[英]Is there a way to refactor this AND and XOR logical branch?
我有 2 个用户 ID,如果我只有一个或两个,我想做不同但非常相似的逻辑。 有没有办法整合这段代码,因为现在它看起来很丑。
function getUserPermission(primaryId, secondaryId, role) {
let permission = userInfo.permissionList.filter( permission => {
//logical AND
if(primaryId && secondaryId){
// have both IDs, use both IDs
(permission.primaryId === primaryId && permission.secondaryId === secondaryId) && permission.role === role
}
//logical XOR
else if((primaryId && !secondaryId) || (!primaryId && secondaryId)) {
// have only 1 ID, use 1 ID
(permission.primaryId === primaryId || permission.secondaryId === secondaryId) && permission.role === role
}
})[0]
return permission
}
像这样的事情呢?
if(!primaryId && !secondaryId) {
throw Error("No ID was provided");
}
let permission = userInfo.permissionList.filter(p => p.role === role);
if(primaryId) {
permission = permission.filter(p => p.primaryId === primaryId);
}
if(secondaryId) {
permission = permission.filter(p => p.secondaryId === secondaryId);
}
return permission[0];
首先,当两个 id 都无效时,这个逻辑似乎无法处理。 您将希望以某种方式处理该问题(到目前为止,其他答案似乎都包括这些方面的内容)。
接下来,由于您只返回第一个匹配权限,我建议使用一旦找到第一个匹配项就不会继续循环的解决方案。 在这方面,对于这个用例, Array.find()
比Array.filter()
好得多。
如果我将其他答案放在一起,那将是这样的:
function getUserPermission(primaryId, secondaryId, role) {
if (!primaryId && !secondaryId) return null // at least one must be populated, why loop at all?
return userInfo.permissionList.find(permission =>
permission.role === role
&& (!primaryId || primaryId === permission.primaryId) // if primary is populated, it needs to match
&& (!secondaryId || secondaryId === permission.secondaryId) // if secondary is populated, it needs to match
);
}
...但这并不能处理完全有效的 ID 恰好为 0 的情况。不过,这可能取决于您的数据。 例如,也许您总是使用基于字符串的 ID。 它还会检查每个循环上的有效 ID(您可能会或可能不会完全满意)
考虑到这些潜在问题,我对您的数据/类型做了一些假设,并提出了一种更非正统的方法来解决它们:
const getUserPermission = (primaryId = -1, secondaryId = -1, role = '') => {
if (primaryId < 0 && secondaryId < 0) return null
const conditions = [ p => p.role === role ]
.concat(primaryId >= 0 ? [ p => p.primaryId === primaryId ] : [])
.concat(secondaryId >= 0 ? [ p => p.secondaryId === secondaryId ] : [])
return userInfo.permissionList.find(p => conditions.every(fn => fn(p)))
}
...最后一个创建了一组函数, conditions
,以检查每个权限。 应该返回匹配每个条件的第一个(理论上,至少 - 我没有测试它)
function getUserPermission(primaryId, secondaryId, role) {
return userInfo.permissionList.find(permission =>
permission.role === role
&& (!!primaryId || !!secondaryId)// at least one is populated
&& (!primaryId || primaryId === permission.primaryId)// if primary is populated, it needs to match
&& (!secondaryId || secondaryId === permission.secondaryId)// if secondary is populated, it needs to match
);
}
这与您的代码略有不同,因为如果没有匹配,它将返回 null,而您的则抛出异常。 如果你想扔,你总是可以把结果放在一个变量中并检查 null。
这个重构怎么样:
function getUserPermission(primaryId = undefined,secondaryId = undefined,role) {
primaryId === undefined && secondaryId === undefined && throw new Error("Ids are required");
const userPermission = userInfo.permissionList
.filter((permission) => role && permission.role === role)
.filter((permission) => primaryId && permission.primaryId === primaryId)
.filter(
(permission) => secondaryId && permission.secondaryId === secondaryId
);
return userPermission[0];
}
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