我试图从我的应用中将数据插入数据库。我可以获取数据,但是当我尝试将变量传递给我的服务器时,我得到"必填字段缺失"错误。
我之前使用其他应用程序完成了此操作,但那是在我的网站上安装SSL之前。 SSL是否有可能阻止变量。
为了测试目的,我尽量保持代码尽可能简单,但我无法弄明白。我已经重新做了几个教程,只是为了确保我没有犯错,但显然我在某个地方出错了。任何帮助非常感谢!
class CreateNewProduct extends AsyncTask<String, String, String> {
@Override
protected void onPreExecute() {
super.onPreExecute();
pDialog = new ProgressDialog(NewProductActivity.this);
pDialog.setMessage("Creating Product..");
pDialog.setIndeterminate(false);
pDialog.setCancelable(true);
pDialog.show();
}
protected String doInBackground(String... args) {
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("name", "name"));
params.add(new BasicNameValuePair("price", "2"));
params.add(new BasicNameValuePair("imgurl", "imgurl"));
JSONObject json = jsonParser.makeHttpRequest("http://myserver.com",
"POST", params);
Log.d("Create Response", json.toString());
try {
int success = json.getInt(TAG_SUCCESS);
if (success == 1) {
Intent i = new Intent(getApplicationContext(), AllProductsActivity.class);
startActivity(i);
finish();
} else {
Log.d("TEST", "Failed to create product");
}
} catch (JSONException e) {
e.printStackTrace();
}
return null;
}
protected void onPostExecute(String file_url) {
pDialog.dismiss();
}
}
PHP代码:
<?php
$response = array();
if (isset($_POST['name']) && isset($_POST['price']) && isset($_POST['imgurl'])) {
$name = "joey";
$price = "3";
$imgurl = "blowey";
require_once __DIR__ . '/db_connect.php';
$db = new DB_CONNECT();
$result = mysqli_query("INSERT INTO products(name, price, imgurl) VALUES('$name', '$price', '$imgurl')");
if ($result) {
$response["success"] = 1;
$response["message"] = "Product successfully created.";
echo json_encode($response);
} else {
$response["success"] = 0;
$response["message"] = "Oops! An error occurred.";
echo json_encode($response);
}
} else {
$response["success"] = 0;
$response["message"] = "Required field(s) is missing";
echo json_encode($response);
}
?>
JSON Parser:
public class JSONParser {
static InputStream is = null;
static JSONObject jObj = null;
static String json = "";
// constructor
public JSONParser() {
}
// function get json from url
// by making HTTP POST or GET mehtod
public JSONObject makeHttpRequest(String url, String method,
List<NameValuePair> params) {
// 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);
httpPost.setEntity(new UrlEncodedFormEntity(params, "utf-8"));
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
}else if(method == "GET"){
// request method is GET
DefaultHttpClient httpClient = new DefaultHttpClient();
String paramString = URLEncodedUtils.format(params, "utf-8");
url += "?" + paramString;
HttpGet httpGet = new HttpGet(url);
HttpResponse httpResponse = httpClient.execute(httpGet);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
}
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
try {
BufferedReader reader = new BufferedReader(new InputStreamReader(
is, "iso-8859-1"), 8);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
is.close();
json = sb.toString();
} catch (Exception e) {
Log.e("Buffer Error", "Error converting result " + e.toString());
}
// try parse the string to a JSON object
try {
jObj = new JSONObject(json);
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
// return JSON String
return jObj;
}
}
答案 0 :(得分:1)
尝试替换:
post.setEntity(new UrlEncodedFormEntity(dataToSend));
与
post.setRequestBody(dataToSend);