我无法从 mysql 数据库中获取数据。我制作了一个 android 应用程序,它连接到 mysql db 并使用 php 脚本获取数据....当我运行代码时,它抛出异常字符串无法转换为 jsonarray.... 还指导我将我的 php 脚本放在 XAMP PHP 脚本中的位置:
$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("login",$con);
$sql=mysql_query("SELECT * FROM category ORDER BY `category`.`category` ASC");
while($row=mysql_fetch_assoc($sql))
$output[]=$row;
print(json_encode($output));
mysql_close($con);
代码:-
tv=(TextView)findViewById(R.id.textView1);
String result = null;
InputStream is = null;
StringBuilder sb = null;
ArrayList nameValuePairs = new ArrayList();
List r = new ArrayList();
try{
//http post
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://ip../xampp/htdocs/db.php");
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();
is = entity.getContent();
}
catch(Exception e){
Toast.makeText(getBaseContext(),e.toString() ,Toast.LENGTH_LONG).show();
}
//Convert response to string
try
{
BufferedReader reader = new BufferedReader(new InputStreamReader(is,"UTF-8"));
sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null)
{
sb.append(line +"\n");
}
is.close();
result = sb.toString();
}
catch(Exception e)
{
Toast.makeText(getBaseContext(),e.toString() ,Toast.LENGTH_LONG).show();
}
//END Convert response to string
try{
JSONArray jArray = new JSONArray(result);
JSONObject json_data=null;
for(int i=0;i<jArray.length();i++) {
json_data = jArray.getJSONObject(i);
tv.append(json_data.getString("category").toString()+"\n");
}
}
catch(JSONException e1){
Toast.makeText(getBaseContext(),e1.toString() ,Toast.LENGTH_LONG).show();
} catch (ParseException e1) {
Toast.makeText(getBaseContext(),e1.toString() ,Toast.LENGTH_LONG).show();
}