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Re: Access to S_* Structure of TABLE crash Application. [message #36259 is a reply to message #36256]

Thu, 17 May 2012 09:59

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Messages: 301
Registered: November 2008

Senior Member

sergeynikitin wrote on Thu, 17 May 2012 07:53

For example, there are two tables:

TABLE_(WORKER)
   INT_     (ID) PRIMARY_KEY
   STRING_  (NAME, 200)
   STRING_  (LASTNAME, 200)
   INT_     (PLANT_ID)
END_TABLE

TABLE_(PLANT)
   INT_     (ID) PRIMARY_KEY
   STRING_  (NAME, 200)
   STRING_  (ADDRESS, 200)
END_TABLE

Correct version of tables:

TABLE_(WORKER)
   INT_     (ID) PRIMARY_KEY
   STRING_  (NAME, 200)
   STRING_  (LASTNAME, 200)
   INT_     (PLANT_ID)
END_TABLE

TABLE_(PLANT)
   INT     (ID) PRIMARY_KEY
   STRING  (NAME, 200)
   STRING_ (ADDRESS, 200)
END_TABLE

The INT_(ID) macro creates new extern SqlId ID;. The same with other *_ macro versions. If you want to use the same SqlId, there are macros without '_' postfix.

Quote:

I think, that S_* structure created to use in this situation, to decide this collision.

S_WORKER w;
S_PLANT p;

SQL * Select (w.NAME,w.LASTNAME,p.NAME,p.ADDRESS).FROM(WORKER).LeftJoin(PLANT).On(w.PLANT_ID==p.ID);

while SQL.Fetch() {
   Cout() << SQL[w.NAME] << SQL[w.LASTNAME] << SQL[p.NAME] << SQL[p.ADDRESS];
}

Correct version of source code, might look like:

LOG("Inserting values:");
SQL * Insert(WORKER)(ID, 0)(NAME, "Joe")(LASTNAME, "Smith")(PLANT_ID, 0);
LOG(SQL.ToString());
SQL * Insert(WORKER)(ID, 1)(NAME, "Mike")(LASTNAME, "Smith")(PLANT_ID, 0);
LOG(SQL.ToString());
SQL * Insert(WORKER)(ID, 2)(NAME, "Jon")(LASTNAME, "Goober")(PLANT_ID, 1);
LOG(SQL.ToString());

SQL * Insert(PLANT)(ID, 0)(NAME, "First Plant")(ADDRESS, "First st.");
LOG(SQL.ToString());
SQL * Insert(PLANT)(ID, 1)(NAME, "Second Plant")(ADDRESS, "Second st.");
LOG(SQL.ToString());

LOG("Selecting values:");
SQL * Select(WORKER(NAME, LASTNAME), PLANT(NAME, ADDRESS)).From(WORKER).LeftJoin(PLANT).On(WORKER(PLANT_ID) == PLANT(ID));
LOG(SQL.ToString());
while (SQL.Fetch()) {
	LOG("-----");
	DUMP(SQL[NAME]); DUMP(SQL[LASTNAME]);
	DUMP(SQL[2]); DUMP(SQL[3]);
}

With following output:

Inserting values:
insert into WORKER(ID, NAME, LASTNAME, PLANT_ID) values (0, 'Joe', 'Smith', 0)
insert into WORKER(ID, NAME, LASTNAME, PLANT_ID) values (1, 'Mike', 'Smith', 0)
insert into WORKER(ID, NAME, LASTNAME, PLANT_ID) values (2, 'Jon', 'Goober', 1)
insert into PLANT(ID, NAME, ADDRESS) values (0, 'First Plant', 'First st.')
insert into PLANT(ID, NAME, ADDRESS) values (1, 'Second Plant', 'Second st.')
Selecting values:
select WORKER.NAME, WORKER.LASTNAME, PLANT.NAME, PLANT.ADDRESS from WORKER left outer join PLANT on WORKER.PLANT_ID = PLANT.ID
-----
SQL[NAME] = Joe
SQL[LASTNAME] = Smith
SQL[2] = First Plant
SQL[3] = First st.
-----
SQL[NAME] = Mike
SQL[LASTNAME] = Smith
SQL[2] = First Plant
SQL[3] = First st.
-----
SQL[NAME] = Jon
SQL[LASTNAME] = Goober
SQL[2] = Second Plant
SQL[3] = Second st.

[Updated on: Thu, 17 May 2012 10:33]

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