当我搜索表中不存在的名称时，我的android sqlite数据库应用程序会关闭

下面是查找表中是否存在某个名称的代码/方法

Contact getContact(String name) {
SQLiteDatabase db = this.getReadableDatabase();

    Cursor cursor = db.query(TABLE_CONTACTS, new String[] { KEY_ID,
            KEY_NAME, KEY_PH_NO }, KEY_NAME + "=?",
            new String[] { String.valueOf(name) }, null, null, null, null);
    if (cursor != null)
        cursor.moveToFirst();

    Contact contact = new Contact(Integer.parseInt(cursor.getString(0)),
            cursor.getString(1), cursor.getString(2));

    db.close();
    cursor.close();
    // return contact
    return contact;
}

---

我已经有了一个函数来获取arrayList中的所有名称。我可以在调用上述函数之前调用它来解决我的问题。但是我想问一下，当你调用

cursor.moveToFirst（）

时，如果有有效的结果，它将返回

true

，如果没有结果，它将返回

false

。如果

cursor.moveToFirst（）

返回

false

，则调用任何

getXXX（）

方法都将失败

if (cursor == null)
Toast.makeText(getApplicationContext(), "No Records Exist", Toast.LENGTH_SHORT).show();

试着这样做：

if( cursor.moveToFirst() )
{
    Contact contact = new Contact(Integer.parseInt(cursor.getString(0)),
        cursor.getString(1), cursor.getString(2));
}
cursor.close();

---

请注意，从

SQLiteDatabase.query

返回的

光标

保证为非空。

以下是从表中获取列表中所有联系人的代码

/**
 * Getting all the contacts in the database
 */
public List<Contact> getAllContacts() {
    List<Contact> contactList = new ArrayList<Contact>();
    // Select All Query
    String selectQuery = "SELECT  * FROM " + TABLE_CONTACTS;

    SQLiteDatabase db = this.getWritableDatabase();
    Cursor cursor = db.rawQuery(selectQuery, null);

    // looping through all rows and adding to list
    if (cursor.moveToFirst()) {
        do {
            Contact contact = new Contact();
            contact.setID(Integer.parseInt(cursor.getString(0)));
            contact.setName(cursor.getString(1));
            contact.setPhoneNumber(cursor.getString(2));
            // Adding contact to list
            contactList.add(contact);
        } while (cursor.moveToNext());
    }

    cursor.close();
    db.close();

    // return contact list
    return contactList;
}

}

---

注意：我们也可以从联系人列表中获取此联系人，两者都可以用于获取所需的联系人（即本例中的“姓名”）

我看到

query

返回一个空光标，尽管我无法确定具体情况+1对于

moveToFirst

返回false，这就是答案。logcat在哪里。。请提供给我们看看发生了什么。

/**
 * Getting all the contacts in the database
 */
public List<Contact> getAllContacts() {
    List<Contact> contactList = new ArrayList<Contact>();
    // Select All Query
    String selectQuery = "SELECT  * FROM " + TABLE_CONTACTS;

    SQLiteDatabase db = this.getWritableDatabase();
    Cursor cursor = db.rawQuery(selectQuery, null);

    // looping through all rows and adding to list
    if (cursor.moveToFirst()) {
        do {
            Contact contact = new Contact();
            contact.setID(Integer.parseInt(cursor.getString(0)));
            contact.setName(cursor.getString(1));
            contact.setPhoneNumber(cursor.getString(2));
            // Adding contact to list
            contactList.add(contact);
        } while (cursor.moveToNext());
    }

    cursor.close();
    db.close();

    // return contact list
    return contactList;
}

    /**
     * checks if name already present in the database
     * @param name
     * @return
     */
    public boolean checkDbData(String name){
        List<Contact> contactList =getAllContacts();
        boolean checkName = false ;

        for(Contact cn: contactList){
            String dbName = cn.getName();
            if(name.equals(dbName)){
                checkName = true ;
            }               
        }

        return checkName;
    }

Contact getContact(String name) {
SQLiteDatabase db = this.getReadableDatabase();

Cursor cursor = db.query(TABLE_CONTACTS, new String[] { KEY_ID,
        KEY_NAME, KEY_PH_NO }, KEY_NAME + "=?",
        new String[] { String.valueOf(name) }, null, null, null, null);
if (cursor != null)
    cursor.moveToFirst();

Contact contact = new Contact(Integer.parseInt(cursor.getString(0)),
        cursor.getString(1), cursor.getString(2));

db.close();
cursor.close();
// return contact
return contact;