Gau*_*pta 2 regex node.js hapijs joi
我需要接受一个手机号码作为网络服务的输入,但我在使用Joi框架验证时面临以下问题.
Joi说:
Error: pattern must be a RegExp
at Object.exports.assert (/home/gaurav/Gaurav-Drive/code/nodejsWorkspace/ragchews/node_modules/joi/node_modules/hoek/lib/index.js:524:11)
at internals.String.regex (/home/gaurav/Gaurav-Drive/code/nodejsWorkspace/ragchews/node_modules/joi/lib/string.js:107:10)
at /home/gaurav/Gaurav-Drive/code/nodejsWorkspace/ragchews/src/validators/userValidator.js:10:40
at Object.<anonymous> (/home/gaurav/Gaurav-Drive/code/nodejsWorkspace/ragchews/src/validators/userValidator.js:13:2)
at Module._compile (module.js:456:26)
at Object.Module._extensions..js (module.js:474:10)
at Module.load (module.js:356:32)
at Function.Module._load (module.js:312:12)
at Module.require (module.js:364:17)
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用于验证:
var userProfileValidation = function(){
return {
payload : {
uid: Joi.string().required().alphanum().length(userConfigs.UID_LENGTH),
mobile_num: Joi.string().required().regex('^[0-9]*$').length(userConfigs.RMN_LENGTH) //for this guy
}
};
}();
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我检查了正则表达式freeformatter,它似乎为atleast一些投入正常工作.我不明白为什么joi会抛出这个错误.
实际上,正如这里所记载的那样(链接),Joi框架在这里不是一个正则表达式模式,而是一个真正的正则表达式.即:你应该使用:
Joi.string().required().regex(/^[0-9]*$/) [...]
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... 代替:
Joi.string().required().regex('^[0-9]*$') [...]
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