c++ - 在数据库中插入空行而不是传递值

Question

我正在尝试使用 c 向 sqlite3 插入值，但不是传递值而是插入表中的空白行

int main() {
sqlite3 *db;
sqlite3_stmt *stmt;
int rc ;
rc = sqlite3_open("mydb.db", &db);

if (rc)
{
    fprintf(stderr, "Can't open database: %s\n", sqlite3_errmsg(db));
}
else
{
    fprintf(stderr, "Opened database successfully\n");
}

sqlite3_prepare_v2(db, "select * FROM `99940` a INNER JOIN (SELECT rowid FROM `Search_99940` ('B Rujesh P Balakrishnan')) b ON b.rowid = a.rowid WHERE upper(a.circle) = ('TAMIL NADU')", -1, &stmt, NULL);

sqlite3_bind_int(stmt, 0, 16);

if ((rc = sqlite3_step(stmt)) == SQLITE_ROW)
{
    string try1 = string(reinterpret_cast<const char*>(sqlite3_column_text(stmt, 1)));
    string try2 = string(reinterpret_cast<const char*>(sqlite3_column_text(stmt, 2)));
    string try3 = string(reinterpret_cast<const char*>(sqlite3_column_text(stmt, 3)));
    string try4 = string(reinterpret_cast<const char*>(sqlite3_column_text(stmt, 4)));
    string try5 = string(reinterpret_cast<const char*>(sqlite3_column_text(stmt, 5)));
    string try6 = string(reinterpret_cast<const char*>(sqlite3_column_text(stmt, 6)));
    string try7 = string(reinterpret_cast<const char*>(sqlite3_column_text(stmt, 7)));
    string try8 = string(reinterpret_cast<const char*>(sqlite3_column_text(stmt, 8)));
    string try9 = string(reinterpret_cast<const char*>(sqlite3_column_text(stmt, 9)));


    char* errorMessage;
    sqlite3_exec(db, "BEGIN TRANSACTION", NULL, NULL, &errorMessage);

    char buffer[] = "INSERT INTO `99946` (Number,Full_Name,Address,Date,Circle,Operator,Alterno,IDProof,SimType)  VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?);";   
    sqlite3_prepare_v2(db, buffer, strlen(buffer), &stmt, NULL);

    for (unsigned i = 0; i < 1; i++)
    {
        sqlite3_bind_text(stmt, 1, try1.c_str(), 0, NULL);
        sqlite3_bind_text(stmt, 2, try2.c_str(), 0, NULL);
        sqlite3_bind_text(stmt, 3, try3.c_str(), 0, NULL);
        sqlite3_bind_text(stmt, 4, try4.c_str(), 0, NULL);
        sqlite3_bind_text(stmt, 5, try5.c_str(), 0, NULL);
        sqlite3_bind_text(stmt, 6, try6.c_str(), 0, NULL);
        sqlite3_bind_text(stmt, 7, try7.c_str(), 0, NULL);
        sqlite3_bind_text(stmt, 8, try8.c_str(), 0, NULL);
        sqlite3_bind_text(stmt, 9, try9.c_str(), 0, NULL);

        if (sqlite3_step(stmt) != SQLITE_DONE)
        {
            printf("Commit Failed!\n");
        }
        else
        {
            printf("Success!\n");
        }
        sqlite3_reset(stmt);
   }
    sqlite3_exec(db, "COMMIT TRANSACTION", NULL,NULL, &errorMessage);
    sqlite3_finalize(stmt);
}

return 0;

我的选择查询工作正常，而且我在 try1 到 try9 中获取值。像 try1 = 'xyz' 我用

得到那个值

string try1 = string(reinterpret_cast<const char*>(sqlite3_column_text(stmt, 1)));

有什么问题我不明白请帮我解决这个问题。我尝试插入的方式是错误的我不明白

Answer 1

截至目前，您正在写入 0 个字节。

解释:

    sqlite3_bind_text(stmt, 1, try1.c_str(), 0, NULL);

语法是

int sqlite3_bind_text(sqlite3_stmt*,int,const char*,int,void(*)(void*));

根据 sqlite3

The third argument is the value to bind to the parameter. If the third parameter to sqlite3_bind_text() or sqlite3_bind_text16() or sqlite3_bind_blob() is a NULL pointer then the fourth parameter is ignored and the end result is the same as sqlite3_bind_null().

In those routines that have a fourth argument, its value is the number of bytes in the parameter. To be clear: the value is the number of bytes in the value, not the number of characters. If the fourth parameter to sqlite3_bind_text() or sqlite3_bind_text16() is negative, then the length of the string is the number of bytes up to the first zero terminator.

---

于是改为如下写直到空字节。

sqlite3_bind_text(stmt, 1, try1.c_str(), -1, NULL);