Question

我一直在阅读有关弹簧请求验证的很多内容。我读了很多关于如何正确实现它的文章，但我有一些问题。这是我的代码：

RestController：

    @Autowired
EmployeeManager employeeManager;

@Autowired
EmployeeValidator employeeValidator;

@InitBinder("employee")
public void setupBinder(WebDataBinder binder) {
    binder.addValidators(employeeValidator);
}

// -------------- CREATE EMPLOYEES --------------

@PostMapping(value = "add")
public ResponseEntity<EmployeeDTO> addEmployee(@Valid @RequestBody EmployeeDTO employee) {

    boolean isCreated = employeeManager.addEmployee(employee);

    if(isCreated) {
        return new ResponseEntity<>(employee, HttpStatus.CREATED);
    }

    return new ResponseEntity(new CustomError("Unable to create, employee with email " +
            employee.getEmail() + " already exist."), HttpStatus.CONFLICT);
}

验证

    package com.employee.api.EmployeeAPI.validator;


import com.employee.api.EmployeeAPI.model.dto.EmployeeDTO;
import org.springframework.stereotype.Component;
import org.springframework.validation.Errors;
import org.springframework.validation.Validator;

import java.util.regex.Matcher;
import java.util.regex.Pattern;

@Component
public class EmployeeValidator implements Validator {

    private Pattern pattern;
    private Matcher matcher;

    private static final String STRING_PATTERN = "[a-zA-Z]+";
    private static final String EMAIL_PATTERN = "^[_A-Za-z0-9-\\+]+(\\.[_A-Za-z0-9-]+)*@"
            + "[A-Za-z0-9-]+(\\.[A-Za-z0-9]+)*(\\.[A-Za-z]{2,})$";

    @Override
    public boolean supports(Class<?> clazz) {
        return EmployeeDTO.class.equals(clazz);
    }

    @Override
    public void validate(Object target, Errors errors) {
        EmployeeDTO employee = (EmployeeDTO) target;
        if (validateInputString(employee.getFirstName(), STRING_PATTERN)) {
            errors.rejectValue("firstName", "firstName.invalid");
        }

        if (validateInputString(employee.getLastName(), STRING_PATTERN)) {
            errors.rejectValue("lastName", "lastName.invalid");
        }

        if (validateInputString(employee.getJob(), STRING_PATTERN)) {
            errors.rejectValue("job", "job.invalid");
        }

        if (validateInputString(employee.getEmail(), EMAIL_PATTERN)) {
            errors.rejectValue("email", "email.invalid");
        }
    }

    private boolean validateInputString(String input, String regexPattern) {
        pattern = Pattern.compile(regexPattern);
        matcher = pattern.matcher(input);

        return (!matcher.matches() || input == null || input.trim().length() == 0);
    }

}

在配置中我添加了bean：

@Bean
public EmployeeValidator beforeAddOrUpdateEmployeeValidator() {
    return new EmployeeValidator();
}

我不确定在添加员工时应该如何调用它，因为它现在肯定不起作用。你能帮我正确实施还是指向正确的方向？

Answer 1

我不熟悉org.springframework.validation.Validator，但会建议您如何使用javax.validation.ConstraintValidator（ JSR-303 ）进行相同的验证。您的控制器类很好，不需要进行任何更改。

您需要创建自定义注释@ValidEmployee并使用它注释您的dto：

@ValidEmployee
public class EmployeeDto {
    ...
}

ValidEmployee注释：

import javax.validation.Constraint;
import javax.validation.Payload;

@Target({TYPE, ANNOTATION_TYPE})
@Retention(RUNTIME)
@Constraint(validatedBy = EmployeeValidator.class)
@Documented
public @interface ValidEmployee {
    String message() default "{ValidEmployee.message}";
    Class<?>[] groups() default {};
    Class<? extends Payload>[] payload() default {};
}

并在isValid方法中实现验证逻辑：

import javax.validation.ConstraintValidator;
import javax.validation.ConstraintValidatorContext;

public class EmployeeValidator implements ConstraintValidator<ValidEmployee, EmployeeDto> {

    @Override
    public void initialize(ValidEmployee constraintAnnotation) {
    }

    @Override
    public boolean isValid(EmployeeDto employee, ConstraintValidatorContext context) {
            // do your validation logic
    }

}

Spring Boot请求验证

1 个答案: