为什么不允许%
为什么不允许%
提问于 2013-02-21 22:14:46
如果我将密码输入为Admin$&%^或Admin$&^%，那么我的ajax调用可以正常工作，但是如果我像这样输入Admin$!%或Admin$!%^。ajax调用抛出以下异常.....
请帮帮我，因为我找不到这个问题的根本原因。
INFO: Character decoding failed. Parameter [txt_password] with value [Admin$!%] has been ignored. No
te that the name and value quoted here may be corrupted due to the failed decoding. Use debug level
logging to see the original, non-corrupted values.
java.io.CharConversionException: EOF
at org.apache.tomcat.util.buf.UDecoder.convert(UDecoder.java:80)
at org.apache.tomcat.util.buf.UDecoder.convert(UDecoder.java:46)
at org.apache.tomcat.util.http.Parameters.urlDecode(Parameters.java:410)
at org.apache.tomcat.util.http.Parameters.processParameters(Parameters.java:370)
at org.apache.tomcat.util.http.Parameters.processParameters(Parameters.java:217)
at org.apache.catalina.connector.Request.parseParameters(Request.java:2647)
at org.apache.catalina.connector.Request.getParameter(Request.java:1106)
at org.apache.catalina.connector.RequestFacade.getParameter(RequestFacade.java:355)
at javax.servlet.ServletRequestWrapper.getParameter(ServletRequestWrapper.java:158)在探索过程中，我发现：
Special characters are not allowed inside the query string. They must be replaced by a "%" followed by the ASCII code in Hex. E.g., "~" is replaced by "%7E", "#" by "%23" and so on. Since blank is rather common, it can be replaced by either "%20" or "+" (the "+" character must be replaced by "%2B"). This replacement process is called URL-encoding, and the result is a URL-encoded query string. 那么这是否意味着我们不能在输入字段中使用%作为值呢？
回答 2
Stack Overflow用户
回答已采纳
发布于 2013-02-25 16:35:00
最终得到了解决方案，
在查询字符串中添加参数之前，我使用了javascript的encodeURIComponent函数。来自W3Schools
The encodeURI() function is used to encode a URI.
This function encodes special characters, except: , / ? : @ & = + $ # (Use encodeURIComponent() to encode these characters).
Tip: Use the decodeURI() function to decode an encoded URI.Stack Overflow用户
发布于 2013-02-21 23:43:42
您正在尝试对已在上一步中进行URL解码的字符串进行URL解码。这就是你的根本原因。
两组已验证和已失效的字符串之间的区别是'&‘字符。不确定为什么带“与”号的验证...解析器可能在到达'%‘之前停止，因为'&’被视为分隔符。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/15004452
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