Question

我正在尝试将HttpRequest发送到位于Web服务器上的php脚本。我想要检索所有特定类别的菜肴。菜肴以JSON格式返回，我的JSONParser类解析数据。

以下是我提出文件请求的doInBackground方法：

    @Override
    protected Boolean doInBackground(String... params) {
        // location of file to retrieve
        String url = "http://www.jdiadt.com/getAllDishes.php";
        JSONObject json = jParser.makeHttpRequest(url, "POST", cat);
        // For testing purposes - Ensure all data was received.
        Log.d("SINGLE DISH DETAILS: ", json.toString());

        try {
            int success = json.getInt(TAG_SUCCESS);
            if (success == 1) {
                dishes = json.getJSONArray(TAG_DISHES);
                for (int i = 0; i < dishes.length(); i++) {
                    JSONObject d = dishes.getJSONObject(i);
                    MenuItem m = new MenuItem();
                    m.setName(d.getString(TAG_NAME));
                    m.setPrice(Double.parseDouble(d.getString(TAG_PRICE)));
                    Log.d("MENUITEM " + i + ":", " NAME:" + m.getName()
                            + " PRICE:" + m.getPrice());
                    starters.add(m);
                }
            } else {
                return false;
            }

            return true;

        } catch (JSONException e) {
            e.printStackTrace();
        }
        return null;
    }

这是makeHttpRequest()类的JSONParser方法：

public JSONObject makeHttpRequest(String url, String method, String cat) {

    // Making HTTP request
    try {
        // check for request method
        if(method == "POST"){
            // request method is POST
            // defaultHttpClient
            DefaultHttpClient httpClient = new DefaultHttpClient();
            HttpPost httpPost = new HttpPost(url);

            List <NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(1);
            nameValuePairs.add(new BasicNameValuePair("cat", cat));
            try {
                httpPost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
            } catch (UnsupportedEncodingException e) {
                e.printStackTrace();
            }

            HttpResponse httpResponse = httpClient.execute(httpPost);
            HttpEntity httpEntity = httpResponse.getEntity();
            is = httpEntity.getContent();

        }

这是我试图检索的脚本：

$response = array();

   //Variables for connecting to your database.
            //These variable values come from your hosting account.
            $hostname = "xxx.xxx.xxx.xxx";
            $username = "xxxxxx";
            $dbname = "xxxxxxx";



            //These variable values need to be changed by you before deploying
            $password = "xxxxxxxxxxx";
            $usertable = "dishes";
            $yourfield = "dish_name";

            //Connecting to your database
            mysql_connect($hostname, $username, $password) OR DIE ("Unable to 
            connect to database! Please try again later.");
            mysql_select_db($dbname);

            $cat = $_POST['cat'];


// get all dishes from dishes table
$result = mysql_query("SELECT * FROM $usertable where dish_cat = '$cat' ") or die(mysql_error());

// check for empty result
if (mysql_num_rows($result) > 0) {
    // looping through all results
    // dishes node
    $response["dishes"] = array();

    while ($row = mysql_fetch_array($result)) {
        // temp user array
        $dishes = array();
        $dishes["dish_id"] = $row["dish_id"];
        $dishes["dish_name"] = $row["dish_name"];
        $dishes["dish_desc"] = $row["dish_desc"];
        $dishes["dish_price"] = $row["dish_price"];

        // push single dish into final response array
        array_push($response["dishes"], $dishes);
    }
    // success
    $response["success"] = 1;

    // echoing JSON response
    echo json_encode($response);
} else {
    // no dishes found
    $response["success"] = 0;
    $response["message"] = "No dishes found";

    // echo no users JSON
    echo json_encode($response);
}

注意这一行：

$cat = $_POST['cat']

那应该是检索菜类，sql语句应该成功执行....目前我收到一条消息说“没有找到菜肴”。

非常感谢任何帮助。我不知道我做错了什么。

Answer 1

我认为问题是你的条件：

if (method == "POST") {
  // execute request
}

在这里，您要将Strings与==进行比较，这将始终返回false（并且永远不会执行正文），因为它会比较引用而不是值。您必须使用equals()方法。

if (method.equals("POST") {
   // do your work
}

Answer 2

嘿尝试打印$ cat以确保首先正确接收它。

HttpPost请求返回错误响应

2 个答案: