Question

我正在PHP中对Web服务执行HTTP GET请求。我作为JSON得到回应。但是我在打印JSON中的一些值时遇到了一些错误，如inkey2＆amp; KEY3。我使用的GET请求是正确的。也没有语法错误。 URL是正确的，即使我在问题中指定的URL有点假。：）

我作为HTTP GET请求的响应获得的JSON

{
   "name":
   {
       "key1": "salala",          
       "key2":
       {              
           "inkey1": "hike",
           "inkey2":
           [
               {
                   "@sunny": "fake",
                   "@leone": "take"
               },
               {
                   "@sunny": "make",
                   "@leone": "bake"
               },
               {
                   "@sunny": "cake",
                   "@leone": "drake"
               },
               {
                   "@sunny": "sake",
                   "@leone": "lake"
               }
          ],
           "inkey3": "bike"
      },
      "key3":
       [
           "batman",
           "superman",
           "spiderman",
           "ironman",
           "hancock"
       ],
       "key4": "nike"
   }
}

// HTTP GET请求

$url='iwonttellyaguys.com';

$jsondata= httpGet($url);
$val_array = json_decode($jsondata, true);

echo'</br>';
echo'</br>';

//To print value salala (successful)

$print_salala=$val_array['name']['key1'];

echo'</br>';
echo'</br>';

//To print value hike & bike (successful)

$print_hike=$val_array['name']['key2']['inkey1'];
$print_bike=$val_array['name']['key2']['inkey3'];

echo'</br>';
echo'</br>';

//To print contents of key4. i.e, to print value nike (successful)

$print_nike=$val_array['name']['key4'];

echo'</br>';
echo'</br>';

//To print contents of inkey2 (Error)
//Incorrect value being printed.

$print_inkey2=$val_array['name']['key2']['inkey2'];

foreach($print_inkey2 as $key=>$value)
       {
          $sunny_leone=$value['@sunny']['@leone'];
          echo $sunny_leone;
       }
This foreach loop returns 3 as output...I am trying to print the contents, not the count.


//To print contents of key3 (Error)
//Incorrect value being printed.

echo'</br>';
echo'</br>';

$hero=$val_array['name']['key3'];
$count_hero=count($hero);

for($i=0;$i<$count_hero;$i++)
{
  echo $hero[$i];
}


// Function Definition of httpGet

function httpGet($url)
{
    $ch = curl_init();

    curl_setopt($ch,CURLOPT_URL,$url);
    curl_setopt($ch,CURLOPT_RETURNTRANSFER,true);
    // curl_setopt($ch,CURLOPT_HEADER, false);

    $output=curl_exec($ch);
    curl_close($ch);
    return $output;
}

为什么我无法打印key3＆amp;的内容？ inkey2 ???

Answer 1

只需调用javaScript函数，该函数将使用ajax lik将corrdiantes传递给PHP脚本

<script>
function sendCoords(_lat,_long){
  $.ajax({
url: "yourdomain.com/coords.php",
data:{lat:_lat,long:_long}
});
}
</script>

您的coords.php脚本应如下所示：

$lat = $_REQUEST['lat'];
$lang = $_REQUEST['long'];
///Do something

Answer 2

你应该使用：

$print_inkey2=$val_array['name']['key2']['inkey2'];

foreach($print_inkey2 as $key=>$value)
{
    $sunny_leone=$value['@sunny'].$value['@leone'];
    echo $sunny_leone;
}

这部分很好，并且在你的代码中工作。

$hero=$val_array['name']['key3'];
$count_hero=count($hero);

for($i=0;$i<$count_hero;$i++)
{
  echo $hero[$i];
}

希望它有所帮助。

Answer 3

这很好用：

foreach($printinkey2 as $key=>$value)
{
    print $key." @sunny ".$value['@sunny'].PHP_EOL;
    print $key." @leone ". $value['@leone'].PHP_EOL;
}

在PHP中打印JSON数据时出错？

3 个答案: