Question

我想从我的表中获取名为'person'的行。我想在指标的帮助下做到这一点，以避免在此人没有名字时发生异常。怎么做？

我写了代码：

try
{
soci::statement st = (sql.prepare << "SELECT firstname FROM person;", soci::into(r, ind));
st.execute();

while (st.fetch())
{
    if(sql.got_data())
    {
        switch(ind)
        {
        case soci::i_ok:
            std::cout << r.get<std::string>(0) << "\n";
            break;
        case soci::i_null:
            std::cout << "Person has no firstname!\n";
            break;
        }
    }else
    {
        std::cout << "There's no such person!\n";
    }
}
}

但是只有当我添加一行时才显示没有行：

std::cout << r.get<std::string>(0) << "\n";

在我的if语句之前，我才看到数据库中的firstnames。

Answer 1

我认为您不需要soci::row用于此目的，而是std::string

//...
std::string firstname;
soci::statement st = (sql.prepare << "SELECT firstname FROM person;"
                    , soci::into(firstname, ind));

//...
        case soci::i_ok:
            std::cout << firstname << std::endl;
            break;
//...

Answer 2

如果要查询整个数据库表行，则只能使用soci :: row。由于您只查询列，＆＃39; firstname＆＃39;你可以直接把它拿到一个字符串中。所以你的代码看起来像;

soci::indicator ind;
std::string sFirstName;
try
{
soci::statement st = (sql.prepare << "SELECT firstname FROM person;", 
    soci::into(sFirstName, ind));
st.execute();

while (st.fetch())
{
    switch(ind)
    {
        case soci::i_ok: {
            std::cout << sFirstName << std::endl;
            break; }
        case soci::i_null: {
            std::cout << "Person has no first name!" << std::endl;
            break;
        }
    }
}

使用SOCI与soci :: indicators [C ++]从表中获取行

2 个答案: