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我正在调用 instagram api 并以 $record 中的数组形式获得响应。

我想知道如何从 $record 中获取 username、first_name、profile_picture、id、last_name 变量并将它们插入到 mysql db 中，如果它们还没有在 mysql db 中？谁能告诉我如何做到这一点？提前致谢。

根据Instagram API Docs以下 api：

https://api.instagram.com/v1/users/XXXX/follows?access_token==XXXX&count=-1

编辑：现在我可以打印数组变量，但如何将它们插入 mysql db 并避免重复？

foreach($record->data as $user) // each user data (JSON array) defined as $user
{
echo "User Name:".$user->username;
echo "<br>Bio:".$user->bio;
echo "<br>Website:".$user->website;
echo "<br>Profile Picture:".$user->profile_picture;
echo "<br>Full Name:".$user->full_name;
echo "<br>id:".$user->id;
echo "<br><br>";
}

回报：

{
    "data": [{
        "username": "washington",
        "first_name": "George",
        "profile_picture": "http://distillery.s3.amazonaws.com/profiles/profile_6623_75sq.jpg",
        "id": "1",
        "last_name": "Washington"
    },
    {
        "username": "SammyDavis",
        "first_name": "Sammy",
        "profile_picture": "http://distillery.s3.amazonaws.com/profiles/profile_29648_75sq_1294520029.jpg",
        "id": "29648",
        "last_name": "Davis"
    },
    {
        "username": "FrankTheTank",
        "first_name": "Frank",
        "profile_picture": "http://distillery.s3.amazonaws.com/profiles/profile_13096_75sq_1286441317.jpg",
        "id": "13096",
        "last_name": "Roberts"
    }]
}

php脚本：

$url = "https://api.instagram.com/v1/users/XXXX/follows?access_token==XXXX&count=-1";
$api_response = get_data(''.$url);
$record = json_decode($api_response); // JSON decode 

//now i want to get username,first_name,profile_picture,id,last_name variables and insert it into mysql db


/* gets the data from a URL */
function get_data($url) {
    $ch = curl_init();
    $timeout = 5;
    curl_setopt($ch, CURLOPT_URL, $url);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
    curl_setopt($ch, CURLOPT_CONNECTTIMEOUT, $timeout);
    $data = curl_exec($ch);
    curl_close($ch);
    return $data;
}
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1 回答 1

1 
foreach ($record->data as $rows){
    //now you can get vars from $rows
    $username = $rows->username;
}
于 2013-09-08T03:02:48.507 回答