php - 将参数从 Android 传递到 php

Question

在我的安卓系统中:

    String cat_id = getIntent().getExtras().getString("category_id");

    ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
    nameValuePairs.add(new BasicNameValuePair("cat_id", cat_id));

    try {
        HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost(
                "http://www.---.com/items.php");
        httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
        HttpResponse response = httpclient.execute(httppost);
        HttpEntity entity = response.getEntity();
        is = entity.getContent();
    } catch (Exception e) {
        Log.e("log_tag", "Error in http connection" + e.toString());
    }

在我的 PHP 中:

// editing out DB connections which have been verified to work


 <?php
$sql=mysql_query("SELECT * FROM items WHERE cat_id = '".$_REQUEST['cat_id']."'");
while($row=mysql_fetch_assoc($sql))
    $output[]=$row;
print(json_encode($output));
mysql_close();
 ?>

所以我的问题是它没有返回任何内容(在 JSON 结果部分 -- Toast 说“找不到任何项目”)。我的代码是否正确，我将参数从 Android 传递到 PHP 的方式是否正确？

Answer 1

不应该是这一行

$sql=mysql_query("SELECT * FROM items WHERE cat_id = '".$_REQUEST['cat_id']."'");

成为

$sql=mysql_query("SELECT * FROM items WHERE cat_id = '".$_POST['cat_id']."'");

即$_REQUEST 到 $_POST？

我建议使用 Firefox 的 Firebug 插件，并跟踪发送到您的服务器的确切表单。