dongwuxie5112
2018-08-02 19:21
浏览 295
已采纳

JSON编码到PHP变量通过Curl失败并返回空响应

i have searched a lot on google and here but im not getting any further with my problem. I am not a coder though I am trying to parse JSON to PHP Variables, but i get an empty response, where i want a table to be shown or at least any jsondata

Here is what my code looks like:

<!DOCTYPE html>
<html>
<body>

<h1>Available Agents </h1>

<?php
$url = 'https://url/livewebservice/imoscontactagentstate?Username=username&Pwd=password&Cmd=GetState&ResultType=JSON';
//  Initiate curl
$ch = curl_init ($url);
$data = json_encode ($data,true);

curl_setopt ($ch, CURLOPT_CUSTOMREQUEST, "POST");
curl_setopt ($ch, CURLOPT_POSTFIELDS, $data);
curl_setopt ($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt ($ch, CURLOPT_HTTPHEADER, array (
' Content-Type: application/x-www-form-urlencoded ',
'charset=utf-8')
);
$result = curl_exec ($ch);
curl_close ($ch);
return $result;
var_dump(json_decode($result, true));
 print_r($result); 
foreach ($result as $key => $value)
              {
          echo '  <td><font  face="calibri"color="red">'.$value[type].'   </font></td><td><font  face="calibri"color="blue">'.$value[category].'   </font></td><td><font  face="calibri"color="green">'.$value[amount].'   </font></tr><tr>';

           }
           echo "</tr></table>";

?>



</body>
</html>

I am grateful for any hints

图片转代码服务由CSDN问答提供 功能建议

我在谷歌和这里搜索过很多但我没有进一步解决我的问题。 我不是编码器，虽然我试图解析JSON到PHP变量，但我得到一个空响应，我想要一个表显示或至少任何jsondata

这是什么 我的代码如下所示： 

 ＆lt;！DOCTYPE html＆gt; 
＆lt; html＆gt; 
＆lt; body＆gt; 
 
＆lt; h1＆gt;可用代理＆lt; / h1＆gt; \  n 
＆lt;？php 
 $ url ='https：// url / livewebservice / imoscontactagentstate？用户名=用户名＆amp; Pwd =密码＆amp; Cmd = GetState＆amp; ResultType = JSON'; 
 //启动curl 
 $ ch =  curl_init（$ url）; 
 $ data = json_encode（$ data，true）; 
 
 ncurl_setopt（$ ch，CURLOPT_CUSTOMREQUEST，“POST”）; 
curl_setopt（$ ch，CURLOPT_POSTFIELDS，$ data）; 
curl_setopt（$  ch，CURLOPT_RETURNTRANSFER，true）; 
curl_setopt（$ ch，CURLOPT_HTTPHEADER，array（
'Content-Type：application / x-www-form-urlencoded'，
'charset = utf-8'）
）; \  n $ result = curl_exec（$ ch）; 
curl_close（$ ch）; 
return $ result; 
var_dump（json_decode（$ result，true））; 
 print_r（$ result）;  
Nacach（$ result as $ key =＆gt; $ value）
 {
 echo'＆lt; td＆gt;＆lt; font face =“calibri”color =“red”＆gt;'。$ value [type]。'  ＆lt; / font＆gt;＆lt; / td＆gt;＆lt; td＆gt;＆lt; font face =“calibri”color =“blue”＆gt;'。$ value [category]。'  ＆lt; / font＆gt;＆lt; / td＆gt;＆lt; td＆gt;＆lt; font face =“calibri”color =“green”＆gt;'。$ value [amount]。'  ＆lt; / font＆gt;＆lt; / tr＆gt;＆lt; tr＆gt;'; 
 
} 
 echo“＆lt; / tr＆gt;＆lt; / table＆gt;”; 
 
？＆gt; 
 
 
  
＆lt; / body＆gt; 
＆lt; / html＆gt; 
   
 
 
我很感激任何提示
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2条回答 默认 最新

  • dongtao4319 2018-08-04 06:14
    已采纳

    The problem was solved my Username did not have the permission to access the data and we made minor changes to the code so it looks like this:

    php


$data = array(

"UserName" => "Username",

"Pwd" => "Password",

"Cmd" => "GetAgentStateList",

"ResultType" => "JSON",

);

$url='https://host/livewebservice/service/?'.http_build_query($data);

 echo $url;

$ch = curl_init();

curl_setopt($ch, CURLOPT_URL, $url);

curl_setopt($ch, CURLOPT_HEADER, false);

curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_HTTPHEADER, array( 
"Content-Type: application/x-www-form-urlencoded", "charset=UTF-8",)); 

$result = curl_exec($ch);

if(curl_errno($ch)){

    throw new Exception(curl_error($ch));

}

curl_close($ch);

$f_result=json_decode($result);

print_r($f_result);

?>

    i hope this helps someone coming from google someday.

    </div>
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  • dsepcxw181184853 2018-08-02 20:12

    Try to remove true from $data = json_encode ($data,true); as far as can i remember true is used only in json_decode to create an associative array

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