Question

我有一个小问题。我在对数据库进行查询后尝试迭代rows，但是我无法再次遍历它以找到females。

我相信这是因为在迭代rows打印出来之后，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)
    }
}

反正我可以把指针放在开头吗？或者我还需要做另一个查询吗？

Answer 1

您无法“回滚”指向前一个值的指针，因此您必须使用另一个（临时）var来进行迭代。

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 loop s？

你不能只使用一个吗？像：

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

GOLang：控制指针

1 个答案: