SQLite3 的 insert、update，和 or replace

没太多时间写正文，就看一个简单的例子程序吧，:)

#include
#include
#include

int main (void)
{
    int ret = -1;
    sqlite3 *db = NULL;

    // 创建数据库和表
    ret = sqlite3_open ("testonconf.db", &db);
    char *c = "CREATE TABLE test (id INTEGER PRIMARY KEY, data INTEGER)";
    ret = sqlite3_exec (db, c, NULL, NULL, NULL);

    // 插入三行数据
    char *i1 = "INSERT INTO test VALUES (1, 2222)";
    ret = sqlite3_exec (db, i1, NULL, NULL, NULL);
    char *i2 = "INSERT INTO test VALUES (5, 5555)";
    ret = sqlite3_exec (db, i2, NULL, NULL, NULL);
    char *i3 = "INSERT INTO test VALUES (9, 1129)";
    ret = sqlite3_exec (db, i2, NULL, NULL, NULL);

    // 试验1: 没什么悬念，id=5的原来data为5555的被替换成333333
    //char *i4 = "INSERT OR REPLACE INTO test (id, data) VALUES (5, 333333)";
    //ret = sqlite3_exec (db, i4, NULL, NULL, NULL);

    // 试验2: 错误19(SQLITE_CONSTRAINT)，约束冲突 己有id=1，且id是unique值
    //char *i5 = "UPDATE test SET id=5, data=545454 WHERE id=1";
    //ret = sqlite3_exec (db, i5, NULL, NULL, NULL);
    //printf ("ret: %d\n", ret); // 返回19, SQLITE_CONSTRAINT // Abort due to constraint violation
    // 试验3: 没什么悬念，id=5的原来data为5555的被替换成333333
    //char *i5 = "UPDATE test SET id=5, data=545454 WHERE id=5";
    //ret = sqlite3_exec (db, i5, NULL, NULL, NULL);

    // 试验4: 先删除 unique 冲突类型的id=1和id=5的数据，再执行update
    //char *i6 = "UPDATE OR REPLACE test SET id=5, data=545454 WHERE id=1";
    //ret = sqlite3_exec (db, i6, NULL, NULL, NULL);

    // 试验5: 没有 where 语句约束, 表示所有数据?
    char *i7 = "UPDATE OR REPLACE test SET id=5, data=666666";
    ret = sqlite3_exec (db, i7, NULL, NULL, NULL);

    system ("sqlite3 testonconf.db \"SELECT * FROM test;\"");
    char *d = "DELECT FROM test";
    ret = sqlite3_exec (db, d, NULL, NULL, NULL);
    ret = sqlite3_close (db);
    return 0;
}