我是开发Android应用程序的初学者,这是我第一次从头开始编码,这就是为什么我觉得这对我很难。基本上,我只想摆脱错误,以便我现在可以开始在我的Android应用程序上显示SQL表。当我从mySQL表中获取数据时,我被这个错误禁止了。我在LogCat上得到这个:
Error parsing data org.json.JSONException: Value <!DOCTYPE of type java.lang.String cannot be converted to JSONObject
05-03 15:45:25.586: W/System.err(552): java.lang.NullPointerException
05-03 15:45:25.626: W/System.err(552): at hs.stockapp.StocksTask.doInBackground(StocksTask.java:64)
如果我理解错误是正确的,那么我的StocksTask.java文件第64行就出错了。
第64行:if (json.getString(KEY_SUCCESS) != null)
StocksTask.java代码摘录:
protected Integer doInBackground(String... params) {
UserFunctions userFunction = new UserFunctions();
JSONObject json = userFunction.findStocks();
try {
if (json.getString(KEY_SUCCESS) != null) {
String res = json.getString(KEY_SUCCESS);
if(Integer.parseInt(res) == 1){
DatabaseHandler db = new DatabaseHandler(activity.getApplicationContext());
JSONObject json_stocks = json.getJSONObject("stocks");
Log.v("company_name", json_stocks.getString(STOCK_COMPANY_NAME));
userFunction.findStocks();
db.findStocks(json_stocks.getString(STOCK_COMPANY_NAME), json_stocks.getString(STOCK_SYMBOL), json_stocks.getString(STOCK_MCAP));
responseCode = 1;
// Close Screen
//finish();
}else{
responseCode = 0;
// Error
}
}
}
catch (NullPointerException e) {
e.printStackTrace();
}
catch (JSONException e) {
e.printStackTrace();
}
return responseCode;
}
UserFunctions.java
public JSONObject findStocks(){
// Building Parameters
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("tag", findstocks_tag));
JSONObject json = jsonParser.getJSONFromUrl(stocksURL, params);
// return json
return json;
}
的index.php
else if ($tag == 'stocks'){
// request type is view stock list
$stocks = $db->search_stocks();
if ($stocks != false) {
// stocks
// echo json with success = 1
$response["success"] = 1;
$response["stocks"]["company_name"] = $stocks["name"];
$response["stocks"]["com_sym"] = $stocks["com_sym"];
$response["stocks"]["mkt_cap"] = $stocks["mkt_cap"];
echo json_encode($response);
} else {
// no stocks found
// echo json with error = 0
$response["error"] = 1;
$response["error_msg"] = "No stocks found";
echo json_encode($response);
}
}
DB_functions.php
public function search_stocks() {
$result = mysql_query("SELECT company_name, com_sym, mkt_cap from stock ORDER BY com_sym") or die(mysql_error());
$no_of_rows = mysql_num_rows($result);
if ($no_of_rows > 0) {
while($row=mysql_fetch_assoc($result)){
$output[] = $row;
}
print(json_encode($output));
return $row;
return true;
} else {
// stocks not existed
return false;
}
mysql_close();
}
产生search_stocks函数的结果
[{"company_name":"Asian Chorva","com_sym":"ASIA","mkt_cap":"4534356"},{"company_name":"Banko de Oro","com_sym":"BDO","mkt_cap":"54434"},{"company_name":"Bank of the Phil Island","com_sym":"BPI","mkt_cap":"5464554"},{"company_name":"College Hive In Boon","com_sym":"CHIB","mkt_cap":"5434654"},{"company_name":"Critical Heart Throb Racing","com_sym":"CHTR","mkt_cap":"53544564654"}]
我应该怎么做Line 64?非常感谢:)任何帮助将不胜感激。有关它的其他信息将非常有帮助。
答案 0 :(得分:7)
错误消息显示:
值&lt;!java.lang.String类型的DOCTYPE无法转换为JSONObject
如果是<!DOCTYPE
,那么它不是[{"company_name":…
。
您需要确切了解正在发出的请求以及服务器返回HTML文档的原因。
检查您的服务器日志。它可能是一个错误文档。
答案 1 :(得分:0)
在UserFunctions.java中正确显示您要显示的网址,例如"http://10.0.2.2/api/"
,其中api文件夹存储在www或htdocs文件夹中