在 PHP 中构建 HTTP POST 查询时,可以使用一个称为 http_build_query 的简单方法,该方法基于传递给函数的数组返回以下内容:
简单数组:
Array
(
[0] => foo
[1] => bar
[2] => baz
[3] => boom
[cow] => milk
[php] => hypertext processor
)
返回值:
flags_0=foo&flags_1=bar&flags_2=baz&flags_3=boom&cow=milk&php=hypertext+processor
一个稍微复杂的数组:
Array
(
[user] => Array
(
[name] => Bob Smith
[age] => 47
[sex] => M
[dob] => 5/12/1956
)
[pastimes] => Array
(
[0] => golf
[1] => opera
[2] => poker
[3] => rap
)
[children] => Array
(
[bobby] => Array
(
[age] => 12
[sex] => M
)
[sally] => Array
(
[age] => 8
[sex] => F
)
)
[0] => CEO
)
返回:
user%5Bname%5D=Bob+Smith&user%5Bage%5D=47&user%5Bsex%5D=M&user%5Bdob%5D=5%2F12%2F1956&pastimes%5B0%5D=golf&pastimes%5B1%5D=opera&pastimes%5B2%5D=poker&pastimes%5B3%5D=rap&children%5Bbobby%5D%5Bage%5D=12&children%5Bbobby%5D%5Bsex%5D=M&children%5Bsally%5D%5Bage%5D=8&children%5Bsally%5D%5Bsex%5D=F&flags_0=CEO
我想问的是,在Java/Android中有没有办法创建后一种实体格式?我尝试了以下方法,但都没有成功:
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(3);
nameValuePairs.add(new BasicNameValuePair("user", null));
nameValuePairs.add(new BasicNameValuePair("firstname", "admin"));
nameValuePairs.add(new BasicNameValuePair("lastname", "admin"));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
希望有人知道如何实现这个 :) 祝好, Morten
编辑:
基本上我需要生成Java版本的这个PHP代码:
$params = array('user' => array(
'firstname' => 'Bob Smith',
'lastname' => 'Johnson'
));
以下是相同请求的JSON格式:
{"user":{"firstname":"Bob Smith","lastname":"Johnson"}}
我只需要以application/x-www-form-urlencoded格式提供Java等效代码;)
顺便说一句,非常感谢您回答伪代码,真的很感激!