UITableView auto resizing row constraint breaking mysteriously on iPhone 6Plus
This warning is telling you there's a conflict in your constraints.
Reduce the priority of the height constraint to 999 and it will go away. Tested it in your Github project and worked perfectly.
Swift UITableViewCell separatorStyle breaking autolayout on iPhone MIni
This only happens when the tableView.separatorStyle = .none
So to fix it I simply leave the separator on, but set the separator color to clear
self.tableView.separatorStyle = .singleLine
self.tableView.separatorColor = UIColor.clear
Thanks to @CloudBalacing for the help. More info about this problem here
Auto-layout: What creates constraints named UIView-Encapsulated-Layout-Width & Height?
Based on a ton of observation I believe (but cannot know for certain) that the constraints named UIView-Encapsulated-Layout-Width
and UIView-Encapsulated-Layout-Height
are created by UICollectionView
and friends, and exist to enforce the size returned by the sizeForItemAtIndexPath
delegate method. I guess it's there to ensure that the UICollectionViewCell
set up by cellForItemAtIndexPath
ends up the size that it was told it would be.
Which answers my initial question here. A second question is why were the constraints unsatisfiable? The cell's intrinsic height should have been the same as UIView-Encapsulated-Layout-Height
. Again, I don't know for certain, but I suspect it was a rounding error (i.e. intrinsic height came to 200.1 pixels, the UIView-Encapsulated-Layout-Height
maybe rounded to 200. The fix I came up with was to just lower the priority of the relevant cell constraint to allow UIView-Encapsulated-Layout-Height
to have the last word.
Set constraint for the last row of Tableview
Use this method func tableView(_ tableView: UITableView, heightForRowAt indexPath: IndexPath) -> CGFloat {
if indexPath.row == ('lastRow') {
return heightNeeded
}
}
And replace 'last row' with the row you need
and heightNeeded with the pixels or height constraints required
Need to get the rest of an iterator in python
Yes, just use iter.next()
Example
iter = xrange(3).__iter__()
iter.next() # this pops 0
for i in iter:
print i
1
2
You can pop off the front of an iterator with .next(). You cannot do any other fancy operations.
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