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如何访问JSON元素

[英]How to Access JSON Element

I have one json i wanted to print "formatted_address" element from that json 我有一个json,我想从该json打印“ formatted_address”元素

Json 杰森

{
   "html_attributions" : [],
   "results" : [
      {
         "formatted_address" : "Narayan Peth, Pune, Maharashtra 411030, India",
         "geometry" : {
            "location" : {
               "lat" : 18.515797,
               "lng" : 73.852335
            }
         },
         "icon" : "https://maps.gstatic.com/mapfiles/place_api/icons/shopping-71.png",
         "id" : "3a25975b3806df28aa79ac4a8d954c307be4aa57",
         "name" : "Aditya Medical",
         "place_id" : "ChIJJxwmOHDAwjsRjRDO4LnGJ-I",
         "reference" : "CmRSAAAA3E7ih55-2BZjRQcw_URQ2gwi8eWb5HU6hdfNUj_TqtDJ7TtASVMowcuWMkohNjp7F6UKuGsMuR-IlzZEt4YUJyzNxzWg-TYy6hyN8P5n2asAO6ztZeU3oHZdH7OBFFW_EhBe4cQbAU99oILcDmvv_gOhGhR7jzP0Z9-mDrncd5Gr9hOY7aOqRg",
         "types" : [ "pharmacy", "health", "store", "point_of_interest", "establishment" ]
      }
   ],
   "status" : "OK"
}

I tired to print using but unable to print it. 我厌倦了使用来打印,但是无法打印。

foreach (json_decode($address[0]->Response) as $obj){
     print_r($obj['results']['formatted_address']);
}

You need to set second param as true to get json as array. 您需要将第二个参数设置为true才能将json作为数组。 Also your formatted_address is agin inarray so need to pass index in it 另外您的formatted_address是agin inarray,因此需要在其中传递索引

foreach (json_decode($address[0]->Response, true) as $obj){
     print_r($obj['results'][0]['formatted_address']);
}

DEMO 演示

That can be a solution : 那可以是一个解决方案:

$jsonAsArray = json_decode($yourJson, true);
$results = $array["results"][0];

var_dump($results['formatted_address']);

Good luck 祝好运

Try 尝试

foreach (json_decode($address[0]->Response, true) as $obj){
     print_r($obj['results']['formatted_address']);
}

json_decode has a second parameter to determine the returned result format - object or array . json_decode具有第二个参数,用于确定返回的结果格式- objectarray

I Try like @Plamen Nikolov, but but it doesn't work. 我尝试像@Plamen Nikolov一样,但不起作用。 And I Try change 'True' to 1 it's work! 然后我尝试将“ True”更改为1,就可以了!

foreach (json_decode($address[0]->Response, 1) as $obj){
     print_r($obj['results']['formatted_address']);
}

Try 尝试

foreach (json_decode($address[0]->Response)->results as $obj){
     print_r($obj->formatted_address);
}

json_decode($address[0]->Response) will give you an object. json_decode($ address [0]-> Response)将为您提供一个对象。 You should not use like array. 您不应该使用like数组。 so you should use "->" instead of array form. 因此,您应该使用“->”而不是数组形式。 Better you can put the result on foreach and get the data of formatted_address from that $obj 更好的是,您可以将结果放在foreach上,并从该$ obj获取formatted_address的数据

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