Question

我有这个网址

try {
    $ossClient = new OssClient($accessKeyId, $accessKeySecret, $endpoint);
    $ossClient->multiuploadFile($bucket, $object, $filePath);
} catch (OssException $e) {
    print $e->getMessage();
}

那是前2个的json输出（例如）

 https://poloniex.com/public?command=returnTicker

现在我希望获得所有数据到变量。喜欢... $ curreny = xxx; $ last = xxx，..

我已经尝试过了..

 {"BTC_BCN":{"id":7,"last":"0.00000018","lowestAsk":"0.00000018","highestBid":"0.00000017","percentChange":"0.00000000","baseVolume":"58.73610647","quoteVolume":"328275043.97652394","isFrozen":"0","high24hr":"0.00000019","low24hr":"0.00000017"},"BTC_BELA":{"id":8,"last":"0.00001191","lowestAsk":"0.00001191","highestBid":"0.00001174","percentChange":"0.13536701","baseVolume":"6.33896473","quoteVolume":"572949.02508918","isFrozen":"0","high24hr":"0.00001200","low24hr":"0.00001033"},

适用于第一级。然后我得到所有货币的清单。但是......我如何才能将所有其他数据输入货币？

第二个foreach？喜欢..

$url = "https://poloniex.com/public?command=returnTicker";
$response = file_get_contents($url);
$obj = json_decode($response,true);
$keys = array_keys((array)$obj);

foreach($keys as $result) {
$marketname = $result; }

我认为

没有意义！？

我想以那种形式出现

$ currency = BTC_BCN; $ last = 0.00003223; 等等..

有什么想法吗？

Answer 1

您不需要使用array_keys 您可以使用foreach来跟踪密钥。

//$json = file_get_contents($url);
$json = '{"BTC_BCN":{"id":7,"last":"0.00000018","lowestAsk":"0.00000018","highestBid":"0.00000017","percentChange":"0.00000000","baseVolume":"58.73610647","quoteVolume":"328275043.97652394","isFrozen":"0","high24hr":"0.00000019","low24hr":"0.00000017"},"BTC_BELA":{"id":8,"last":"0.00001191","lowestAsk":"0.00001191","highestBid":"0.00001174","percentChange":"0.13536701","baseVolume":"6.33896473","quoteVolume":"572949.02508918","isFrozen":"0","high24hr":"0.00001200","low24hr":"0.00001033"}}';

$arr = json_decode($json, true);

Foreach($arr as $key => $val){
    Echo "in ".$key . " last is ". $val['last'] ."\n";
}

输出是：

in BTC_BCN last is 0.00000018
in BTC_BELA last is 0.00001191

https://3v4l.org/UpkOm

<小时/> 如果要将数组中的项目用作变量，可以提取变量我自己不使用这种方法，因为值无论如何都在数组中我添加了这种方法，因为你的问题意味着它是你想要的在每次迭代时，变量都将被覆盖，这就是为什么不推荐这种方法的原因如果其中一个项目没有＆＃34; last＆＃34;所有其他变量都将被迭代中的新数据覆盖，除了＆＃34; last＆＃34;。
这意味着您同时在变量中同时拥有新旧数据我的建议是使用数组而不是提取变量。

$json = '{"BTC_BCN":{"id":7,"last":"0.00000018","lowestAsk":"0.00000018","highestBid":"0.00000017","percentChange":"0.00000000","baseVolume":"58.73610647","quoteVolume":"328275043.97652394","isFrozen":"0","high24hr":"0.00000019","low24hr":"0.00000017"},"BTC_BELA":{"id":8,"last":"0.00001191","lowestAsk":"0.00001191","highestBid":"0.00001174","percentChange":"0.13536701","baseVolume":"6.33896473","quoteVolume":"572949.02508918","isFrozen":"0","high24hr":"0.00001200","low24hr":"0.00001033"}}';

$arr = json_decode($json, true);

Foreach($arr as $key => $val){
    Extract($val);
    Echo $key ."\n";
    Echo $last ."\n";
    Echo $baseVolume ."\n\n";
}

https://3v4l.org/aZv1s

Answer 2

auth0.users.updateUserMetadata

您已经完成了获取信息的所有工作，只需要使用它：D

使用<?php $url = "https://poloniex.com/public?command=returnTicker"; $response = file_get_contents($url); $obj = json_decode($response,true); $keys = array_keys((array)$obj); for($i=0;$i<count($keys);$i++) { echo 'Currency '.$keys[$i].', last '.$obj[$keys[$i]]['last'].' '; } ?>查看数组中包含的内容及其具有的索引。

PHP - 从url到变量的json_decode

2 个答案: