166

我有一个返回 JSON 对象的 URL，如下所示：

{
    "expires_in":5180976,
    "access_token":"AQXzQgKTpTSjs-qiBh30aMgm3_Kb53oIf-VA733BpAogVE5jpz3jujU65WJ1XXSvVm1xr2LslGLLCWTNV5Kd_8J1YUx26axkt1E-vsOdvUAgMFH1VJwtclAXdaxRxk5UtmCWeISB6rx6NtvDt7yohnaarpBJjHWMsWYtpNn6nD87n0syud0"
}

我想从 URL 中获取 JSON 对象，然后是access_token值。

那么如何通过 PHP 检索它呢？

4

11 回答 11

395 
$json = file_get_contents('url_here');
$obj = json_decode($json);
echo $obj->access_token;

为此，file_get_contents需要allow_url_fopen启用它。这可以在运行时通过包括：

ini_set("allow_url_fopen", 1);

您也可以使用curl获取网址。要使用 curl，您可以使用此处找到的示例：

$ch = curl_init();
// IMPORTANT: the below line is a security risk, read https://paragonie.com/blog/2017/10/certainty-automated-cacert-pem-management-for-php-software
// in most cases, you should set it to true
curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_URL, 'url_here');
$result = curl_exec($ch);
curl_close($ch);

$obj = json_decode($result);
echo $obj->access_token;
于 2013-03-25T14:33:30.103 回答
28 
$url = 'http://.../.../yoururl/...';
$obj = json_decode(file_get_contents($url), true);
echo $obj['access_token'];

PHP 也可以使用带有破折号的属性：

garex@ustimenko ~/src/ekapusta/deploy $ psysh
Psy Shell v0.4.4 (PHP 5.5.3-1ubuntu2.6 — cli) by Justin Hileman
>>> $q = new stdClass;
=> <stdClass #000000005f2b81c80000000076756fef> {}
>>> $q->{'qwert-y'} = 123
=> 123
>>> var_dump($q);
class stdClass#174 (1) {
  public $qwert-y =>
  int(123)
}
=> null
于 2013-03-25T14:35:27.997 回答
18

你可以使用 PHP 的json_decode函数：

$url = "http://urlToYourJsonFile.com";
$json = file_get_contents($url);
$json_data = json_decode($json, true);
echo "My token: ". $json_data["access_token"];
于 2013-03-25T14:36:32.713 回答
8

你需要阅读关于 json_decode 函数http://php.net/manual/en/function.json-decode.php

干得好

$json = '{"expires_in":5180976,"access_token":"AQXzQgKTpTSjs-qiBh30aMgm3_Kb53oIf-VA733BpAogVE5jpz3jujU65WJ1XXSvVm1xr2LslGLLCWTNV5Kd_8J1YUx26axkt1E-vsOdvUAgMFH1VJwtclAXdaxRxk5UtmCWeISB6rx6NtvDt7yohnaarpBJjHWMsWYtpNn6nD87n0syud0"}';
//OR $json = file_get_contents('http://someurl.dev/...');

$obj = json_decode($json);
var_dump($obj-> access_token);

//OR 

$arr = json_decode($json, true);
var_dump($arr['access_token']);
于 2013-03-25T14:37:31.813 回答
8 
// Get the string from the URL
$json = file_get_contents('https://maps.googleapis.com/maps/api/geocode/json?latlng=40.714224,-73.961452');

// Decode the JSON string into an object
$obj = json_decode($json);

// In the case of this input, do key and array lookups to get the values
var_dump($obj->results[0]->formatted_address);
于 2015-11-19T23:06:31.650 回答
3 
$ch = curl_init();
curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_URL, 'url_here');
$result = curl_exec($ch);
curl_close($ch);

$obj = json_decode($result);
echo $obj->access_token;
于 2016-01-11T07:31:04.673 回答
2

file_get_contents()没有从 url 获取数据，然后我尝试curl了它，它工作正常。

于 2017-04-28T05:27:14.093 回答
1

我们的解决方案，为响应添加了一些验证，因此我们确定我们在$json变量中有一个格式良好的 json 对象

$ch = curl_init();
curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_URL, $url);
$result = curl_exec($ch);
curl_close($ch);
if (! $result) {
    return false;
}

$json = json_decode(utf8_encode($result));
if (empty($json) || json_last_error() !== JSON_ERROR_NONE) {
    return false;
}
于 2018-12-18T17:28:47.330 回答
1

我的解决方案仅适用于以下情况：如果您将多维数组误认为是一个数组

$json = file_get_contents('url_json'); //get the json
$objhigher=json_decode($json); //converts to an object
$objlower = $objhigher[0]; // if the json response its multidimensional this lowers it
echo "<pre>"; //box for code
print_r($objlower); //prints the object with all key and values
echo $objlower->access_token; //prints the variable

我知道答案已经得到了回答，但是对于那些来这里寻找东西的人，我希望这可以帮助你

于 2019-04-01T20:02:39.190 回答
0 
$curl_handle=curl_init();
curl_setopt($curl_handle, CURLOPT_URL,'https://www.xxxSite/get_quote/ajaxGetQuoteJSON.jsp?symbol=IRCTC&series=EQ');
//Set the GET method by giving 0 value and for POST set as 1
//curl_setopt($curl_handle, CURLOPT_POST, 0);
curl_setopt($curl_handle, CURLOPT_CUSTOMREQUEST, "GET");
curl_setopt($curl_handle, CURLOPT_CONNECTTIMEOUT, 2);
curl_setopt($curl_handle, CURLOPT_RETURNTRANSFER, 1);
$query = curl_exec($curl_handle);
$data = json_decode($query, true);
curl_close($curl_handle);

//print complete object, just echo the variable not work so you need to use print_r to show the result
echo print_r( $data);
//at first layer
echo $data["tradedDate"];
//Inside the second layer
echo $data["data"][0]["companyName"];

有时您可能会得到 405，请正确设置方法类型。

于 2019-11-01T11:47:39.047 回答
0

当您使用curl时有时会给您 403（禁止访问）通过添加此行来模拟浏览器来解决。

curl_setopt($ch, CURLOPT_USERAGENT, 'Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1; .NET CLR 1.0.3705; .NET CLR 1.1.4322)');

希望这对某人有所帮助。

于 2018-07-06T09:46:08.413 回答