英文:
How to trim and escape a string pointer in golang?
问题
用户名不是必需的，可以为nil，所以我将该属性指针化为字符串。
type User struct {Username *string `json:"username"`Email string `json:"email"`Password string `json:"password"`}func (u *User) PrepareUser() {if u.Username != nil {*u.Username = html.EscapeString(strings.TrimSpace(*u.Username))}u.Email = html.EscapeString(strings.TrimSpace(u.Email))u.Password = strings.TrimSpace(u.Password)}
尝试修剪和转义时，出现错误："无法将html.EscapeString(strings.TrimSpace(u.Username))（类型为string的值）用作*string赋值的值"。
英文:
Username is not required and it can be nil, so I made this property pointer to string.
type User struct {Username *string `json:"username"`Email string `json:"email"`Password string `json:"password"`}func (u *User) PrepareUser() {if u.Username != nil {u.Username = html.EscapeString(strings.TrimSpace(u.Username))}u.Email = html.EscapeString(strings.TrimSpace(u.Email))u.Password = strings.TrimSpace(u.Password)}
When trying to trim and escape I see "cannot use html.EscapeString(strings.TrimSpace(u.Username)) (value of type string) as *string value in assignment"
答案1
得分: 2
你需要通过*运算符访问指针的值。
html.EscapeString(strings.TrimSpace(*u.Username))
更新
还要记得使用*运算符来赋值
*u.Username = html.EscapeString(strings.TrimSpace(*u.Username))
英文:
You need to access the value of pointer via * operator.
html.EscapeString(strings.TrimSpace(*u.Username))
Update
Also don't forget to use * operator for assigning value
*u.Username = html.EscapeString(strings.TrimSpace(*u.Username))
答案2
得分: 1
func trimEscapeStrPtr(s *string) *string {
t := html.EscapeString(strings.TrimSpace(*s))
return &t
}
英文:
func trimEscapeStrPtr(s *string) *string {t := html.EscapeString(strings.TrimSpace(*s))return &t}
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