Question

我的过滤器出了问题。我这样定义了它：

function time_elapsed_string($difference)
{       

    //Days
    $days = round(($difference / 86400), 2);

    //Hours
    $hours = floor($difference / 3600);
    if($hours >= 24) {
        $remainderHours = fmod($hours, 24); // Get the remainder from the days.

        if ($remainderHours < 10) {
            $remainderHours = '0' . $remainderHours;
        }
    } else {
        $remainderHours = $hours; 
        $days = 0;


        if ($remainderHours < 10) {
            $remainderHours = '0' . $hours;
        }
    }   

    //Minutes
    $mins = floor($difference / 60);

    if ($mins >= 60){
        $remainderMins = fmod($mins, 60);

        if ($remainderMins < 10) {
            $remainderMins = '0' . $remainderMins;
        }
    } else {
        $remainderMins = $mins;
        if ($remainderMins < 10) {
            $remainderMins = '0' . $remainderMins;
        }
    }

    //Seconds
    $seconds = floor($difference);
    if($seconds >= 60) {
        $remainderSeconds = fmod($seconds, 60); 

        if ($remainderSeconds < 10) {
            $remainderSeconds = '0' . $remainderSeconds;
        }
    } else {
        $remainderSeconds = $seconds;
        if ($remainderSeconds < 10) {
            $remainderSeconds = '0' . $remainderSeconds;
        }
    }

    //Format day due to days being reset to 0 format in hours fuction
    $days = (floor($days) < 10 ? ('0' . floor($days)) : floor($days));

    return  $days . ':' . $remainderHours . ':' . $remainderMins     . ':' . $remainderSeconds;
}

在我的web-fragment.xml中，我有：

@WebFilter("springSecurityFilterChain")
public class JwtAuthenticationFilter extends AbstractAuthenticationProcessingFilter {...}

模式应该有效，但我一直得到这个例外：

<filter-mapping>
<filter-name>springSecurityFilterChain</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>

我还尝试在XML和@WebFilter注释中使用“/”，“/ security / *”，“”来更改模式，但我一直得到这个例外。有人可以解释一下我做错了吗？

谢谢。

修改

这是Spring Security配置：

java.lang.IllegalArgumentException: Invalid <url-pattern> springSecurityFilterChain in filter mapping

Answer 1

您的@WebFilter注释存在问题。如果您未在@WebFilter注释中指定任何内容，则默认情况下，它会将param作为网址格式，在您的情况下为springSecurityFilterChain。

因此，对于您的情况，您可以使用

@WebFilter("/*") 
public class JwtAuthenticationFilter extends AbstractAuthenticationProcessingFilter {...}

没有在xml文件中或下面指定任何内容：

@WebFilter(filterName = "abc") 
public class JwtAuthenticationFilter extends AbstractAuthenticationProcessingFilter {...}

并在您的xml文件中，您可以指定URL模式，如下所示：

<filter-mapping>
    <filter-name>abc</filter-name>
    <url-pattern>/*</url-pattern>
</filter-mapping>

@WebFilter（“springSecurityFilterChain”）抛出java.lang.IllegalArgumentException：无效<url-pattern>

1 个答案: