Java ESAPI getValidInput 方法的使用
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原文地址: http://stackoverflow.com/questions/22664006/
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Use of ESAPI getValidInput method
提问by abc
I was not able to use the of method present under ESAPIclass'
我无法使用ESAPI类下的方法'
java.lang.String getValidInput(java.lang.String context,
java.lang.String input,
java.lang.String type,
int maxLength,
boolean allowNull)
throws ValidationException,
IntrusionException
Parameters:
type - The regular expression name that maps to the actual regular expression from "ESAPI.properties".
How to pass parameter type from ESAPI.propertiesfile?
Any example to use properties file value from which I can refer?
如何从ESAPI.properties文件传递参数类型？任何使用我可以参考的属性文件值的示例？
采纳答案by avgvstvs
Here's an example call where I'm validating the "to" address field:
这是我验证“收件人”地址字段的示例调用：
validator.getValidInput("toAddress", it.next(), "Email", Email.MAX_ADDRESS_SIZE, true)
ESAPI assumes you're using an IDE or have access to the direct source. If you're using Eclipse, just mouse-hover over the method name, and the parameter types will be displayed.
ESAPI 假定您使用的是 IDE 或可以访问直接源。如果您使用 Eclipse，只需将鼠标悬停在方法名称上，就会显示参数类型。
===UPDATED===
===更新===
Here's the rip directly from the javadoc:
这是直接来自 javadoc 的 rip：
/**
* Returns canonicalized and validated input as a String. Invalid input will generate a descriptive ValidationException,
* and input that is clearly an attack will generate a descriptive IntrusionException.
*
* @param context
* A descriptive name of the parameter that you are validating (e.g., LoginPage_UsernameField). This value is used by any logging or error handling that is done with respect to the value passed in.
* @param input
* The actual user input data to validate.
* @param type
* The regular expression name that maps to the actual regular expression from "ESAPI.properties".
* @param maxLength
* The maximum post-canonicalized String length allowed.
* @param allowNull
* If allowNull is true then an input that is NULL or an empty string will be legal. If allowNull is false then NULL or an empty String will throw a ValidationException.
*
* @return The canonicalized user input.
*
* @throws ValidationException
* @throws IntrusionException
*/
回答by srinivas gowda
ESAPI.validator().getValidInput("Validationofinput", StringInput, "Onlycharacters",200, true);
