I have a UIView which is placed on the screen via several constraints. Some of the constraints are owned by the superview, others are owned by other ancestors (e.g. perhaps the view property of a UIViewController).
I want to remove all of these old constraints, and place it somewhere new using new constraints.
How can I do this without creating an IBOutlet for every single constraint and having to remember which view owns said constraint?
To elaborate, the naive approach would be to create a bunch of IBOutlets for each of the constraints, and would then involve calling code such as:
[viewA removeConstraint:self.myViewsLeftConstraint];
[viewB removeConstraint:self.myViewsTopConstraint];
[viewB removeConstraint:self.myViewsBottomConstraint];
[self.view removeConstraint:self.myViewsRightConstraint];
The problem with this code is that even in the simplest case, I would need to create 2 IBOutlets. For complex layouts, this could easily reach 4 or 8 required IBOutlets. Furthermore, I would need to ensure that my call to remove the constraint is being called on the proper view. For example, imagine that myViewsLeftConstraint is owned by viewA. If I were to accidentally call [self.view removeConstraint:self.myViewsLeftConstraint], nothing would happen.
Note: The method constraintsAffectingLayoutForAxis looks promising, but is intended for debugging purposes only.
Update: Many of the answers I am receiving deal with self.constraints, self.superview.constraints, or some variant of those. These solutions won't work since those methods return only the constraints owned by the view, not the ones affecting the view.
To clarify the problem with these solutions, consider this view hierarchy:
Now imagine we create the following constraints, and always attach them to their nearest common ancestor:
- C0: Me: same top as Son (owned by Me)
- C1: Me: width = 100 (owned by Me)
- C2: Me: same height as Brother (owned by Father)
- C3: Me: same top as Uncle (owned by Grandfather)
- C4: Me: same left as Grandfather (owned by Grandfather)
- C5: Brother: same left as Father (owned by Father)
- C6: Uncle: same left as Grandfather (owned by Grandfather)
- C7: Son: same left as Daughter (owned by Me)
Now imagine we want to remove all constraints affecting Me. Any proper solution should remove [C0,C1,C2,C3,C4] and nothing else.
If I use self.constraints (where self is Me), I will get [C0,C1,C7], since those are the only constraints owned by Me. Obviously it wouldn't be enough to remove this since it is missing [C2,C3,C4]. Furthermore, it is removing C7 unnecessarily.
If I use self.superview.constraints (where self is Me), I will get [C2,C5], since those are the constraints owned by Father. Obviously we cannot remove all these since C5 is completely unrelated to Me.
If I use grandfather.constraints, I will get [C3,C4,C6]. Again, we cannot remove all of these since C6 should remain intact.
The brute force approach is to loop over each of the view's ancestors (including itself), and seeing if firstItem or secondItem are the view itself; if so, remove that constraint. This will lead to a correct solution, returning [C0,C1,C2,C3,C4], and only those constraints.
However, I'm hoping there is a more elegant solution than having to loop through the entire list of ancestors.
This approach worked for me:
@interface UIView (RemoveConstraints)
- (void)removeAllConstraints;
@end
@implementation UIView (RemoveConstraints)
- (void)removeAllConstraints
{
UIView *superview = self.superview;
while (superview != nil) {
for (NSLayoutConstraint *c in superview.constraints) {
if (c.firstItem == self || c.secondItem == self) {
[superview removeConstraint:c];
}
}
superview = superview.superview;
}
[self removeConstraints:self.constraints];
self.translatesAutoresizingMaskIntoConstraints = YES;
}
@end
After it's done executing your view remains where it was because it creates autoresizing constraints. When I don't do this the view usually disappears. Additionally, it doesn't just remove constraints from superview but traversing all the way up as there may be constraints affecting it in ancestor views.
Swift 4 Version
extension UIView {
public func removeAllConstraints() {
var _superview = self.superview
while let superview = _superview {
for constraint in superview.constraints {
if let first = constraint.firstItem as? UIView, first == self {
superview.removeConstraint(constraint)
}
if let second = constraint.secondItem as? UIView, second == self {
superview.removeConstraint(constraint)
}
}
_superview = superview.superview
}
self.removeConstraints(self.constraints)
self.translatesAutoresizingMaskIntoConstraints = true
}
}
You can remove all constraints in a view by doing this:
self.removeConstraints(self.constraints)
EDIT: To remove the constraints of all subviews, use the following extension in Swift:
extension UIView {
func clearConstraints() {
for subview in self.subviews {
subview.clearConstraints()
}
self.removeConstraints(self.constraints)
}
}
The only solution I have found so far is to remove the view from its superview:
[view removeFromSuperview]
This looks like it removes all constraints affecting its layout and is ready to be added to a superview and have new constraints attached. However, it will incorrectly remove any subviews from the hierarchy as well, and get rid of [C7] incorrectly.
In Swift:
import UIKit
extension UIView {
/**
Removes all constrains for this view
*/
func removeConstraints() {
let constraints = self.superview?.constraints.filter{
$0.firstItem as? UIView == self || $0.secondItem as? UIView == self
} ?? []
self.superview?.removeConstraints(constraints)
self.removeConstraints(self.constraints)
}
}
There are two ways of on how to achieve that according to Apple Developer Documentation
1. NSLayoutConstraint.deactivateConstraints
This is a convenience method that provides an easy way to deactivate a
set of constraints with one call. The effect of this method is the
same as setting the isActive property of each constraint to false.
Typically, using this method is more efficient than deactivating each
constraint individually.
// Declaration
class func deactivate(_ constraints: [NSLayoutConstraint])
// Usage
NSLayoutConstraint.deactivate(yourView.constraints)
2. UIView.removeConstraints (Deprecated for >= iOS 8.0)
When developing for iOS 8.0 or later, use the NSLayoutConstraint
class’s deactivateConstraints: method instead of calling the
removeConstraints: method directly. The deactivateConstraints: method
automatically removes the constraints from the correct views.
// Declaration
func removeConstraints(_ constraints: [NSLayoutConstraint])`
// Usage
yourView.removeConstraints(yourView.constraints)
Tips
Using Storyboards or XIBs can be such a pain at configuring the constraints as mentioned on your scenario, you have to create IBOutlets for each ones you want to remove. Even so, most of the time Interface Builder creates more trouble than it solves.
Therefore when having very dynamic content and different states of the view, I would suggest:
- Creating your views programmatically
- Layout them and using NSLayoutAnchor
- Append each constraint that might get removed later to an array
- Clear them every time before applying the new state
Simple Code
private var customConstraints = [NSLayoutConstraint]()
private func activate(constraints: [NSLayoutConstraint]) {
customConstraints.append(contentsOf: constraints)
customConstraints.forEach { $0.isActive = true }
}
private func clearConstraints() {
customConstraints.forEach { $0.isActive = false }
customConstraints.removeAll()
}
private func updateViewState() {
clearConstraints()
let constraints = [
view.leadingAnchor.constraint(equalTo: parentView.leadingAnchor),
view.trailingAnchor.constraint(equalTo: parentView.trailingAnchor),
view.topAnchor.constraint(equalTo: parentView.topAnchor),
view.bottomAnchor.constraint(equalTo: parentView.bottomAnchor)
]
activate(constraints: constraints)
view.layoutIfNeeded()
}
References
- NSLayoutConstraint
- UIView
Details
- Xcode 10.2.1 (10E1001), Swift 5
Solution
import UIKit
extension UIView {
func removeConstraints() { removeConstraints(constraints) }
func deactivateAllConstraints() { NSLayoutConstraint.deactivate(getAllConstraints()) }
func getAllSubviews() -> [UIView] { return UIView.getAllSubviews(view: self) }
func getAllConstraints() -> [NSLayoutConstraint] {
var subviewsConstraints = getAllSubviews().flatMap { $0.constraints }
if let superview = self.superview {
subviewsConstraints += superview.constraints.compactMap { (constraint) -> NSLayoutConstraint? in
if let view = constraint.firstItem as? UIView, view == self { return constraint }
return nil
}
}
return subviewsConstraints + constraints
}
class func getAllSubviews(view: UIView) -> [UIView] {
return view.subviews.flatMap { [$0] + getAllSubviews(view: $0) }
}
}
Usage
print("constraints: \(view.getAllConstraints().count), subviews: \(view.getAllSubviews().count)")
view.deactivateAllConstraints()
Based on previous answers (swift 4)
You can use immediateConstraints when you don't want to crawl entire hierarchies.
extension UIView {
/**
* Deactivates immediate constraints that target this view (self + superview)
*/
func deactivateImmediateConstraints(){
NSLayoutConstraint.deactivate(self.immediateConstraints)
}
/**
* Deactivates all constrains that target this view
*/
func deactiveAllConstraints(){
NSLayoutConstraint.deactivate(self.allConstraints)
}
/**
* Gets self.constraints + superview?.constraints for this particular view
*/
var immediateConstraints:[NSLayoutConstraint]{
let constraints = self.superview?.constraints.filter{
$0.firstItem as? UIView === self || $0.secondItem as? UIView === self
} ?? []
return self.constraints + constraints
}
/**
* Crawls up superview hierarchy and gets all constraints that affect this view
*/
var allConstraints:[NSLayoutConstraint] {
var view: UIView? = self
var constraints:[NSLayoutConstraint] = []
while let currentView = view {
constraints += currentView.constraints.filter {
return $0.firstItem as? UIView === self || $0.secondItem as? UIView === self
}
view = view?.superview
}
return constraints
}
}
I use the following method to remove all constraints from a view:
.h file:
+ (void)RemoveContraintsFromView:(UIView*)view
removeParentConstraints:(bool)parent
removeChildConstraints:(bool)child;
.m file:
+ (void)RemoveContraintsFromView:(UIView *)view
removeParentConstraints:(bool)parent
removeChildConstraints:(bool)child
{
if (parent) {
// Remove constraints between view and its parent.
UIView *superview = view.superview;
[view removeFromSuperview];
[superview addSubview:view];
}
if (child) {
// Remove constraints between view and its children.
[view removeConstraints:[view constraints]];
}
}
You can also read this post on my blog to better understand how it works behind the hood.
If you need more granular control, I'd strongly advise switching to Masonry, a powerful framework class you could use whenever you need to properly handle constraints programmatically.
A Swift solution:
extension UIView {
func removeAllConstraints() {
var view: UIView? = self
while let currentView = view {
currentView.removeConstraints(currentView.constraints.filter {
return $0.firstItem as? UIView == self || $0.secondItem as? UIView == self
})
view = view?.superview
}
}
}
It's important to go through all the parents, since the constraints between two elements are holds by the common ancestors, so just clearing the superview as detailed in this answer is not good enough, and you might end up having bad surprise later on.
With objectiveC
[self.superview.constraints enumerateObjectsUsingBlock:^(__kindof NSLayoutConstraint * _Nonnull obj, NSUInteger idx, BOOL * _Nonnull stop) {
NSLayoutConstraint *constraint = (NSLayoutConstraint *)obj;
if (constraint.firstItem == self || constraint.secondItem == self) {
[self.superview removeConstraint:constraint];
}
}];
[self removeConstraints:self.constraints];
}
The easier and efficient approach is to remove the view from superView and re add as subview again.
this causes all the subview constraints get removed automagically.
