php - 如何使用参数和回调从 PHP 调用 JSON Web 服务？

Question

我从 JQuery $.getJSON 函数调用网络服务，它工作正常。

    var p = {
       'field1': 'value1',
       'field2': 'value2',
       'field3': 'value3'
    };

    $.getJSON('https://service:xxx@xxx.xxx.yyyyy.com.xx/service/search?callback=?', p, function(data) {
    if (data[0]) {      
        // print results
    } else  {
        // no results found
    }
});

我正在尝试从 PHP 和 CURL 进行连接，但是它不起作用，它总是返回 false。

//第一次尝试

$params = array(  'field1' => 'value1',  'field2' => 'value2', 'field3'=> 'value3');
$ch = curl_init();
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_POST, 1);
curl_setopt($ch, CURLOPT_URL, 'https://service:xxx@xxx.xxx.yyyyy.com.xx/service/search?callback=?');
curl_setopt($ch, CURLOPT_POSTFIELDS, $params);
$result = curl_exec($ch); // return false instead of my JSON

//第二次尝试

    $data_string = json_encode($params);                                                                                   
    $ch = curl_init('https://https://service:xxx@xxx.xxx.yyyyy.com.xx/service/search?callback=?');                                                                      
    curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "POST");                                                                     
    curl_setopt($ch, CURLOPT_POSTFIELDS, $data_string);                                                                  
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);                                                                      
    curl_setopt($ch, CURLOPT_HTTPHEADER, array(                                                                          
    'Content-Type: application/json',                                                                                
    'Content-Length: ' . strlen($data_string))                                                                       
);                                                                                                                   

    $result2 = curl_exec($ch); //return false instead of my JSON

我做错了什么？

非常感谢，

Answer 1

尝试将您的代码更改为:

$params = array(  'field1' => 'value1',  'field2' => 'value2', 'field3'=> 'value3');

$data_string = implode('&',$params);
//NB: you may need to urlencode the each of your params

$ch = curl_init('https://service:xxx@xxx.xxx.yyyyy.com.xx/service/search?  callback=?&' .$data_string);                                                                      
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);    
$result2 = curl_exec($ch);

未经测试的代码，希望对您有所帮助。