Question

此代码提供AFAIK正确的JSON输出[{}，{}]，但每行都会附加并替换所有以前的行，因此结果仅显示最后一行的副本。

var rows *sql.Rows
rows, err = db.Query(query)
cols, _ := rows.Columns()
colnames, _ := rows.Columns()
vals := make([]interface{}, len(cols))

for i, _ := range cols {
   vals[i] = &cols[i]
}

m := make(map[string]interface{})

for i, val := range vals {
  m[colnames[i]] = val
}

list := make([]map[string]interface{}, 0)
for rows.Next() {
err = rows.Scan(vals...)
   list = append(list, m)
}
json, _ := json.Marshal(list)
fmt.Fprintf(w,"%s\n", json)

这是在循环遍历行的幕后发生的事情：

循环1：{“ID”：“1”，“NAME”：＆＃34; John}

循环2：{“ID”：“2”，“NAME”：“Jane Doe”} {“ID”：“2”，“NAME”：“Jane Doe”}

循环3：{“ID”：“3”，“NAME”：“唐老鸭”} {“ID”：“3”，“NAME”：“唐老鸭”} {“ID”：“3” ，“NAME”：“唐老鸭”}

rows.Scan获取正确的值，但它会附加AND替换所有以前的值。

最终输出是

[{“ID”：“3”，“NAME”：“Donald Duck”}，{“ID”：“3”，“NAME”：“Donald Duck”}，{“ID”：“3” ，“NAME”：“唐老鸭”}]

但应该是这样的：

[{“ID”：“1”，“NAME”：“John Doe”}，{“ID”：“2”，“NAME”：“Jane Doe”}，{“ID”：“3” ，“NAME”：“唐老鸭”}]

我做错了什么？

你可能会投票，但请解释原因。我仍然是Golang的新手，想要学习。

Answer 1

我修复了它并用评论解释你做错了什么：

// 1. Query
var rows *sql.Rows
rows, err = db.Query(query)
cols, _ := rows.Columns()

// 2. Iterate
list := make([]map[string]interface{}, 0)
for rows.Next() {
    vals := make([]interface{}, len(cols))
    for i, _ := range cols {
        // Previously you assigned vals[i] a pointer to a column name cols[i].
        // This meant that everytime you did rows.Scan(vals),
        // rows.Scan would see pointers to cols and modify them
        // Since cols are the same for all rows, they shouldn't be modified.

        // Here we assign a pointer to an empty string to vals[i],
        // so rows.Scan can fill it.
        var s string
        vals[i] = &s

        // This is effectively like saying:
        // var string1, string2 string
        // rows.Scan(&string1, &string2)
        // Except the above only scans two string columns
        // and we allow as many string columns as the query returned us — len(cols).
    }

    err = rows.Scan(vals...)

    // Don't forget to check errors.
    if err != nil {
        log.Fatal(err)
    }

    // Make a new map before appending it.
    // Remember maps aren't copied by value, so if we declared
    // the map m outside of the rows.Next() loop, we would be appending
    // and modifying the same map for each row, so all rows in list would look the same.
    m := make(map[string]interface{})
    for i, val := range vals {
        m[cols[i]] = val
    }
    list = append(list, m)
}

// 3. Print.
b, _ := json.MarshalIndent(list, "", "\t")
fmt.Printf("%s\n", b)

不用担心，当我还是初学者时，我很难理解。

现在，有趣的事情：

var list []map[string]interface{}
rows, err := db.Queryx(query)
for rows.Next() {
    row := make(map[string]interface{})
    err = rows.MapScan(row)
    if err != nil {
      log.Fatal(err)
    }
    list = append(list, row)
}

b, _ := json.MarshalIndent(list, "", "\t")
fmt.Printf("%s\n", b)

这与上面的代码相同，但使用sqlx。有点简单，没有？

sqlx是database/sql之上的扩展程序，其方法是将行直接扫描到地图和结构中，因此您不必手动执行此操作。

我认为你的模型作为结构看起来更好：

type Person struct {
    ID int
    Name string
}

var people []Person
rows, err := db.Queryx(query)
for rows.Next() {
    var p Person
    err = rows.StructScan(&p)
    if err != nil {
        log.Fatal(err)
    }
    people = append(people, p)
}

你不这么认为吗？

如何将最后的sql行附加到列表而不替换Golang中的前一行

1 个答案: