Go Gin 将 json 响应转换为 base64
Go Gin converting json response to base64
我正在尝试将数据库查询数据作为 json 响应发送。这是我的控制器:
import (
"fmt"
"github.com/json-iterator/go"
"log"
)
func GetNewsPapers() []byte{
db := GetDB()
var json = jsoniter.ConfigCompatibleWithStandardLibrary
rows, err := db.Queryx(`SELECT title, language, ranking, slug, search_term, logo_url FROM public.news_newspaper`)
if err != nil {
log.Println(err)
}
defer rows.Close()
tableData := make([]map[string]interface{}, 0)
for rows.Next() {
entry := make(map[string]interface{})
err := rows.MapScan(entry)
if err != nil {
log.Println(err)
}
tableData = append(tableData, entry)
}
jsonData, _ := json.Marshal(tableData)
fmt.Println(string(jsonData)) // printing expected json
err = rows.Err()
if err != nil {
panic(err)
}
return jsonData
}
和
func (n *NewsPaperController) GetList(c *gin.Context) {
value := database.GetNewsPapers()
c.JSON(http.StatusOK, value)
}
问题是我得到的是 base64 字符串作为响应,而不是我期望的 json 对象。如果我将 value
转换为如下所示的字符串,我将获得人类可读的值 .
c.JSON(http.StatusOK, string(value))
但整个响应编码为这样的字符串:
"[{\"language\":\"en\",\"logo_url\":\"..\",\"ranking\":2,\"search_term\":\"..\",\"slug\":\"..\",\"title\":\"....\"}]
如何获得如下 json 响应:
[{"language":"en","logo_url":"..","ranking":2,"search_term":"..","slug":"..","title":".."} ]
func (c *Context) JSON(code int, obj interface{})
JSON serializes the
given struct as JSON into the response body. It also sets the
Content-Type as "application/json".
c.JSON()
序列化为 JSON 你不需要在使用前解组。在 c.JSON()
中使用 tableData
func GetNewsPapers() []map[string]interface{}{
// your existing code
return tableData
}
func (n *NewsPaperController) GetList(c *gin.Context) {
value := database.GetNewsPapers()
c.JSON(http.StatusOK, value)
}
并且使用 %#v
你可以看到值的 Go 语法表示,你也可以在其中找到转义字符
fmt.Printf("%#v", string(jsonData))
我正在尝试将数据库查询数据作为 json 响应发送。这是我的控制器:
import (
"fmt"
"github.com/json-iterator/go"
"log"
)
func GetNewsPapers() []byte{
db := GetDB()
var json = jsoniter.ConfigCompatibleWithStandardLibrary
rows, err := db.Queryx(`SELECT title, language, ranking, slug, search_term, logo_url FROM public.news_newspaper`)
if err != nil {
log.Println(err)
}
defer rows.Close()
tableData := make([]map[string]interface{}, 0)
for rows.Next() {
entry := make(map[string]interface{})
err := rows.MapScan(entry)
if err != nil {
log.Println(err)
}
tableData = append(tableData, entry)
}
jsonData, _ := json.Marshal(tableData)
fmt.Println(string(jsonData)) // printing expected json
err = rows.Err()
if err != nil {
panic(err)
}
return jsonData
}
和
func (n *NewsPaperController) GetList(c *gin.Context) {
value := database.GetNewsPapers()
c.JSON(http.StatusOK, value)
}
问题是我得到的是 base64 字符串作为响应,而不是我期望的 json 对象。如果我将 value
转换为如下所示的字符串,我将获得人类可读的值 .
c.JSON(http.StatusOK, string(value))
但整个响应编码为这样的字符串:
"[{\"language\":\"en\",\"logo_url\":\"..\",\"ranking\":2,\"search_term\":\"..\",\"slug\":\"..\",\"title\":\"....\"}]
如何获得如下 json 响应:
[{"language":"en","logo_url":"..","ranking":2,"search_term":"..","slug":"..","title":".."} ]
func (c *Context) JSON(code int, obj interface{})
JSON serializes the given struct as JSON into the response body. It also sets the Content-Type as "application/json".
c.JSON()
序列化为 JSON 你不需要在使用前解组。在 c.JSON()
tableData
func GetNewsPapers() []map[string]interface{}{
// your existing code
return tableData
}
func (n *NewsPaperController) GetList(c *gin.Context) {
value := database.GetNewsPapers()
c.JSON(http.StatusOK, value)
}
并且使用 %#v
你可以看到值的 Go 语法表示,你也可以在其中找到转义字符
fmt.Printf("%#v", string(jsonData))