錯誤屬性未定義傳遞匿名函數（如何咖喱）？

Question

我認爲這種技巧叫做咖喱。

問題：

https://repl.it/E4iG

ReferenceError: type is not defined 
    at hasType:14:27 
    at allApprovalTypes.map:18:26 
    at eval:17:21 
    at eval 

問： 如何使用hasType所以最後我可以代替未命名的匿名功能approvals.find(hasType)？

var approvals = [ 
     {type: 'media'}, 
     {type: 'scope'}, 
    ] 

    var allApprovalTypes = [ 
     'media', 
     'scope', 
     'finance', 
     'compliance', 
    ]; 

    var hasType = (el) => { 
     return el.type === type; 
    } 

    allApprovalTypes.map((type) => { 
     return approvals.find((el)=> el.type === type) || {type: type} // this works. 
    }); 

    allApprovalTypes.map((type) => { 
     return approvals.find(hasType) || {type: type} // this wont work.   
    }); 

Answer 1

hasType正在引用其範圍中不存在的「type」變量。您需要內聯函數爲其拿起型（或封閉包裹，但是這幾乎是同樣的事情）

allApprovalTypes.map((type) => { 
    return approvals.find((el) => { 
     return el.type === type; 
    }) || {type: type} 
}); 


    // closure 
    var hasType = (type) => (el) => { 
     return el.type === type; 
    } 

附： - 無關，但在ES6中，「{type：type}」可以縮寫爲「{type}」

Answer 2

您可以通過使用.bind()。您可以將type（額外的參數）傳遞給回調函數。

var approvals = [ 
 
     {type: 'media'}, 
 
     {type: 'scope'}, 
 
    ] 
 

 
    var allApprovalTypes = [ 
 
     'media', 
 
     'scope', 
 
     'finance', 
 
     'compliance', 
 
    ]; 
 

 
    var hasType = (el, type) => { 
 
     return el.type === type; 
 
    } 
 

 
    console.log(allApprovalTypes.map((type) => { 
 
     return approvals.find(hasType.bind(null, type)) || {type: type} // this works. 
 
    })); 
 

 
    
 

 
    //without bind(); 
 
    console.log(allApprovalTypes.map((type) => { 
 
     return approvals.find((el) => return hasType(el, type)) || {type: type} // this works. 
 
    }));