[英]Is there a way to refactor this AND and XOR logical branch?
我有 2 個用戶 ID,如果我只有一個或兩個,我想做不同但非常相似的邏輯。 有沒有辦法整合這段代碼,因為現在它看起來很丑。
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
}
像這樣的事情呢?
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,當兩個 id 都無效時,這個邏輯似乎無法處理。 您將希望以某種方式處理該問題(到目前為止,其他答案似乎都包括這些方面的內容)。
接下來,由於您只返回第一個匹配權限,我建議使用一旦找到第一個匹配項就不會繼續循環的解決方案。 在這方面,對於這個用例, Array.find()
比Array.filter()
好得多。
如果我將其他答案放在一起,那將是這樣的:
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
);
}
...但這並不能處理完全有效的 ID 恰好為 0 的情況。不過,這可能取決於您的數據。 例如,也許您總是使用基於字符串的 ID。 它還會檢查每個循環上的有效 ID(您可能會或可能不會完全滿意)
考慮到這些潛在問題,我對您的數據/類型做了一些假設,並提出了一種更非正統的方法來解決它們:
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)))
}
...最后一個創建了一組函數, conditions
,以檢查每個權限。 應該返回匹配每個條件的第一個(理論上,至少 - 我沒有測試它)
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
);
}
這與您的代碼略有不同,因為如果沒有匹配,它將返回 null,而您的則拋出異常。 如果你想扔,你總是可以把結果放在一個變量中並檢查 null。
這個重構怎么樣:
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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