如何从PHP中的JSON服务器响应中获取数据

如何从PHP中的JSON服务器响应中获取数据,php,json,Php,Json,我用PHP发送了一个JSON请求,从服务器得到一个响应,如下所示: {"success":true,"result": {"items": [{"woj":"łódzkie","powiat":"kutnowski","gmina":"Bedlno","kod":"99- 311","miasto":"Adamów","id":"99-311Adamów167271172700"}]}, "error":null, "unAuthorizedRequest":false} $c

我用PHP发送了一个JSON请求,从服务器得到一个响应,如下所示:

{"success":true,"result":
{"items":
[{"woj":"łódzkie","powiat":"kutnowski","gmina":"Bedlno","kod":"99-        
311","miasto":"Adamów","id":"99-311Adamów167271172700"}]},
"error":null,
"unAuthorizedRequest":false}
$code = 'https://hetman.e4b.com.pl/api/services/app/kodPoczt/KodPocztInfo? 
 kod='.$kod;
 $ch = curl_init();
 //set the url, number of POST vars, POST data
 curl_setopt($ch,CURLOPT_URL, $code);
 curl_setopt($ch,CURLOPT_POST, true);
 curl_setopt($ch,CURLOPT_POSTFIELDS, $fields_string);
 curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); 
 //execute post
 $result1 = curl_exec($ch);
 //close connection
 curl_close($ch);
 echo $result1;
 $wynik  = json_decode($result1, true);
 echo $wynik['result'][0]['woj'];   
我想从“woj”和“Poviat”中挑选一个内容,但它不起作用

我的代码如下所示:

{"success":true,"result":
{"items":
[{"woj":"łódzkie","powiat":"kutnowski","gmina":"Bedlno","kod":"99-        
311","miasto":"Adamów","id":"99-311Adamów167271172700"}]},
"error":null,
"unAuthorizedRequest":false}
$code = 'https://hetman.e4b.com.pl/api/services/app/kodPoczt/KodPocztInfo? 
 kod='.$kod;
 $ch = curl_init();
 //set the url, number of POST vars, POST data
 curl_setopt($ch,CURLOPT_URL, $code);
 curl_setopt($ch,CURLOPT_POST, true);
 curl_setopt($ch,CURLOPT_POSTFIELDS, $fields_string);
 curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); 
 //execute post
 $result1 = curl_exec($ch);
 //close connection
 curl_close($ch);
 echo $result1;
 $wynik  = json_decode($result1, true);
 echo $wynik['result'][0]['woj'];   

显示您的完整代码。。。在您的示例中,curl没有初始化,return\u transfer没有设置。$result1的输出是什么

反正是

$wynik['result']['items'][0]['woj'];
$temp='1{ “成功”:没错, “结果”:{ “项目”:[{ “woj”:“łódzkie”, “波维亚特”:“库特诺夫斯基”, “gmina”:“Bedlno”, “kod”:“99-311”, “miasto”:“Adamów”, “id”:“99-311Adamów167271172700” }] }, “错误”:空, “未经授权的请求”:错误 }';

    $result = json_decode($temp,true);
    echo $result['result']['items'][0]['woj'];

您的字符串不是有效的json。它是-您只需删除一些空格@Andreas-无论如何,如果您需要删除某些内容以使其工作/有效,则其结尾的回显是错误的,这不意味着在更改之前它是无效的吗?为什么在发布Json时首先使用空格使其无效。