[英]How to collect data from JSON in android WITHOUT key value pair?
我試圖在android中以JSON的形式從服務器收集數據。 但是有一些技術故障,因為我必須解析不是key:value對形式的JSON數據。 或如何解析在JSON中創建的用戶創建的關聯數組? 需要幫忙? 預先感謝您。
我想解析以下兩種情況之一中收到的JSON對象。
情況1:
PHP腳本如下:
<?php
session_start();
$arraygive = $_SESSION['arraygive'];
$con=mysql_connect("localhost","root","");
mysql_select_db("rrugd");
$output = array();
$output1 = array();
foreach ($arraygive as $lid)
{
echo "<br>";echo "new pass";echo "<br>";
$query = "SELECT * FROM places WHERE(LID = '$lid');";
$result = mysql_query($query) or die(mysql_error());
$output = mysql_fetch_row($result);
array_push($output1, $output);
}
print(json_encode($output1));
?>
它創建以下格式的JSON對象:
[["1","shopknock","0","0","22","18.5123","73.8563"],["2","Food Shopei","231","1","17","18.5122","73.8562"],["10","Ccd","0","0","22","18.5211","73.857"]]
我根本不知道如何在Android中解析此格式。
情況2:
PHP腳本如下:
<?php
session_start();
$arraygive = $_SESSION['arraygive'];
$con=mysql_connect("localhost","root","");
mysql_select_db("rrugd");
$output = array();
$output1 = array();
foreach ($arraygive as $lid)
{
echo "<br>";echo "new pass";echo "<br>";
$query = "SELECT * FROM places WHERE(LID = '$lid');";
$result = mysql_query($query) or die(mysql_error());
$output = mysql_fetch_assoc($result);
array_push($output1, $output);
}
print(json_encode($output1));
?>
它創建以下格式的JSON對象:
[{"lid":"1","name":"shopknock","rid":"0","cid":"0","ccnt":"22","locx":"18.5123","locy":"73.8563"},{"lid":"2","name":"Food Shopei","rid":"231","cid":"1","ccnt":"17","locx":"18.5122","locy":"73.8562"},{"lid":"10","name":"Ccd","rid":"0","cid":"0","ccnt":"22","locx":"18.5211","locy":"73.857"}]
請注意,Case1和Case 2 PHP腳本之間的唯一區別是 mysql_fetch_row 和 mysql_fetch_assoc
我無法使用我使用的代碼(在情況3下給出)解析此JSON對象,盡管它可以與情況3一起使用。
情況3: BUT每當從下面的另一個PHP腳本創建這種格式的JSON對象(情況2)時:
<?php
//used for populating list of catagories at different instances
$con=mysql_connect("localhost","root","");
mysql_select_db("rrugd");
$query = "SELECT * FROM places ORDER BY ccnt DESC;";
$result=mysql_query($query) or die(mysql_error());
$output=array();
while($row=mysql_fetch_assoc($result))
{
$output[]=$row;
}
print(json_encode($output));
?>
JSON對象有效(從上述腳本創建時,但與案例2中給出的對象相同)
在案例3中,我用來解析JSON對象的Android代碼如下:
JSONArray jsonArray = new JSONArray(result);
int length = jsonArray.length();
for (int i = 0; i < length; i++)
{
JSONObject jObj = jsonArray.getJSONObject(i);
String name = jObj.getString(TAG_NAME);
String rid = jObj.getString(TAG_RID);
HashMap<String, String> map = new HashMap<String, String>();
map.put(TAG_RID, rid);
map.put(TAG_NAME, name);
oslist.add(map);
ListAdapter adapter = new SimpleAdapter(User_Home_List_Activity.this, oslist,
R.layout.list_v, new String[] { TAG_NAME }, new int[] { R.id.name});
l1.setAdapter(adapter);
}
我嘗試使用Toasts進行調試,但我意識到在上面的代碼(在Try塊中)中,控件本身未到達第一行。 即
JSONArray jsonArray = new JSONArray(result);
如果我在上面這一行之前的Try塊內應用Toast,則顯示它(即,控制到達該行)。 但是,此行之后的Toast不會顯示。
像這樣嘗試
for (int i = 0; i < mJsonArray.length(); i++) {
JSONObject mJsonObject = new JSONObject();
mJsonObject = mJsonArray.getJSONObject(i);
for (int j = 0; j < mJsonObject .length(); j++) {
int Id = mJsonObject.getString("lid");
String Name = mJsonObject.getString("name");
.
.
.
}
}
我已經嘗試過使用您的Json響應和Wola及其工作方式。 這是可用於將Case 2
JSON輸出與該Json字符串一起解析的代碼段
StringBuilder output = new StringBuilder();
JSONArray jArr;
try
{
jArr = new JSONArray(jString);
for (int i = 0; i < jArr.length(); i++)
{
JSONObject jObj = jArr.getJSONObject(i);
output.append("\n\n");
output.append("\n lid : " + jObj.getInt("lid"));
output.append("\n name : " + jObj.getString("name"));
output.append("\n rid : " + jObj.getInt("rid"));
output.append("\n cid : " + jObj.getInt("cid"));
output.append("\n ccnt : " + jObj.getInt("ccnt"));
output.append("\n locx : " + jObj.getDouble("locx"));
output.append("\n locy : " + jObj.getDouble("locy"));
}
}
catch (JSONException e1)
{
// TODO Auto-generated catch block
e1.printStackTrace();
}
tvText.setText(output.toString());
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