I need such a condition in ts:
([treeLevel = 0] && [globalVar = GP:R > 0]) || [PIDinRootline = {$pages.2018}]
I wanna show block if page has treelevel=0 and the get var R > 0, or if page id = $pages.2018
It looks like the similar code in php:
if(($treeLevel == 0 && $r > 0) || (pid == number))
The all expression in first brackets should be right, or in second.
Is it exist the method to group it like the previous record or I can only use userfunc?
There is no grouping in TS conditions, but if you need this particular condition from your post I think it is not needed because brackets around && are useless in this case.
(p && q) || r
is exactly the same as
p && q || r
Did you tested it?
Does anyone know if there is a way to use some kind shorthand in swift? more specifically, leaving out the braces in things like IF statements... eg
if num == 0
// Do something
instead of
if num == 0
{
// Do something
}
Those braces become rather space consuming when you have a few nested IF's.
PS. I do know I can do the following:
if num == 0 {
// Do something }
But I'm still curious if that sort of thing is possible
You can do that :
let x = 10, y = 20;
let max = (x < y) ? y : x ; // So max = 20
And so much interesting things :
let max = (x < y) ? "y is greater than x" : "x is greater than y" // max = "y is greater than x"
let max = (x < y) ? true : false // max = true
let max = (x > y) ? func() : anotherFunc() // max = anotherFunc()
(x < y) ? func() : anotherFunc() // code is running func()
This following stack : http://codereview.stackexchange.com can be better for your question ;)
Edit : ternary operators and compilation
By doing nothing more than replacing the ternary operator with an if else statement, the build time was reduced by 92.9%.
https://medium.com/#RobertGummesson/regarding-swift-build-time-optimizations-fc92cdd91e31#.42uncapwc
In swift you have to add braces even if there is just one statement in if:
if num == 0 {
// Do something
}
You cannot leave the braces, that how swift if statement work.
You could use a shorthand if statement like you would in objective-c:
num1 < num2 ? DO SOMETHING IF TRUE : DO SOMETHING IF FALSE
Swift 2.0 update
Method 1:
a != nil ? a! : b
Method 2: Shorthand if
b = a ?? ""
Referance: Apple Docs: Ternary Conditional Operator
and it does work,
u.dob = (userInfo["dob"] as? String) != nil ? (userInfo["dob"] as! String):""
I am replacing a json string with blank string if it is nil.
Edit: Adding Gerardo Medina`s suggestion...we can always use shorthand If
u.dob = userInfo["dob"] as? String ?? ""
It is called shorthand if-else condition. If you are into iOS development in Swift, then you can also manipulate your UI objects' behaviour with this property.
For e.g. - I want my button to be enabled only when there is some text in the textfield. In other words, should stay disabled when character count in textfield is zero.
button.enabled = (textField.characters.count > 0) ? true : false
its very simple :
in Swift 4
playButton.currentTitle == "Play" ? startPlay() : stopPlay()
Original Code is
if playButton.currentTitle == "Play"{
StartPlay()
}else{
StopPlay()
}
You could always put the entire if on one line:
if num == 0 { temp = 0 }
This question already has answers here:
Is there a Matlab conditional IF operator that can be placed INLINE like VBA's IIF
(10 answers)
Closed 9 years ago.
I would like to use a C-like ternary operator but in Matlab, how can this be done?
For example:
a = (test == 'yes') ? c : d
where a,c and d are numeric but in Matlab?
A couple of options:
Inline if statement
a = (test == 'yes') * c;
Inline if else statement
a = (test == 'yes') * c + (test ~= 'yes') * d;
or more neatly:
t = test == 'yes'; a = t * c + ~t * d;
This works in the numeric case since test == 'yes' is cast to 0 or 1 depending on whether it's true or not - which can then be multiplied by the desired outcomes (if they are numeric).
To provide an alternative:
t = xor([false true], isequal(test, 'yes')) * [c; d]
or if you want
ternary = #(condition, trueValue, falseValue)...
xor([false true], condition) * [trueValue; falseValue];
...
t = ternary(isequal(test, 'yes'), c, d);
> db.c.remove()
> db.c.insert( { x : 10} )
> a1 = db.c.findOne( {x : 100} )
null
> a1 == null
true
> a2 = db.c.find( {x : 100} )
>
Q1> How do I check if a2 is assigned with nothing inside the MongoDB interactive shell?
Q2> How to I check the same thing in Python 3?
> a2 = db.c.find( {x : 100} )
> a2.hasNext()
false
Because you a2 variable is indeed an MongoCursor.
And a lot of language drivers have this kind of method.
Hint: you can use variable.help() to have some function on mongo client command line.
Edit about Python: I don't know exactly how it works but regarding the documentation Python driver return an iterable element. So it's probably the same way than other languages.
I need to merge two tables, with the contents of the second overwriting contents in the first if a given item is in both. I looked but the standard libraries don't seem to offer this. Where can I get such a function?
for k,v in pairs(second_table) do first_table[k] = v end
Here's what i came up with based on Doug Currie's answer:
function tableMerge(t1, t2)
for k,v in pairs(t2) do
if type(v) == "table" then
if type(t1[k] or false) == "table" then
tableMerge(t1[k] or {}, t2[k] or {})
else
t1[k] = v
end
else
t1[k] = v
end
end
return t1
end
Wouldn't this work properly?
function merge(t1, t2)
for k, v in pairs(t2) do
if (type(v) == "table") and (type(t1[k] or false) == "table") then
merge(t1[k], t2[k])
else
t1[k] = v
end
end
return t1
end
For numeric-index table merging:
for k,v in pairs(secondTable) do table.insert(firstTable, v) end
Doug Currie's answer is the simplest for most cases. If you need more robust merging of tables, consider using the merge() method from the Penlight library.
require 'pl'
pretty.dump(tablex.merge({a=1,b=2}, {c=3,d=4}, true))
-- {
-- a = 1,
-- d = 4,
-- c = 3,
-- b = 2
-- }
Here's iterative version for deep merge because I don't like potential stack overflows of recursive.
local merge_task = {}
function merge_to_left_o(orig, new)
merge_task[orig] = new
local left = orig
while left ~= nil do
local right = merge_task[left]
for new_key, new_val in pairs(right) do
local old_val = left[new_key]
if old_val == nil then
left[new_key] = new_val
else
local old_type = type(old_val)
local new_type = type(new_val)
if (old_type == "table" and new_type == "table") then
merge_task[old_val] = new_val
else
left[new_key] = new_val
end
end
end
merge_task[left] = nil
left = next(merge_task)
end
end
I preferred James version for its simplicity and use it in my utils.lua - i did add a check for table type for error handling.
function merge(a, b)
if type(a) == 'table' and type(b) == 'table' then
for k,v in pairs(b) do if type(v)=='table' and type(a[k] or false)=='table' then merge(a[k],v) else a[k]=v end end
end
return a
end
Thanks for this nice function which should be part of the table class so you could call a:merge(b) but doing table.merge = function(a, b) ... did not work for me. Could even be compressed to a one liner for the real nerds :)
Like Doug Currie said, you can use his function, but there is a problem with his method. If first_table has things in it's k index, the function will over write it.
I'm assuming you're trying to merge these tables, not overwrite index's and value's. So this would be my method, it's very similar but is used for merging tables.
for _, v in pairs(second_table) do table.insert(first_table, v) end
The only problem with this solution is that the index is set as numbers, not as strings. This will work with tables with numbers as the index, and for tables with strings as their index, use Doug Currie's method.
Doug Currie's method:
for k,v in pairs(second_table) do first_table[k] = v end
Extending this great answer, https://stackoverflow.com/a/1283399/1570165, I would like to go with a (pure) functional approach like this one below:
-- example values
local t1 = { a = 0, b = 2 }
local t2 = { a = 1, c = 3 }
-- merge function that takes functional approach
local merge = function(a, b)
local c = {}
for k,v in pairs(a) do c[k] = v end
for k,v in pairs(b) do c[k] = v end
return c
end
-- t1 and t2 value still same after merge
print(merge(t1, t2)) -- { a = 1, b = 2, c = 3 }
print(t2) -- { a = 1, c = 3 }
print(t1) -- { a = 0, b = 2 }
for k,v in pairs(t2) do t1[k] = v end
key for string solution