php + curlでhttpリクエストを含むjsonデコード

Question

私はPHPで作業しています。私はhttp要求をibeaconクラウドに送信したい&クラウドからいくつかのデータを取得&は、それをmysqlデータベースに保存します。

Answer 1

はこのような何かしてみてください：以下のような

<?php include "connection.php"; 
    $ch = curl_init();//instead of $ch = curl init(); 
    curl_setopt($ch,CURLOPT_URL, "https://cloud.estimote.com/v1/beacons"); 
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, true); 
    curl_setopt($ch, CURLOPT_USERPWD, "achopra-smu-edu-sg-s-your iuw:10e891270177480e7443148908d0afa3"); 
    curl_setopt($ch, CURLOPT_HTTPHEADER, array("Accept: application/json")); 
    $output = curl_exec($ch); 

    print_r($output); 
    //perform task based on response 


     $json=json_decode($output,true));//based on key available in response. 
     $major = $json['major']; 
      $battery = $json['battery']; 
      $AdvertisingInterval = $json['Advertising Interval']; 
      $BroadcastingPower = $json['BroadcastingPower']; 
      $value = "INSERT INTO 
     battery_level(major,battery,AdvertisingInterval,BroadcastingPower)VALUES('".$major."''".$battery."''".$AdvertisingInterval."''".$BroadcastingPower."')"; 
    echo $value; 
    $result = mysqli_query($conn,$value); 
    if(!result) { die ("could not insert data : " . PHP_EOL.mysqli_connect_errno()); 
        }else{ 

echo "Insert data successfully\n"; curl_close($ch); ?> 
} 

Answer 2

するTry代を...あなたの正確な質問が

$data = array("name" => "Hagrid", "age" => "36");                  
$data_string = json_encode($data);                     

$ch = curl_init('http://api.local/rest/users');                  
curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "POST");                  
curl_setopt($ch, CURLOPT_POSTFIELDS, $data_string);                 
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);                  
curl_setopt($ch, CURLOPT_HTTPHEADER, array(                   
    'Content-Type: application/json',                     
    'Content-Length: ' . strlen($data_string))                  
);                             

$result = curl_exec($ch);