如何从PHP中的JSON服务器响应中获取数据
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: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
{"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。