将字符串拆分为两个变量 - 未定义不是函数

Question

我正在使用一个 for 循环，里面有一个 foreach 循环。在 foreach 循环中，我得到我需要的信息并将其回显到这样的字符串：19:00^-2,2° C。我想将时间（19:00）放在一个变量中，将温度（-2,2° C）放在另一个变量中，这样我就可以将它们用于the chart。但是我不知道如何将这两个拼接起来。这是目前的样子：

var yr = localStorage.yr.split('|');
time = yr[16].split('~');
temp = time.split('^');

lineChartData = {
    labels: time,

    datasets: [{
        label: "My Second dataset",
        fillColor: "rgba(151,187,205,0.2)",
        strokeColor: "rgba(151,187,205,1)",
        pointColor: "rgba(151,187,205,1)",
        pointStrokeColor: "#fff",
        pointHighlightFill: "#fff",
        pointHighlightStroke: "rgba(151,187,205,1)",
        data: temp
    }]
}

使用此代码，我在temp = time.split('^'); 上得到Uncaught TypeError: undefined is not a function。 yr[16] 打印以下内容：

19:00^-2,2° C~20:00^-2° C~21:00^-2° C~22:00^-1,9° C~23:00^-1,6° C~00:00^-1,5° C~01:00^-1,3° C~02:00^-1,1° C~03:00^-0,9° C~04:00^-0,9° C~05:00^-1,1° C~06:00^-1,3° C~07:00^-1,6° C~

PHP 代码如下所示：

for($i = 0; $i < 13; $i++) {
    $datetime           = date('Y-m-d\TH', strtotime('+'.$i.' hour'));
    $forecast_period    = $forecast->xpath('(//product/time[contains(@datatype, "forecast")][contains(@from, "'.$datetime.':00:00Z")][contains(@to, "'.$datetime.':00:00Z")])');

    foreach($forecast_period AS $period) {
        $period_datetime                    = date('H:i', strtotime($period->attributes()->from));
        $period_temperature             = $period->location->temperature->attributes()->value;
        $period_temperature_dewpoint        = $period->location->dewpointTemperature->attributes()->value;
        $period_temperature_unit            = $period->location->temperature->attributes()->unit;
        $period_wind_direction              = $period->location->windDirection->attributes()->name;
        $period_wind_direction_degrees      = $period->location->windDirection->attributes()->deg;
        $period_wind_speed                  = $period->location->windSpeed->attributes()->mps;
        $period_fog                     = $period->location->fog->attributes()->percent;
        $period_cloudiness                  = $period->location->cloudiness->attributes()->percent;
        $period_cloudiness_low              = $period->location->lowClouds->attributes()->percent;
        $period_cloudiness_medium           = $period->location->mediumClouds->attributes()->percent;
        $period_cloudiness_high         = $period->location->highClouds->attributes()->percent;
        $period_pressure                    = $period->location->pressure->attributes()->value;
        $period_pressure_unit               = $period->location->pressure->attributes()->unit;
        $period_humidity                    = $period->location->humidity->attributes()->value;

        echo $period_datetime.'^'.temp($period_temperature, $period_temperature_unit, 'not-normal').'~';
    }
}

我通过simplexml_load_file() 从YR's API 获取信息（$forecast 是使用simplexml_load_file() 从 API 获取 XML 数据的人）。

话虽如此，我的问题是：如何将yr[16] 中的时间和温度放入两个变量中而不会出现任何错误，从而构建图表？

Answer 1

你可以使用Array.prototype.map:

var time_and_temp = time.map(function(el) {
    var temp = el.split('^');
    return { time: temp[0], temperature: temp[1] };
});

这会将time_and_temp 设置为如下数组：

[ { time: "19:00",
    temperature: "-2,2° C"
  },
  { time: "20:00",
    temperature: "-2° C"
  }
  ...
]

Answer 2

yr16 = yr[16].toString();

time = yr16.split('^')[0];

temp = yr16.split('^')[1];

您需要在.split('^') 之后包含[0] 和[1]，以指示将哪个部分（^ 之前或之后）分配给变量。