Question

我有这个包含JSON数据的php文件

<?php
header('Content-type: application/json');

echo'{"ID":"9999","RSSI":-48,"Time":"","sensors":[
{"Type":"AirFlow","Unit":"Analog","Val":0},
{"Type":"Temperature","Unit":"C","Val":28.65},
{"Type":"SkinConduct","Unit":"microSiemens","Val":-1.00},
{"Type":"SkinResist","Unit":"Ohm","Val":-1.00},
{"Type":"SkinConductVolt","Unit":"V","Val":0.49},
{"Type":"HeartRate","Unit":"BPM","Val":0},
{"Type":"02 Saturation","Unit":"%","Val":0},
{"Type":"BodyPosition","Unit":"^<>_|","Value":3}]}
{"Type":"ElectroCardioGram","Unit":"Analog","Val":3.78},
';

?>

我希望它被另一个php文件读取，请求其数据，解码JSON并将值HeartRate和ElectroCardioGram（Val）存储在两个变量中。我找不到任何领导 - 我应该从哪里开始研究？

我设法使用此代码获取数据 `

$myfile = fopen("sensorjson.php", "r") or die("Unable to open file!");
echo fread($myfile,filesize("sensorjson.php"));
fclose($myfile);

我想我可以解析fopen和fclose之间的json来显示更可理解的结果？

Answer 1

查看file_get_contents从服务器（或同一服务器）获取文件并将其放入字符串，然后您可以使用json_decode来操作数据。

PHP - 解析JSON数据/ HTTP请求

1 个答案: