Question

SQLite结构

Table: Class

 Column1: classID (primary key)
 Column2: className


Table: Students

 Column1: studentID (primary key)
 Column2: classID (foreign key)
 Column3: studentName

问题

我已按照Adding Custom Suggestions开发者指南进行操作。当我输入操作栏搜索小部件时，系统正确调用我的内容提供者的查询功能，我可以看到搜索建议。我的问题是查询返回整个学生表中的建议（这是预期的行为）。但我需要限制给定classID的建议。但是，如开发人员指南中所述搜索建议的工作方式，只有一个参数传递给查询函数，在我的例子中是selectionArgs [0]。

searchable.xml

<?xml version="1.0" encoding="utf-8"?>
<searchable xmlns:android="http://schemas.android.com/apk/res/android"
    android:hint="@string/student_search_hint"
    android:label="@string/app_name"
    android:searchSuggestThreshold="2"
    android:searchSuggestAuthority="com.androidapp.providers.database"
    android:searchSuggestSelection="studentName LIKE ? "
    android:searchSuggestIntentAction="android.intent.action.VIEW"
    android:searchMode="queryRewriteFromText" >

摘自内容提供商

@Override
public Cursor query(Uri uri, String[] projection, String selection,String[], selectionArgs, String sortOrder) {

    ...

    queryBuilder.setTables(StudentsTable.TABLE);

    HashMap<String, String> columnMap = new HashMap<String, String>();
    columnMap.put(BaseColumns._ID, StudentsTable.STUDENT_ID + " AS " + BaseColumns._ID);
    columnMap.put(SearchManager.SUGGEST_COLUMN_TEXT_1, StudentsTable.NAME + " AS " + SearchManager.SUGGEST_COLUMN_TEXT_1);
    columnMap.put(SearchManager.SUGGEST_COLUMN_INTENT_DATA, StudentsTable.STUDENT_ID + " AS " + SearchManager.SUGGEST_COLUMN_INTENT_DATA);
    queryBuilder.setProjectionMap(columnMap);

    limit = uri.getQueryParameter(SearchManager.SUGGEST_PARAMETER_LIMIT);
    selectionArgs[0] = "%"+selectionArgs[0] + "%";

    SQLiteDatabase db = database.getWritableDatabase();
    Cursor cursor = queryBuilder.query(db, projection, selection, selectionArgs, null, null, sortOrder, limit);

    return cursor;
}

如果查询是由我发起的，我可以简单地在selectionArgs [1]中传递第二个参数（classID）并相应地修改选择字符串以形成正确的SQLite语句。

但查询是由系统启动的。那我该怎么做？是否可以拦截对我的内容提供商的调用并修改对查询的调用？

Answer 1

我希望这对某人有帮助，我花了好几天试图解决这个问题，感谢pskink提供的link我能够实现以下解决方案：

我在内容提供商中创建了一个变量，并使用了相应的setter方法：

public class MyContentProvider extends ContentProvider {

    ...

    private long id;

    ...

    public void setId(long id) {
        this.id = id;     
    }

}

然后在我的活动中，我想为特定班级的学生提供搜索建议，我这样做：

ContentProviderClient client = getContentResolver().acquireContentProviderClient("com.androidapp.providers.database");
MyContentProvider provider = (MyContentProvider)client.getLocalContentProvider();
provider.setId(rowid); //rowid is the classId from which students will be searched
client.release();

修改查询功能：

@Override
public Cursor query(Uri uri, String[] projection, String selection,String[], selectionArgs, String sortOrder) {

    ...

    queryBuilder.setTables(StudentsTable.TABLE);

    HashMap<String, String> columnMap = new HashMap<String, String>();
    columnMap.put(BaseColumns._ID, StudentsTable.STUDENT_ID + " AS " + BaseColumns._ID);
    columnMap.put(SearchManager.SUGGEST_COLUMN_TEXT_1, StudentsTable.NAME + " AS " + SearchManager.SUGGEST_COLUMN_TEXT_1);
    columnMap.put(SearchManager.SUGGEST_COLUMN_INTENT_DATA, StudentsTable.STUDENT_ID + " AS " + SearchManager.SUGGEST_COLUMN_INTENT_DATA);
    queryBuilder.setProjectionMap(columnMap);

    limit = uri.getQueryParameter(SearchManager.SUGGEST_PARAMETER_LIMIT);
    selectionArgs[0] = "%" + selectionArgs[0] + "%";

    /*AMMENDMENT*/
    selection = selection + "AND " + StudentsTable.CLASS_ID + "=" + id;
    //This forms the SQL statement "WHERE studentName LIKE ? AND classId=3" (Assuming that rowid was 3).
    //As per normal Android behaviour the "?" mark will be replaced by selectionArgs[0].
    /*END*/

    SQLiteDatabase db = database.getWritableDatabase();
    Cursor cursor = queryBuilder.query(db, projection, selection, selectionArgs, null, null, sortOrder, limit);

    return cursor;
}

Answer 2

我刚为内容提供商的第二个参数创建了一个静态变量，并从活动中设置了它。我一直在考虑它，这是我找到的最干净的解决方法！

Android自定义搜索建议：查询两个参数

2 个答案: