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Fix a bug with adding a word

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It was possible to insert a word without a shortcut while the same
word used to be present with a shortcut. This change fixes that.

Bug: 6026080
Change-Id: I3be98bf450aad8e2eb38336e8f77aedab39d5797
This commit is contained in:
Jean Chalard
2012-05-10 15:08:17 +09:00
parent 271543876d
commit eed02dd2d5
1 changed files with 33 additions and 0 deletions
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33
src/com/android/settings/inputmethod/UserDictionaryAddWordContents.java
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@@ -19,6 +19,7 @@ package com.android.settings.inputmethod;
import android.app.Activity; import android.app.Activity;
import android.content.ContentResolver; import android.content.ContentResolver;
import android.content.Context; import android.content.Context;
import android.database.Cursor;
import android.os.Bundle; import android.os.Bundle;
import android.provider.UserDictionary; import android.provider.UserDictionary;
import android.text.TextUtils; import android.text.TextUtils;
@@ -118,6 +119,12 @@ public class UserDictionaryAddWordContents {
// If the word is somehow empty, don't insert it. // If the word is somehow empty, don't insert it.
return; return;
} }
// If there is no shortcut, and the word already exists in the database, then we
// should not insert, because either A. the word exists with no shortcut, in which
// case the exact same thing we want to insert is already there, or B. the word
// exists with at least one shortcut, in which case it has priority on our word.
if (hasWord(newWord, context)) return;
// Disallow duplicates. If the same word with no shortcut is defined, remove it; if // Disallow duplicates. If the same word with no shortcut is defined, remove it; if
// the same word with the same shortcut is defined, remove it; but we don't mind if // the same word with the same shortcut is defined, remove it; but we don't mind if
// there is the same word with a different, non-empty shortcut. // there is the same word with a different, non-empty shortcut.
@@ -134,6 +141,32 @@ public class UserDictionaryAddWordContents {
TextUtils.isEmpty(mLocale) ? null : Utils.createLocaleFromString(mLocale)); TextUtils.isEmpty(mLocale) ? null : Utils.createLocaleFromString(mLocale));
} }
private static final String[] HAS_WORD_PROJECTION = { UserDictionary.Words.WORD };
private static final String HAS_WORD_SELECTION_ONE_LOCALE = UserDictionary.Words.WORD
+ "=? AND " + UserDictionary.Words.LOCALE + "=?";
private static final String HAS_WORD_SELECTION_ALL_LOCALES = UserDictionary.Words.WORD
+ "=? AND " + UserDictionary.Words.LOCALE + " is null";
private boolean hasWord(final String word, final Context context) {
final Cursor cursor;
// mLocale == "" indicates this is an entry for all languages. Here, mLocale can't
// be null at all (it's ensured by the updateLocale method).
if ("".equals(mLocale)) {
cursor = context.getContentResolver().query(UserDictionary.Words.CONTENT_URI,
HAS_WORD_PROJECTION, HAS_WORD_SELECTION_ALL_LOCALES,
new String[] { word }, null /* sort order */);
} else {
cursor = context.getContentResolver().query(UserDictionary.Words.CONTENT_URI,
HAS_WORD_PROJECTION, HAS_WORD_SELECTION_ONE_LOCALE,
new String[] { word, mLocale }, null /* sort order */);
}
try {
if (null == cursor) return false;
return cursor.getCount() > 0;
} finally {
if (null != cursor) cursor.close();
}
}
public static class LocaleRenderer { public static class LocaleRenderer {
private final String mLocaleString; private final String mLocaleString;
private final String mDescription; private final String mDescription;