非法参数异常的解决方案：Last encoded character is a valid base 64 alphabet but not a possible value

冷不防 2022-11-16 05:57 916阅读 0赞

# 问题 #

## 来源 ##

今天在开发的过程中，前端传过来的是base64的图片地址，需要后端解析出该地址，之前使用的是commonsc-codec进行解密，但报出来如下的错误：  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2x2b2VsaWZl_size_16_color_FFFFFF_t_70]

## 分析原因 ##

于是，到源码里面看报错信息的出处，原来问题出现在这里，commonsc-codec在升级之后，其内部做了一个validateCharacter校验，草报出这样的错误。

/** * Validates whether decoding the final trailing character is possible in the context * of the set of possible base 64 values. * * <p>The character is valid if the lower bits within the provided mask are zero. This * is used to test the final trailing base-64 digit is zero in the bits that will be discarded. * * @param emptyBitsMask The mask of the lower bits that should be empty * @param context the context to be used * * @throws IllegalArgumentException if the bits being checked contain any non-zero value */
        private static void validateCharacter(final int emptyBitsMask, final Context context) { 
            if ((context.ibitWorkArea & emptyBitsMask) != 0) { 
                throw new IllegalArgumentException(
                    "Last encoded character (before the paddings if any) is a valid base 64 alphabet but not a possible value. " +
                    "Expected the discarded bits to be zero.");
            }
        }

## 解决方案 ##

可以使用谷歌的guava来替换common-codec来解密base64格式的文件

<dependency>
          <groupId>com.google.guava</groupId>
          <artifactId>guava</artifactId>
          <version>30.1.1-jre</version>
     </dependency>

如下代码解析代码：

byte[] imageByte = BaseEncoding.base64().decode(imageString);

结果能正常解析。

[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2x2b2VsaWZl_size_16_color_FFFFFF_t_70]: /images/20221022/fd252a9a163c4c5696a25aec09753ab2.png