Question

这是php代码，我想在Android中使用HttpPost方法。我想检查与post方法一起发送的json对象是否与数据库中存在的数据匹配

if(!isset($_POST['data'])) {
    echo "FAIL1";   
    return 'false';
}

$jsonObject = $_POST['data'];
$data = json_decode($jsonObject, true);

if(!is_array($data)) {
    print_r($data);
    exit;  
    echo "FAIL2";
    return 'false';
}

$UNIQUE_ID = $data['UNIQUE_ID'];
$BAR_CODE = $data['BARCODE'];
$BATCH_CODE = $data['BATCHCODE'];

if(!isset($UNIQUE_ID) || trim($UNIQUE_ID) == '' || !isset($BAR_CODE) || trim($BAR_CODE) == '' || !isset($BATCH_CODE) || trim($BATCH_CODE) == '') {
    echo "FAIL3";
    return 'false';
}

$link = mysql_connect($db_server,$db_username,$db_pasword) or die("Can not connect to server $db_server.");

mysql_select_db("drugverifier",$link);

$query = "SELECT * FROM drugverifier where unique_id = $UNIQUE_ID AND bar_code = $BAR_CODE AND batch_code = $BATCH_CODE";
echo " TEST4";
$result = mysql_query($query,$link) or die("Can not able to run the query ' $query ' on server $db_server.");
echo "TEST5";
if(mysql_num_rows($result) > 0){
    echo "SUCCESS6";
    echo 'true';
}
else {
    echo "FAILURE7";
    echo 'false';
}

我的android代码是：

JSONObject j = new JSONObject();
        try {
            j.put("BARCODE", "8904017500190");
            j.put("BATCHCODE", "220");  
            j.put("UNIQUE_ID", "9876"); 

        } catch (JSONException e) {
            Log.v("@@@@@JSONException", e.getMessage());
        }
String url = "http://www.smartcloudinfotech.com/SmartDrugVerifier/verifyBarCode.php";
 HttpClient httpclient = new DefaultHttpClient();
         HttpPost httpPost = new HttpPost(url);

List<NameValuePair> nameValuePair = new ArrayList<NameValuePair>(1);
                nameValuePair.add(new BasicNameValuePair("data",j.toString()));
                httpPost.setEntity(new UrlEncodedFormEntity(nameValuePair));

                 response = httpclient.execute(httpPost);
                 Log.v("RESPONSE",""+response.toString());
                 Toast.makeText(this, "RESPONSE in doPOST()"+response.toString(), Toast.LENGTH_SHORT).show();


            } catch (UnsupportedEncodingException e) {
                Log.v("@@@@@@@@@@@@@@UnsupportedEncodingException", e.getMessage());
            } catch (ClientProtocolException e) {
                Log.v("@@@@@@@@@ClientProtocolException",e.getMessage());
            } catch (IOException e) {
                Log.v("@@@@@@@@@@IOException116", e.getMessage());
            } 

String temp = EntityUtils.toString(response.getEntity());

当我运行此程序时，变量temp的值不包含任何值。请帮我解决我出错的地方。

Answer 1

尝试使用BufferedReader来阅读HttpEntity的内容，如下所示：

void printResponse(HttpResponse response) throws IOException {
    BufferedReader in = new BufferedReader(new InputStreamReader(response.getEntity().getContent()));
    String line = "";
    while ((line = in.readLine()) != null) {
        System.out.println(line);
    }
}

修改
当我使用Curl向您的服务器发出POST data=test时，我没有得到回复。我认为这是由于这段代码：

if(!is_array($data)) { print_r($data); exit; echo "FAIL2"; return 'false'; }

首先，如果它不是数组，则要打印数组。这是没有意义的。其次，退出，并回显你的“FAIL2”。

Answer 2

您正在阅读Android中的错误响应，就像这样

HttpResponse response = httpclient.execute(httppost);
String result = EntityUtils.toString(response.getEntity());

此外，我不是一个伟大的PHP程序员，但我相信你正在读PHP中错误的值，你没有传递JSONArray据我所知，你传递的是JSONObject。 <\ n>删除

print_r($data);
exit;

如何将HttpPost请求发送到服务器上的php脚本

2 个答案: