Question

我有一个pickerView，其中有很多主题。每个科目都是这样的一个班级：

GET https://my_url.aspx?
callback=jQuery32103473…
age%20School&address=Ruby&website=www.hs.com&twitter=%23hs&_=1501421197069 
send @ jquery-3.2.1.min.js:4
ajax @ jquery-3.2.1.min.js:4
(anonymous) @ magic.js:20
dispatch @ jquery-3.2.1.min.js:3
q.handle @ jquery-3.2.1.min.js:3

我以这种方式设置我的pickerView：

class subjects {
    var idSubject: String = "";
    var nameSubject: String = "";
    var notesSubject: String = "";
    var colorSubject: String = "";

    init(subjectId: String, subjectName: String, subjectNotes: String, subjectColor: String) {
        idSubject = subjectId
        nameSubject = subjectName
        notesSubject = subjectNotes
        colorSubject = subjectColor
    }

    func printSubject(){
        print(idSubject," - ",nameSubject," - ",notesSubject," - ",colorSubject)
    } 
}

我想选择一个特定主题的行，但我不能因为public func numberOfComponents(in pickerView: UIPickerView) -> Int{ return 1 } public func pickerView(_ pickerView: UIPickerView, numberOfRowsInComponent component: Int) -> Int{ return MenuViewController.subjectsArray.count } func pickerView(_ pickerView: UIPickerView, titleForRow row: Int, forComponent component: Int) -> String? { self.view.endEditing(true) return MenuViewController.subjectsArray[row].nameSubject }它“无法将类型'String'的值转换为预期的参数类型'（主题）抛出 - ＆gt; Bool' “在这行代码中：

indexOf

有人可以帮助我吗？

Answer 1

if let index = MenuViewController.subjectsArray.index(where: {
    $0.nameSubject == "Matematica"
}) {
    self.subjectsMenu.selectRow(index, inComponent: 0, animated: true)
}

Answer 2

而不是使用index(of:)，这种情况的适当方法是index(where:)：

返回集合元素的第一个索引 满足给定的谓词。

这是适用的，因为您有一个自定义对象数组（subjects），如下所示：

if let index = MenuViewController.subjectsArray.index(where: { (subjectsObject) -> Bool in
    subjectsObject.nameSubject == "Matematica"
}) {
    print("found the desired index: \(index)")
    self.subjectsMenu.selectRow(index, inComponent: 0, animated: true)
}

补充说明：

自定义类的名称应为“Subject”而不是“subject”。通常，类的名称引用单个对象，upper camel case。
比较字符串：subjectsObject.nameSubject == "Matematica"时，最好修剪并比较其 lower / upper 案例版本，如下所示：

subjectsObject.nameSubject.lowercased().trimmingCharacters(in: .whitespaces) == "Matematica".lowercased().trimmingCharacters(in: .whitespaces)

无法选择pickerView的默认值

2 个答案: