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java - 解析 whereEqualTo

转载 作者:行者123 更新时间:2023-12-01 12:39:35 24 4
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我正在尝试从解析中检索信息。特别是,我添加了一个条件,它只会返回选择相同 Activity 的用户列表。

我尝试了下面的代码,但它似乎确实有效,因为该条件返回一个空列表。

ParseQuery<ParseUser> query = ParseUser.getQuery();
query.whereNotEqualTo("objectId", currentUserId);
query.whereEqualTo("ActivityName","");

它本质上会看到哪些用户选择了特定的 Activity 名称并返回

下面是完整的代码

 public class MatchingActivity extends Activity {

private String currentUserId;
private ArrayAdapter<String> namesArrayAdapter;
private ArrayList<String> names;
private ListView usersListView;
private Button logoutButton;

@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.matching);

logoutButton = (Button) findViewById(R.id.logoutButton);
logoutButton.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View view) {
ParseUser.logOut();
Intent intent = new Intent(getApplicationContext(), LoginActivity.class);
startActivity(intent);
}
});

setConversationsList();
}

private void setConversationsList() {
currentUserId = ParseUser.getCurrentUser().getObjectId();
names = new ArrayList<String>();


ParseQuery<ParseUser> query = ParseUser.getQuery();
query.whereNotEqualTo("objectId", currentUserId);
query.whereEqualTo("ActivityName",userActivitySelectionName);

query.findInBackground(new FindCallback<ParseUser>() {
public void done(List<ParseUser> userList, ParseException e) {
if (e == null) {
for (int i=0; i<userList.size(); i++) {
names.add(userList.get(i).getUsername().toString());
}

usersListView = (ListView)findViewById(R.id.usersListView);
namesArrayAdapter =
new ArrayAdapter<String>(getApplicationContext(),
R.layout.user_list_item, names);
usersListView.setAdapter(namesArrayAdapter);

usersListView.setOnItemClickListener(new AdapterView.OnItemClickListener() {
@Override
public void onItemClick(AdapterView<?> a, View v, int i, long l) {
openConversation(names, i);
}
});

} else {
Toast.makeText(getApplicationContext(),
"Error loading user list",
Toast.LENGTH_LONG).show();
}
}
});
}

public void openConversation(ArrayList<String> names, int pos) {
ParseQuery<ParseUser> query = ParseUser.getQuery();
query.whereEqualTo("username", names.get(pos));
query.findInBackground(new FindCallback<ParseUser>() {
public void done(List<ParseUser> user, ParseException e) {
if (e == null) {
Intent intent = new Intent(getApplicationContext(), MessagingActivity.class);
intent.putExtra("RECIPIENT_ID", user.get(0).getObjectId());
startActivity(intent);
} else {
Toast.makeText(getApplicationContext(),
"Error finding that user",
Toast.LENGTH_SHORT).show();
}
}
});
}
}

提前致谢。

最佳答案

在数据浏览器中仔细检查,但我猜测该列具有“未定义”而不是空字符串。

在这种情况下,您需要使用以下内容:

query.whereDoesNotExist("ActivityName");

whereDoesNotExist()whereExists() 函数用于检查列上是否设置了任何值。

关于java - 解析 whereEqualTo,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/25242947/

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