我正在尝试快照UITableViewController
我有一个简单的测试类
class FeedSnapshotTests: XCTestCase {
// Fails
func test_empty_feed_one() {
let sut = UITableViewController(style: .grouped)
sut.loadViewIfNeeded()
record(snapshot: sut.snapshot(), named: "EMPTY_FEED_ONE")
}
// Passes
func test_empty_feed_two() {
let sut = UIViewController()
sut.loadViewIfNeeded()
record(snapshot: sut.snapshot(), named: "EMPTY_FEED_TWO")
}
}
private extension FeedSnapshotTests {
func record(snapshot: UIImage, named name: String, file: StaticString = #filePath, line: UInt = #line) {
guard let imageData = snapshot.pngData() else {
return XCTFail("Failed to generate PNG data representation from snapshot", file: file, line: line)
}
let snapshotURL = URL(fileURLWithPath: String(describing: file))
.deletingLastPathComponent()
.appendingPathComponent("snapshots")
.appendingPathComponent("\(name).png")
do {
try FileManager.default.createDirectory(at: snapshotURL.deletingLastPathComponent(), withIntermediateDirectories: true)
try imageData.write(to: snapshotURL)
} catch {
XCTFail("Failed to record snapshot with error: \(error)", file: file, line: line)
}
}
}
extension UIViewController {
func snapshot() -> UIImage {
let renderer = UIGraphicsImageRenderer(bounds: view.bounds)
return renderer.image(actions: { action in
view.layer.render(in: action.cgContext )
})
}
}
此时我只是试图创建一个UIImage
并保存到磁盘。
然而,出于某种原因,在UITableViewController
我无法对视图进行快照的情况下。如果我切换到UIViewController
这个工作。
尝试创建我UImage
的UITableViewController
测试时总是在这里失败
guard let imageData = snapshot.pngData() else {
return XCTFail("Failed to generate PNG data representation from snapshot", file: file, line: line)
}