如何从Golang中的MongoDB bson.m中制作多行字符串值？

发布于 2025-02-06 14:24:42 字数 740 浏览 1 评论 0原文

我正在使用Golang创建电报机器人，我需要一些有关如何从Golang中检索多行字符串值的建议，该函数具有与此Python String相同的逻辑，

answer = """1 John 95
2 Sam 89
3 Mike 72"""

我具有一个函数，该函数创建了一个mongoDB请求，该请求使我获取数据在bson.m数据类型中。之后，我需要使用此表格将此查询数据作为单个字符串值发送给用户：

msg := tgbotapi.NewMessage(
    update.Message.Chat_ID,
    answer,
)
bot.Send(msg)

我真的不知道如何将此bson.m数据转换为单个多行字符串。

我从功能中获得的bson.m响应：

[map[_id:ObjectID("62a4acf2a494a2814238c6e1") bandMember:John name:School12 points:95] 
map[_id:ObjectID("62a4acf2a494a2814238c6e2") bandMember:Sam name:School15 points:89] 
map[_id:ObjectID("62a4acf2a494a2814238c6e3") bandMember:Mike name:School7 points:72]]

我必须将其插入“答案”的字符串变量（请参见上面的示例），

谢谢！

原文

I'm creating a telegram bot using Golang, and I need some advice on how to retrieve a multiline string value from function in Golang, that has same logic like this Python string

answer = """1 John 95
2 Sam 89
3 Mike 72"""

I have a function that creates a MongoDB request that gets me a data in bson.M datatype. And after that I need to send this queried data to a user as a single string value using this form:

msg := tgbotapi.NewMessage(
    update.Message.Chat_ID,
    answer,
)
bot.Send(msg)

I don't really know how to transform this bson.M data into a single multiline string.

bson.M response that I get from the function:

[map[_id:ObjectID("62a4acf2a494a2814238c6e1") bandMember:John name:School12 points:95] 
map[_id:ObjectID("62a4acf2a494a2814238c6e2") bandMember:Sam name:School15 points:89] 
map[_id:ObjectID("62a4acf2a494a2814238c6e3") bandMember:Mike name:School7 points:72]]

And I have to insert it in a string variable of "answer" (see above example)

Thanks in advance!

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嘿看小鸭子会跑 2025-02-13 14:24:42

循环浏览文档，并为每个文档提供一条线。加入Newline以获得最终结果。

var lines []string
for i, m := range data {
    lines = append(lines, fmt.Sprintf("%d %s %d", i+1, m["bandMember"], m["points"]))
}
result := strings.Join(lines, "\n")

Loop through the documents and sprintf a line for each document. Join the lines with newline to get the final result.

var lines []string
for i, m := range data {
    lines = append(lines, fmt.Sprintf("%d %s %d", i+1, m["bandMember"], m["points"]))
}
result := strings.Join(lines, "\n")

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