Question

我对关系数据库比较陌生，我不确定我想做什么是可能的。如果有人可以告诉我该怎么做或者至少指出我正确的方向，

我有table_1

NameId       SelectedName 
1                 A
3                 C
6                 F

NameId是FOREIGN KEY REFERENCES table_2 (Id)

我有table_2

Id              Name 
1                A
2                B
3                C
4                D
5                E
6                F

我想设置我的数据库，以便只要table_2中的条目更新，table_1将更改其Name列的值，如果条目来自table_2 table_2就在其中。例如，如果我在Id Name 3 C_EDITED

中更改此值

table_1

此更改也会反映在public class LanguageChoice extends AppCompatActivity { private Spinner spWords, spTrans; private Button btnApply; String language_choice; String translation_choice; String dict_source; @Override protected void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.language_choice); btnApply = (Button) findViewById(R.id.aplBtn); spWords = (Spinner) findViewById(R.id.vocabSpin); spTrans = (Spinner) findViewById(R.id.transSpin); //when apply button is clicked, move to choose mode activity btnApply.setOnClickListener(new View.OnClickListener() { @Override public void onClick(View v) { importFile(); } }); chooseLanguage(); } private void chooseLanguage(){ // Create an ArrayAdapter using the string array and a default spinner layout ArrayAdapter<CharSequence> WAdapter = ArrayAdapter.createFromResource(this, R.array.languages, android.R.layout.simple_spinner_item); ArrayAdapter<CharSequence> TAdapter = ArrayAdapter.createFromResource(this, R.array.translations, android.R.layout.simple_spinner_item); // Specify the layout to use when the list of choices appears WAdapter.setDropDownViewResource(android.R.layout.simple_spinner_dropdown_item); TAdapter.setDropDownViewResource(android.R.layout.simple_spinner_dropdown_item); // Apply the adapter to the spinner spWords.setAdapter(WAdapter); spTrans.setAdapter(TAdapter); //spinner for input language spWords.setOnItemSelectedListener(new AdapterView.OnItemSelectedListener() { @Override public void onItemSelected(AdapterView<?> parent, View view, int position, long id) { language_choice = spWords.getSelectedItem().toString(); } @Override public void onNothingSelected(AdapterView<?> parent) { // do nothing } }); //spinner for output language spTrans.setOnItemSelectedListener(new AdapterView.OnItemSelectedListener() { @Override public void onItemSelected(AdapterView<?> parent, View view, int position, long id) { translation_choice = spTrans.getSelectedItem().toString(); } @Override public void onNothingSelected(AdapterView<?> parent) { // do nothing } }); //define the name of the file with the words-translation pairs dict_source = ("C:\\B1\\" + language_choice + "-" + translation_choice).toLowerCase() + ".txt"; } public void importFile() { //first import the correspondent dictionary String filePath = dict_source; Log.i("import from", filePath); try { String line; BufferedReader reader = new BufferedReader(new FileReader(filePath)); while ((line = reader.readLine()) != null) { String[] parts = line.split(" - ", 3); if (parts.length == 3) { String key = parts[0]; String value = parts[2]; this.wordsMap.put(key, value); } else { Log.i("Import:", "ignoring line: " + line); } } reader.close(); } catch (IOException e) { e.printStackTrace(); } } }

中

Answer 1

你可能不希望这样做......

表格看起来没有标准化，因此如果您只是在table_2中重复table_1的值，则可能会出现数据冗余和完整性问题。因此，如果没有table_1列，SelectedName可能会更好，因为您可以随时在表格_2中查找Name的值，例如。

SELECT t1.NameId, t2.Name
FROM table_1 t1 INNER JOIN table_2 t2 ON (t1.NameId = t2.Id);

但是，如果我们假设Id和Name是您的主要密钥（即您有（1，'A'），（1，'B'）等记录，表2中的（1，'Anything'）然后PK将在两个列上定义，引用该表的FK也是如此。这个SQLfiddle显示它是如何工作的，但它确实要求你更改table_2上的PK和table_1上的FK，并使用其他一些评论中提到的ON UPDATE CASCADE方法。

http://sqlfiddle.com/#!9/80daee/1

Answer 2

create table table_2(
    id int,
    name varchar unique,
    primary key (id)
);

create table table_1(
    nameid int ,
    selectedname varchar foreign key references table_2(name) on update cascade,
);

insert into table_2 values
(1,'A'),
(2,'B'),
(3,'C'),
(4,'D'),
(5,'E'),
(6,'F')

insert into table_1 values
(1,'A'),
(3,'C'),
(6,'F')

select * from table_1;
select * from table_2;

update table_2 set name = 'X' where id = 1;

select * from table_2;
--Note X

Answer 3

我不明白为什么你在table_1中保留选定的列。但仍然

update table_1 inner join table_2
    on tabel_2.id=table_1.nameid
    set selectedname =table_2.name

SQL中自动更新外键

3 个答案: