我在vs2012中使用c-interface实现SQLite。
我有三张桌子,其中两张是父母,他们没有任何钥匙连在一起。第三个是子节点,它应该有两个父表中的两个外键。我尝试了以下但是没有给我以下错误:
foreign key contraints failed
这是我的实施: 第一个父表:
CREATE TABLE Persons (
ID INTEGER PRIMARY KEY AUTOINCREMENT NOT NULL,
Name CHAR NULL ,
Age INT NULL
);
第二个父表:
CREATE TABLE Jobs (
Job_ID INTEGER PRIMARY KEY AUTOINCREMENT NOT NULL,
Description CHAR NULL ,
Country CHAR
);
子表
CREATE TABLE Persons_Jobs (
Title CHAR NULL,
country CHAR NULL,
ID INT ,
Job_ID INT ,
FOREIGN KEY (ID) REFERENCES Persons(ID) ,
FOREIGN KEY (Job_ID) REFERENCES Jobs(Job_ID)
);
请注意,我的表已成功创建,前两个表中的数据也已成功插入。
更新2:
插入声明:
void db_prepareInsertSql(sqlite3 *db){
sqlite3_int64 rowPersonID,rowJobID;
int i =0;
char *sql;
char str[100];
do{
i = i+1;
sprintf_s(str, "INSERT INTO Persons VALUES(NULL,'liena',%d);",i);
sql = str;
db_execute_sql(db,sql);
fprintf(stdout,"Persons insertion");
rowPersonID = sqlite3_last_insert_rowid(db);
sprintf_s(str, "INSERT INTO Jobs VALUES(NULL,'Doc','SDN');");
sql = str;
db_execute_sql(db,sql);
fprintf(stdout,"Jobs insertion");
rowJobID = sqlite3_last_insert_rowid(db);
\\the error occurs here
sprintf_s(str, "INSERT INTO Persons_Jobs VALUES('A','krt',%d,%d);",rowPersonID,rowJobID);
sql = str;
db_execute_sql(db,sql);
fprintf(stdout,"Persons_Jobs insertion");
}
while(i!=10);
}
创建表:
void db_prepareCreateTablesSql(sqlite3 *db){
char *sql;
char str[500];
sprintf_s(str, "PRAGMA foreign_keys = ON;");
sql = str;
db_execute_sql(db,sql);
fprintf(stdout,"Enable foriegn-keys feature");
sprintf_s(str, "CREATE TABLE IF NOT EXISTS Persons("
"ID INTEGER PRIMARY KEY AUTOINCREMENT NOT NULL, "
"Name CHAR NULL , "
"Age INT NULL ); ");
sql = str;
db_execute_sql(db,sql);
fprintf(stdout,"table Persons");
sprintf_s(str, "CREATE TABLE IF NOT EXISTS Jobs("
"Job_ID INTEGER PRIMARY KEY AUTOINCREMENT NOT NULL, "
"Description CHAR NULL , "
"Country CHAR )");
sql = str;
db_execute_sql(db,sql);
fprintf(stdout,"table Jobs");
sprintf_s(str, "CREATE TABLE IF NOT EXISTS Persons_Jobs("
"Title CHAR NULL , "
"country CHAR NULL , "
"ID INT , "
"Job_ID INT , "
"FOREIGN KEY (ID) REFERENCES Persons(ID),"
"FOREIGN KEY (Job_ID) REFERENCES Jobs(Job_ID)) ;");
sql = str;
db_execute_sql(db,sql);
fprintf(stdout,"table Persons_Jobs");
db_prepareInsertSql(db);
}
当我调试时,我发现了这个:
所以,我的问题是在子表中,如何从两个不同的父表中指定两个外键?
答案 0 :(得分:3)
两个表中的两个插入(Persons
和Jobs
)可能会导致两个不同的自动递增值。所以你应该用(伪代码)抓住它们:
INSERT INTO Persons VALUES(NULL,'liena',1); // 1st table
rowPersonID = sqlite3_last_insert_rowid(db) // catch PersonID
INSERT INTO Jobs VALUES(NULL,'Doc','SDN'); // 2nd table
rowJobID = sqlite3_last_insert_rowid(db) // catch JobID
然后:
INSERT INTO Career VALUES('A','krt',%d,%d)",rowPersonID,rowJobID;
答案 1 :(得分:1)
问题出在这一行:
sprintf_s(str, "INSERT INTO Persons_Jobs VALUES('A','krt',%d,%d);",rowPersonID,rowJobID);
rowPersonID
和rowJobID
的类型为sqlite3_int64
:
sqlite3_int64 rowPersonID,rowJobID;
是64位类型,而%d
需要一个32位整数。您需要将其强制转换为整数或使用I64d
格式化64位数字:
sprintf_s(str, "INSERT INTO Persons_Jobs VALUES('A','krt',%I64d,%I64d);", rowPersonID,rowJobID);