sql - 在 Go 中控制指针

Question

在对数据库进行查询后，我试图遍历 行，但我在第二次遍历它以找到 女性 时遇到了问题。

我相信这是因为在它遍历 rows 打印出 men 之后，指针留在最后并且无法返回。

rows2, rowErr :=db.Query("SELECT GIVENNAME,gender, count(givenname) as Frequency from people group by givenname order by givenname asc")

for rows2.Next() {
    nextErr := rows2.Scan(&givenName,&gender, &frequency)
    if nextErr != nil{
        log.Fatal(nextErr.Error())
    }
    if gender == "male" {
        fmt.Println(givenName, gender, frequency)
    }
}
fmt.Println("")
for rows2.Next() {
    nextErr := rows2.Scan(&givenName,&gender, &frequency)
    if nextErr != nil{
        log.Fatal(nextErr.Error())
    }
    if gender == "female" {
        fmt.Println(givenName, gender, frequency)
    }
}

如何将指针重新放在开头？还是我必须再做一次查询？

Answer 1

您不能“回滚”指向先前值的指针，因此您必须使用另一个(临时)变量来迭代。

rows2, rowErr := db.Query("SELECT GIVENNAME, gender, count(givenname) as Frequency from people group by givenname order by givenname asc")

var list1, list2 MyPeopleStruct 
temp := rows2

for temp.Next() {
    nextErr := rows2.Scan(&givenName,&gender, &frequency)
    if nextErr != nil{
        log.Fatal(nextErr.Error())
    }

    if gender == "male" {
        fmt.Println(givenName, gender, frequency)
    }
}

fmt.Println("")

temp = rows2 // reinitialise it to the actual value you need

for temp.Next() {
    nextErr := rows2.Scan(&givenName,&gender, &frequency)
    if nextErr != nil{
        log.Fatal(nextErr.Error())
    }

    if gender == "female" {
        fmt.Println(givenName, gender, frequency)
    }
}

---

无论如何，为什么你甚至需要制作两个不同的 for 循环？

你不能只用一个做所有事情吗？喜欢:

type MyPeopleStruct struct{
    Gender string
    Name string
    Frequency int
}

rows2, rowErr := db.Query("SELECT GIVENNAME, gender, count(givenname) as Frequency from people group by givenname order by givenname asc")

var listM, listF []MyPeopleStruct

temp := rows2

for temp.Next() {
    nextErr := rows2.Scan(&givenName,&gender, &frequency)
    if nextErr != nil{
        log.Fatal(nextErr.Error())
    }

    if gender == "male" {
        fmt.Println(givenName, gender, frequency)
        listM = append(listM, MyPeopleStruct{gender,givenName,frequency})
    }

    if gender == "female" {
        fmt.Println(givenName, gender, frequency)
        listF = append(listF, MyPeopleStruct{gender,givenName,frequency})
    }
}

// Use listM and listF