在获取json数据时,我收到错误:
JSONArray无法转换为JSONObject
生成json的代码:
JSONObject parent = new JSONObject();
DatabaseHandler dbh = new DatabaseHandler(getApplicationContext());
for(int i=0; i < allEds.size(); i++){
String edsText = allEds.get(i).getText().toString();
//spinner = allSpns.get(i);
String spinSelected=allSpns.get(i).getSelectedItem().toString();
try
{
JSONObject json = new JSONObject();
json.put("Id", i);
json.put("FieldName", edsText);
json.put("FieldType",spinSelected);
parent.accumulate("data", json);
}
catch (JSONException e)
{
// TODO Auto-generated catch block
e.printStackTrace();
}
}
Generated json is
{"data":
[{"FieldType":"Account Number","FieldName":"r","Id":0},
{"FieldType":"Net Banking Id","FieldName":"tt","Id":1}
]}
code for json read
------------------
JSONObject jsonObj = new JSONObject(folderStructure);
JSONObject data = jsonObj.getJSONObject("data");
String …嗨我很困惑与android中的图像按钮选项.我用以下xml创建了三个图像按钮
<ImageButton
android:id="@+id/imageButton1"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:layout_alignParentRight="true"
android:src="@drawable/folder_add" />
<ImageButton
android:id="@+id/imageButton4"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:layout_alignTop="@+id/imageButton1"
android:layout_toLeftOf="@+id/imageButton1"
android:src="@drawable/search" />
<ImageButton
android:id="@+id/imageButton2"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:layout_alignTop="@+id/imageButton4"
android:layout_toLeftOf="@+id/imageButton4"
android:src="@drawable/advancedsettings"
style="?android:attr/borderlessButtonStyle" />
但是外面就像是带有图像的按钮..请帮我将图像仅显示为按钮..提前完成!