結構陣列排序陣列[數組] [結構]

Question

我有一個3D數組，通過一個枚舉for循環執行本地SQL Lite表的後臺查詢。 99％的時間都是按順序進行的。 1％的時間事情有點失序。

由於執行後臺任務的代碼包含在更接近的完成處理程序中，所以我不確定爲什麼序列會變得有趣。以下是結構和數組是如何形成的。

我在想排序數組，但不能確定語法，不知道這是否是最好的方法？這感覺就像一個工作。

struct CollectionStruct { 
    var name : String 
    var description : String 
    var title : String 
    var image : PFFile 
    var id: String 
} 

var packArray : [CollectionStruct] = [] 
var partArray : [CollectionStruct] = [] 
var multiPartArray : [[CollectionStruct]] = [] 
var tempPartArray : [[CollectionStruct]] = [] 

override func viewDidLoad() { 
    super.viewDidLoad() 

    // Built the self.packArray here in simalr fashion to below. 
    // Always in correct order as the query to build it is sorted 


    // Now build the partArray for each packArray 

    // access with 
    // self.multiPartArray[0][0].name 

    for (index, item) in self.packArray.enumerated() { 

     print(index, item) 

     // is it the index that is out of sequence? 
     self.packId = self.packArray[index].id 

     BuildArray.buildArrayFromQuery(queryForCollection: "Part", selectedPackID: self.packId, delegateSender: "DownloadPart", completeBlock: { (result) in 

      if result.isEmpty == false { 

       self.partArray = result 
       // this is how i was appending the result. But it will occasionaly append the results in the wrong order. 
       // self.packArray[1], self.packArray[0]. Not sure if its because of the index from the enumeration or what?? 

       // self.multiPartArray.append(self.partArray) 

       // this is where i was thinking of sorting the array (or outside for loop) this will do for not to test 
       // set it to a temp so it can be sorted... 
       self.tempPartArray.append(self.partArray) 

       // now sort it.. dont know wtf to do here cant see anything online 
       self.multiPartArray = self.tempPartArray.sorted(by: { ([CollectionStruct], [CollectionStruct]) -> Bool in 
        <#code#> 
       }) 

      } 
     }) 
    } 
} 

我無法弄清楚如何排序這句法。在另一個位置我已經使用下面的代碼進行排序的較小結構陣列：

self.packArray = result.sorted { 
    $0.name < $1.name 
} 

---- ---- EDIT

的提示輸入.sorted功能碼如下，其中示出有是[CollectionStruct], [CollectionStruct]但是，我的Struct中的屬性（.name等）在排序中不可用。

self.multiPartArray = self.tempPartArray.sorted(by: { ([CollectionStruct], [CollectionStruct]) -> Bool in 
    <#code#> 
}) 

典型$ 0 < $ 1 ECT不可用或者作爲我不能二進制運算符適用於2周的Structs ?? 理想情況下，我可以按.name排序，因爲這是一個順序標識符。所以排序

self.multiPartArray[0][0].name < self.multiPartArray[0][1].name < .... so on 

這是數據的樣本輸出，當它是無序的，你可以看到，pk010是pk020後到來，其他的一切都是按順序排列。

[[Proj.CollectionStruct(name: "pk000-01", description: "This is an intro session. ", title: "Session 1", image: <PFFile: 0x7b91dff0>, id: "JKE5K4xOQA"), Proj.CollectionStruct(name: "pk000-02", description: "This is a lot of text.", title: "Session 2", image: <PFFile: 0x7b892a30>, id: "ft4o3EWyxX"), Proj.CollectionStruct(name: "pk000-03", description: "This is session 3", title: "Session 3", image: <PFFile: 0x7a6fa5f0>, id: "jebz0c8Bq1"), Proj.CollectionStruct(name: "pk000-04", description: "This is a test", title: "Session 4", image: <PFFile: 0x7b88e390>, id: "E2rGHJMcR7"), Proj.CollectionStruct(name: "pk000-05", description: "This is a set", title: "Session 5", image: <PFFile: 0x7b892320>, id: "dkRAUEbdUb"), Proj.CollectionStruct(name: "pk000-06", description: "Test", title: "Session 6", image: <PFFile: 0x7a6f68a0>, id: "ODTJd3gIml")], 
[Proj.CollectionStruct(name: "pk020-01", description: "This is an intro session", title: "Session 1", image: <PFFile: 0x7a6f7600>, id: "6uG2cJGxrZ"), Proj.CollectionStruct(name: "pk020-02", description: "Session 2", title: "Session 2", image: <PFFile: 0x7b878d60>, id: "6T6r8e8dS3"), Proj.CollectionStruct(name: "pk020-03", description: "Session 3", title: "Session 3", image: <PFFile: 0x7a7a0ff0>, id: "2hDSALAG0a"), Proj.CollectionStruct(name: "pk020-04", description: "Session 4", title: "Session 4", image: <PFFile: 0x7b910bd0>, id: "vd0eBYb3gr"), Proj.CollectionStruct(name: "pk020-05", description: "Session 5", title: "Session 5", image: <PFFile: 0x7b88f790>, id: "MS8ece8dr8"), Proj.CollectionStruct(name: "pk020-06", description: "Session 6", title: "Session 6", image: <PFFile: 0x7a7765c0>, id: "aDfmH8bFKU"), Proj.CollectionStruct(name: "pk020-07", description: "Session 7", title: "Session 7", image: <PFFile: 0x7b9eaf50>, id: "obfW0pZsBH"), Proj.CollectionStruct(name: "pk020-08", description: "Session 8", title: "Session 8", image: <PFFile: 0x7a783c20>, id: "cVmEefiirT")], 
[Proj.CollectionStruct(name: "pk010-01", description: "This is an intro session", title: "Session 1", image: <PFFile: 0x7a6f9b20>, id: "BPtmDXvzDF"), Proj.CollectionStruct(name: "pk010-02", description: "Session 2", title: "Session 2", image: <PFFile: 0x7a79ac00>, id: "B3MAHoFEYs"), Proj.CollectionStruct(name: "pk010-03", description: "Session 3", title: "Session 3", image: <PFFile: 0x7a627e40>, id: "Yg6TxTuyHi"), Proj.CollectionStruct(name: "pk010-04", description: "Session 4", title: "Session 4", image: <PFFile: 0x7a627f90>, id: "XTRUoKVH9P"), Proj.CollectionStruct(name: "pk010-05", description: "Session 5", title: "Session 5", image: <PFFile: 0x7a627b80>, id: "eO9xS05qsW"), Proj.CollectionStruct(name: "pk010-06", description: "Session 6", title: "Session 6", image: <PFFile: 0x7b879bd0>, id: "ePZdmFnnoQ"), Proj.CollectionStruct(name: "pk010-07", description: "Session 7", title: "Session 7", image: <PFFile: 0x7be58c50>, id: "BO6HGa7w1R"), Proj.CollectionStruct(name: "pk010-08", description: "Session 8", title: "Session 8", image: <PFFile: 0x7a69bd20>, id: "dwIxpmD2y8")], 
[Proj.CollectionStruct(name: "pk030-01", description: "Session 3", title: "Session 3", image: <PFFile: 0x7b87da60>, id: "lU1rm5cnbb")]] 

-----編輯----

我還一直無法理清這一點，但我一直在玩至少捕獲了錯誤，它不是太斯文了其在另一枚枚舉循環中，但它是目前我的大腦中最好的。

// add a group for the for enumeration check 
let insideGroup = DispatchGroup() 

for (index, item) in self.tempPartArray.enumerated() { 
    insideGroup.enter() 
    if index >= 1 { 
     // if the currentItem.name > the previousItem.name 
     if item[0].name > self.tempPartArray[index - 1][0].name { 
      print("The array is in sequence") 
      insideGroup.leave() 

     } else { 
      print("the array is out of sequence") 

      // exit gracefully 
     } 
    } 
} 
// the enumeration has finished you can set the variables 
insideGroup.notify(queue: .main) { 
    self.multiPartArray = self.tempPartArray 
} 

-----編輯----

甚至重新編寫查詢是同步的任務仍然返回數據失序作爲一個查詢可能需要比另一個更長。

我想我需要在這裏做的就是一個操作的查詢部分有axConcurrentOperationCount = 1

這應該迫使成果轉化序列。

編號仍想知道如何將這個類型的數組雖然進行排序....

Answer 1

假設你buildArrayFromQuery是不是異步的，我簡化您的循環中應防止任何古怪的方式：

func findAllItems() { 

    var allFoundItems: [[CollectionStruct]] = [] 
    for item in packArray { 

     let packId = item.id 

     BuildArray.buildArrayFromQuery(queryForCollection: "Part", selectedPackID: packId, delegateSender: "DownloadPart", completeBlock: { (result) in 
      if result.isEmpty == false { 
       allFoundItems.append(result) 
      } 
     }) 
    } 
    multiPartArray = allFoundItems 
} 

如果buildArrayFromQuery是異步的，你應該重新評估你是如何做這件事的，並且做一些異步的事情，簡單地獲取所有的數組，並調用一個閉包。但如果你需要排序，這應該工作：

self.multiPartArray = tempPartArray.sorted { 
    // $0 is the left item, $1 is the right item. 
    // However you're sorting your items, just put the logic in here. 
    return $0 < $1 
}