我正在尝试使用PHP从以下JSON文件中获取数据。我特别想要“temperatureMin”和“temperatureMax”。这可能很简单,但我不知道该怎么做。我坚持在file_get_contents(“file.json”)之后做什么。一些帮助将不胜感激!{ "daily": { "summary": "No precipitation for the week; temperatures rising to 6° on Tuesday.", "icon": "clear-day", "data": [ { "time": 1383458400, "summary": "Mostly cloudy throughout the day.", "icon": "partly-cloudy-day", "sunriseTime": 1383491266, "sunsetTime": 1383523844, "temperatureMin": -3.46, "temperatureMinTime": 1383544800, "temperatureMax": -1.12, "temperatureMaxTime": 1383458400, } ] }}
3 回答
绝地无双
TA贡献1946条经验 获得超4个赞
使用json_decode将您的JSON转换为PHP数组。例:
$json = '{"a":"b"}';
$array = json_decode($json, true);
echo $array['a']; // b
慕仙森
TA贡献1827条经验 获得超7个赞
Try:
$data = file_get_contents ("file.json");
$json = json_decode($data, true);
foreach ($json as $key => $value) {
if (!is_array($value)) {
echo $key . '=>' . $value . '<br/>';
} else {
foreach ($value as $key => $val) {
echo $key . '=>' . $val . '<br/>';
}
}
}
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