此问题可能是

This question is a possible duplicate of How to assign string to bytes array, but still answering it as there is a better, alternative solution:

使用简单的转化:

与字符串类型之间的转换

[...]

1. 将字符串类型的值转换为字节类型的切片会产生一个切片，其连续元素是字符串的字节.

所以您可以简单地做:

s := "some text"
b := []byte(s) // b is of type []byte

但是， string =>[] byte 转换会生成字符串内容的副本(必须复制，因为 string 是不可变的，而 [] byte 值不是不可更改的)，并且在 string 较大的情况下，效率不高.相反，您可以使用 io.Reader href ="https://golang.org/pkg/strings/#NewReader" rel ="nofollow noreferrer"> strings.NewReader() 将从传递的中读取字符串而不复制它.并且您可以将此 io.Reader 传递给 json.NewDecoder() 并使用 Decoder进行解组.Decode() 方法:

However, the string => []byte conversion makes a copy of the string content (it has to, as strings are immutable while []byte values are not), and in case of large strings it's not efficient. Instead, you can create an io.Reader using strings.NewReader() which will read from the passed string without making a copy of it. And you can pass this io.Reader to json.NewDecoder() and unmarshal using the Decoder.Decode() method:

s := `{"somekey":"somevalue"}`

var result interface{}
err := json.NewDecoder(strings.NewReader(s)).Decode(&result)
fmt.Println(result, err)

输出(在游乐场上尝试):

map[somekey:somevalue] <nil>

注意:调用 strings.NewReader()和 json.NewDecoder()确实有一些开销，因此，如果您使用的是小型JSON文本，则可以放心将其转换为 [] byte 并使用 json.Unmarshal() ，它不会变慢:

Note: calling strings.NewReader() and json.NewDecoder() does have some overhead, so if you're working with small JSON texts, you can safely convert it to []byte and use json.Unmarshal(), it won't be slower:

s := `{"somekey":"somevalue"}`

var result interface{}
err := json.Unmarshal([]byte(s), &result)
fmt.Println(result, err)

输出是相同的.在游乐场上尝试.

Output is the same. Try this on the Go Playground.

注意:如果通过读取一些 io.Reader (例如文件或网络连接)来获取JSON输入 string ，则可以直接传递 io.Reader 转换为 json.NewDecoder()，而无需先从中读取内容.

Note: if you're getting your JSON input string by reading some io.Reader (e.g. a file or a network connection), you can directly pass that io.Reader to json.NewDecoder(), without having to read the content from it first.