "delay_time": "172",
        "new_lat": "28.846516",
        "new_long": "77.085357",
        "new_station": "UMB",
        "new_station_name": "AMBALA CANT JN",
        "time_delay": 70,
        "station_left": 64
    },
    {
        "train_no": "11077",
        "train_name": "PUNE - JAMMU TAWI Jhelum Express",
        "dep_date": "2013-05-06",
        "dep_station": "BHA",
        "dep_log": "27.06879445",
        "dep_lat": "77.96653748",
        "arr_station": "BHANDAI",
        "delay_time": "59",
        "new_lat": "27.157722",
        "new_long": "77.989883",
        "new_station": "AGC",
        "new_station_name": "AGRA CANTT",
        "time_delay": 45,
        "station_left": 92
    }
]

您可以选择使用不同类型的 JSON 输出（例如，一个根对象，每个火车都以名称为键，或其他任何内容）。只需扩展如上所示的相同原理即可生成有效的 JSON。

于 2013-05-07T12:33:53.927 回答
1

使用 json_encode 将数据编码为 json 

      $data[$fields[$i]]= $dataField;
      echo json_encode($data);

您将从这里获得更多帮助

于 2013-05-07T06:33:20.610 回答