Question

为简单起见，我有两个课程Location和Program。 Location有objectId和name，而Program有objectId，name和location（这是Location的指针1}} ObjectId）。当用户选择位置时，如何在查询中返回关联的程序？以下代码现在不起作用。

let pointer = PFObject(withoutDataWithClassName: "location", objectId: passedLocationID!) //passedLocationID is a string containing the ObjectId from the selected location    

let query = PFQuery(className:"Program")
query.whereKey("location", equalTo: pointer.objectId!)
query.findObjectsInBackgroundWithBlock {
    (objects: [PFObject]?, error: NSError?) -> Void in
    if error == nil {
        if let objects = objects {
            for object in objects {
                self.programName.append(object["name"] as! String)
                self.programID.append(object.objectId! as String)
            }
        }
    } else {
}

此查询将以下错误输出到控制台“指针字段位置需要指针值”

返回相关程序的正确代码是什么？

修改

我找到了一个解决方案，但似乎有点矫枉过正。对Parse的两个查询，我觉得这应该只使用1个查询来完成。

let firstQuery = PFQuery(className: "Location")
firstQuery.whereKey("objectId", equalTo: passedLocationID!)

let secondQuery = PFQuery(className: "Program")
secondQuery.includeKey("location")
secondQuery.whereKey("location", matchesQuery: firstQuery)
secondQuery.findObjects.......

Answer 1

所以我找到了答案。

当我将位置传递给程序视图控制器时，我需要将其作为PFObject传递。

var passedLocationID: PFObject?

然后只是一个简单的查询。

let newQuery = PFQuery(className: "Program")
newQuery.whereKey("location", equalTo: passedLocationID!)

Answer 2

添加针对 Swift 3 + 更新的正确答案，以防有人搜索：

guard let query = ProgramModel.query(), let pointerQuery = LocationModel.query() else {
   // Your parse class isn't set up correctly. Handle error.
   return
}

// You can also use other parameters like `notEqualTo` or `containedIn`.
pointerQuery.whereKey("objectId", equalTo: passedLocationId)

// The important line is the one below, which calls the pointer query:
query.whereKey("location", matchesQuery: pointerQuery)
query.includeKey("location")
query.findObjects... or query.countObjects...

使用指针解析查询

修改

2 个答案: