我无法使用ESAPI
class'
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".
如何从ESAPI.properties
文件传递参数类型?
有什么例子可以使用我可以参考的属性文件值吗?
答案 0 :(得分:5)
这是一个示例调用,我正在验证“到”地址字段:
validator.getValidInput("toAddress", it.next(), "Email", Email.MAX_ADDRESS_SIZE, true)
ESAPI假设您正在使用IDE或可以访问直接来源。如果您正在使用Eclipse,只需将鼠标悬停在方法名称上,即可显示参数类型。
===修订===
这是直接来自javadoc的翻录:
/**
* 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
*/
答案 1 :(得分:0)
ESAPI.validator().getValidInput("Validationofinput", StringInput, "Onlycharacters",200, true);