As I said in the headline: I do not want the user to type x xxxx in just one textbox (x a number
I tried a masked text box and changed it to the mask under properties:
// mask: 9.9.9.9 / Previw: _
But when I get the value of the maskedtexbox, I get:
// value: 1,1,1, 1
<< P>
// value: 1.1.1.1 The post text "itemprop =" text ">
contains:
. < / Code> The decimal placeholder will be the appropriate decimal symbol suitable for the actual display character format provider, as the format of the control is determined by the provider property.
...
\ Escape. Avoid a mask character, changing it to a literal form, "\\" is an escape sequence for a backslash.
should not use. If you always want a period regardless of area settings: in that case you . should be used.
Popular posts from this blog
I have a text that includes text input and selection box. I was trying to apply Laravel validation. I now want to retain user input values if the verification does not succeed. I'm able to get it on the input box, but not on the selection box. This is my code {{form :: select ('vehicles_year'), how the first selected value (in the selection box) will show. $ Model year, '-1', ['id' = & gt; 'Vehicles_year'])}} & lt; Span class = "help-block" id = "vehicles_year_error" & gt; @if ($ errors-> ('vehicles_year')) {{$ errors- & gt; First ('vehicles_year')}} @endif & lt; / Span & gt; -1 I am the default value key when I am showing the form load. What I do is that I add the "default" option which is "nothing" Is set to. In the validity rules, I say this value is necessary. If the user does not choose any of the other options, the verification fails. $ op...
I want to be able to iterate over a group object: grouped = Df Groupby ('zzz') question_list = [a, b, c] How do I get a list of values for: grouped .grouped.b grouped.c but using a for-loop? You can either remove such a key print Df Groupby ('zzz'). Groups.keys () To capture the grouped. You can do this ... print df.groupby ('Zzz') group ['A'] or you can open the key, For this type of group key, in df.groupby group ('Zzz'): print key # group key Is a dataframe related to
How can we create matrix x, 1000 by 1000 in the matteback, which is sparse, say Are, 5% of independent Bernoulli + 1 nonzero entries? I.e. In such a matrix, rho = = X Choose 5% of irregular elements n = Numel (x); Ind = Randie (n, round (.05 * n), 1); Assign these elements with random variables x (ind) = binorend (1, .5, length), 1) * 2-1; Check for more information To prevent duplicate rand numbers, from the Statistics Toolbox, or randperm as mentioned in some, or something like this Ind = []; T = round (.05 * n); T1 = length (IND); While T1 & LT; T Ind (end + 1: t0) = rand (n, tt-1, 1); Ind = Unique (Ind); T1 = length (IND); End
Comments
Post a Comment