[英]Form input validation with JAX-RS
I want to use JAX-RS REST services as a back-end for a web application used directly by humans with browsers. 我想使用JAX-RS REST服务作为人类使用浏览器直接使用的Web应用程序的后端。 Since humans make mistakes from time to time I want to validate the form input and redisplay the form with validation message, if something wrong was entered.
由于人类不时会犯错误,我想验证表单输入并使用验证消息重新显示表单,如果输入错误。 By default JAX-RS sends a 400 or 404 status code if not all or wrong values were send.
默认情况下,如果未发送全部或错误的值,JAX-RS将发送400或404状态代码。
Say for example the user entered a "xyz" in the form field "count": 例如,用户在表单字段“count”中输入了“xyz”:
@POST
public void create(@FormParam("count") int count) {
...
}
JAX-RS could not convert "xyz" to int
and returns "400 Bad Request". JAX-RS无法将“xyz”转换为
int
并返回“400 Bad Request”。
How can I tell the user that he entered an illegal value into the field "count"? 如何告诉用户他在字段“count”中输入了非法值? Is there something more convenient than using Strings everywhere and perform conversation by hand?
有什么比在任何地方使用字符串更方便,并手动进行对话?
I would recommend to use AOP, JSR-303, and JAX-RS exception mappers for example: 我建议使用AOP,JSR-303和JAX-RS异常映射器,例如:
import javax.validation.constraints.Pattern;
@POST
public void create(@FormParam("count") @Pattern("\\d+") String arg) {
int count = Integer.parseInt(arg);
}
Then, define a JAX-RS exception mapper that will catch all ValidationException
-s and redirect users to the right location. 然后,定义一个JAX-RS异常映射器,它将捕获所有
ValidationException
-s并将用户重定向到正确的位置。
I'm using something similar in s3auth.com form validation with JAX-RS: https://github.com/yegor256/s3auth/blob/master/s3auth-rest/src/main/java/com/s3auth/rest/IndexRs.java 我在使用JAX-RS的s3auth.com表单验证中使用类似的东西: https : //github.com/yegor256/s3auth/blob/master/s3auth-rest/src/main/java/com/s3auth/rest/IndexRs的.java
Use @FormParam("count") Integer count 使用@FormParam(“count”)整数计数
that will work. 那可行。
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