我如何将此数组转换为可用的javascript数组?

时间:2012-12-04 17:46:42

标签: php javascript mysql highcharts

我有从数据库中返回的PHP代码,

[
{
    "Time": "2012-11-27 16:10:35",
    "NumPlayers": "1"
},
{
    "Time": "2012-11-27 16:24:55",
    "NumPlayers": "1"
},
{
    "Time": "2012-11-27 16:25:37",
    "NumPlayers": "2"
},
{
    "Time": "2012-11-27 16:29:33",
    "NumPlayers": "2"

时间是MySQL时间戳。我需要使用这种格式与Highcharts javascript图表一起使用。

data: [
[Date.UTC(1970,  9,  9), 0   ],
[Date.UTC(1970,  9, 14), 0.15],
[Date.UTC(1970, 10, 28), 0.35],
[Date.UTC(1970, 11, 12), 0.46],

我无法弄清楚如何遍历MySQL结果并将时间戳转换为javascript Date.UTC对象。我需要将NumPlayers值放在Date.UTC对象之后,并以下面的格式输出。我是一个PHP和javascript noob:\

看起来应该是这样的:

data: [
[Date.UTC(2012, 11, 27, 16, 10, 35), 1],
[Date.UTC(2012, 11, 27, 16, 24, 55), 1],
[Date.UTC(2012, 11, 27, 16, 25, 37), 2],
[Date.UTC(2012, 11, 27, 16, 29, 33), 2],

5 个答案:

答案 0 :(得分:2)

您应该意识到Date.UTC(2012, 11, 27, 16, 10, 35)只返回自纪元(1356624635000)以来的毫秒数。因此,您可以将对象转换为UNIX时间戳(因为JS使用毫秒时间戳,因此PHP的工作时间为1000秒)。

<强> Sample Code

$data = '[{"Time": "2012-11-27 16:10:35", "NumPlayers": "1"}, {"Time": "2012-11-27 16:24:55", "NumPlayers": "1"}]';

// Make sure date is parsed as UTC
date_default_timezone_set("UTC");

// Convert items into the desired format
$mapper = function($item) {
    return array(strtotime($item->Time)*1000, $item->NumPlayers);
}

echo json_encode(array_map($mapper, json_decode($data)));

<强>输出

[[1354032635000,"1"],[1354033495000,"1"]]

答案 1 :(得分:1)

您似乎从数据库中直接获取JSON,您可以随时将其转换为数组:

$data = '[{"Time": "2012-11-27 16:10:35", "NumPlayers": "1"}, {"Time": "2012-11-27 16:24:55", "NumPlayers": "1"}]';
$arrayData = json_decode($data, true);

之后,您可以简单地遍历数组并以您需要的JS格式打印出数组的内容。这样的事情:

echo 'data: [' . PHP_EOL;
foreach ($arrayData as $item) {
    echo '[Date.UTC(';
    $tmp = preg_split( '/(-| |\:)/', $item['Time'] );
    echo implode(', ', $tmp);
    echo '), ' . $item['NumPlayers'] . '],';
    echo PHP_EOL;
}
echo PHP_EOL . '];';

答案 2 :(得分:1)

您可以split使用/[^\d]+/g

将时间字符串分成几个部分
var finalData = [], numbers=/[^\d]+/g, item, dateParts, newDate;
for (var i=0, l=data.length; i<l; i++) {
     item = data[i];
     dateParts = item.Time.split(numbers);
     newDate = Date.UTC.apply(Date, dateParts);
     finalData.push([newDate, item.NumPlayers]);
}

另请参阅:MDN's documentation on JavaScript regular expressions/[^\d]/g是正则表达式。)

答案 3 :(得分:1)

向Highcharts提供数据的最佳方式是从数据库中获取数据,这样您就不必在客户端进行数据。

要做到这一点,您只需将query更改为以下内容。

<强>后端
$query = "SELECT UNIX_TIMESTAMP(date) * 1000 AS 'date', value FROM ...";

然后将结果发送到前端:
echo json_encode($result);

<强>前端
要在前端获得结果,这个案例使用jQuery并假设url是你的网址:

$.getJSON(url, function(json) {
    // decode json
    var data = $.parseJSON(json);
    // Then you just have to pass it to your series
});

或者更好,以UTC格式存储日期值。

答案 4 :(得分:0)

    var php_data = [
        {
            "Time": "2012-11-27 16:10:35",
            "NumPlayers": "1"
        },
        {
            "Time": "2012-11-27 16:24:55",
            "NumPlayers": "1"
        }
    ];

    var length = php_data.length;
    hc_data = [];

    for(var i = 0; i< length; i++){
      php_date = new Date(php_data[i]["Time"]);
      hc_data.push(
        [
            Date.UTC(
                php_date.getFullYear(), 
                php_date.getMonth() + 1, 
                php_date.getDate()
            ),
            php_data[i]["NumPlayers"]
        ]
      );
    }

    // hc_data is your Array
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