Swift2.3:来自sqlite的NSMutableDictionary存储整数 [英] Swift2.3: NSMutableDictionary store integer from sqlite

问题描述

实际上，我不知道什么是NSMutableDictionary，但似乎我需要它

Actually, I have no idea what is NSMutableDictionary, but it seems I need it

顺便说一句，我在Swift 2.3中使用了xcode 8.

By the way, I use xcode 8 with swift 2.3.

我正在从sqlite向此对象添加一个整数，但这是错误的 另一个都是字符串，所以在这里没问题.

I am adding a integer to this arry from sqlite, but it's error The other are all the string, so they are not problem here.

然后，错误在这里

http://imgur.com/a/Gbrp0

let recipeIsFavor = sqlite3_column_int(statement, 3)
let recipe_isFavor = Int.fromCString(UnsafePointer<CInt>(recipeIsFavor))  //<--------I wnat to use int here, but I have no idea.

我应该怎么做才能编辑此代码?我希望这个错误可以消失.

What should I do to edit this code? I hope this error can disappear.

func loadData() {

    let db_path = NSBundle.mainBundle().pathForResource("Recipes", ofType: "db")
    var db = COpaquePointer()
    let status = sqlite3_open(db_path!,&db)
    if (status == SQLITE_OK) {
        print("Open the sqlite success!\n")
    }else {
        print("Open the sqlite failed!\n")
    }

    let query_stmt = "SELECT * FROM recipe"
    if(sqlite3_prepare_v2(db , query_stmt, -1, &statement, nil) == SQLITE_OK) {
        self.data.removeAllObjects()
        while (sqlite3_step(statement) == SQLITE_ROW) {
            let recipeArray = NSMutableDictionary()

            let recipeName = sqlite3_column_text(statement, 0)
            let recipe_name = String.fromCString( UnsafePointer<CChar>(recipeName))

            let recipeType = sqlite3_column_text(statement, 1)
            let recipe_type.......

            let recipeImage = sqlite3_column_text(statement, 2)
            let recipe_image.......

            let recipeIsFavor = sqlite3_column_int(statement, 3)
            let recipe_isFavor = Int.fromCString(UnsafePointer<CInt>(recipeIsFavor))  //<--------I wnat to use int here, but I have no idea.

            let recipeUserPhoto = sqlite3_column_text(statement, 4)
            let recipe_userPhoto....

            let recipeUserName = sqlite3_column_text(statement, 5)
            let recipe_userName.......

            recipeArray.setObject(recipe_name!, forKey: "recipeName")
            recipeArray.setObject(recipe_type!, forKey: "recipeType")
            recipeArray.setObject(recipe_image!, forKey: "recipeImage")
            recipeArray.setObject(recipe_isFavor!, forKey: "recipIsFavor")
            recipeArray.setObject(recipe_userPhoto!, forKey: "recipUserPhoto")
            recipeArray.setObject(recipe_userName!, forKey: "recipUserName")

            data.addObject(recipeArray)
        }

        sqlite3_finalize(statement)
    }else {
        print("read the sqlite data failed")
    }

}

推荐答案

sqlite3_column_int()返回一个"C整数"，在Swift中为Int32. 要将其转换为Int，只需使用

sqlite3_column_int() returns a "C integer" which is Int32 in Swift. To convert that to an Int, use just

let recipeIsFavor = sqlite3_column_int(statement, 3)
let recipe_isFavor = Int(recipeIsFavor)

---

您可以代替NSMutableDictionary和NSMutableArray 也使用类型的Swift字典数组 [String: AnyObject](或Swift 3中的[String: Any])， 那么它会看起来像这样:

---

Instead of NSMutableDictionary and NSMutableArray you can also use an array of Swift dictionaries of type [String: AnyObject] (or [String: Any] in Swift 3), then it would look like this:

var data: [[String: AnyObject]] = []

let query_stmt = "SELECT * FROM recipe"
if sqlite3_prepare_v2(db , query_stmt, -1, &statement, nil) == SQLITE_OK {
    data = []
    while (sqlite3_step(statement) == SQLITE_ROW) {
        var dict: [String: AnyObject] = [:]

        let recipeName = sqlite3_column_text(statement, 0)
        let recipe_name = String.fromCString(UnsafePointer<CChar>(recipeName)) ?? ""

        let recipeIsFavor = sqlite3_column_int(statement, 3)
        let recipe_isFavor = Int(recipeIsFavor)

        // ...

        dict["recipeName"] = recipe_name
        dict["recipIsFavor"] = recipe_isFavor
        // ...

        data.append(dict)
    }
    sqlite3_finalize(statement)
} else {
    print("read the sqlite data failed")
}

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