object - Javascript 'property inheritance' -

-

Is there any way to specify the common element for object literals in an array? For example: <"Hello", Formatter: deleteCheckboxFormatter}, {{Key: "Hello", label: "hi", formatter: deleteCheckboxFormatter}, {key: "hello", label: key : "Wut", label: "what?", Formatter: deleteCheckboxFormatter}];

All three records use the same formatter How would you refactor it?

A couple of options come to my mind:

For the general field An assistant function with the default value: 

  function (key, label) {return {'key': key, 'label': label, format: deleteCheckboxFormatter}; } Var array = [create ("hi", "hi"), ("hello", "hello"), ("knit", "what?")];

or more generic function that accepts an argument for the formatting property: 

  create the form (formatter) { Return function (key, label) {return {'key': key, 'label': label, 'formatter': formatter}; }} // is a function to create an object that contains 'fooFormatter' var foo = make ('fooFormatter'); Var array = [EFU ("Hi", "Hi"), EFU ("Hello", "Hello"), EFU ("woot", "what?")];

and the last thing in my mind is to repeat the common field with the specified array: 

  var array = [{key: "hi" , Label: "hi"}, {key: "hello", label: "hello"}, {key: "wut", label: "what?"}]; Var i = array.length; While (i--) {array [i] .formatter = 'deleteCheckboxFormatter'; }

I have used a loop for a while in the reverse order, because the order of repetition is not important and such loop


Comments

Post a Comment

Popular posts from this blog

MySql variables and php -

-
I am getting an error with php in order to format mysql_query () in php to format this string What is the right way? SELECT SELECT SELECT (*) Agents @AgentCount; User_Agent_English, User_Angent_Oursal, US.___Gent_Aginal, COUNT (User_Agent_Percade) ACC, COUNT (*) / (AgentCount) Percentage by agents user_agent_preser order DEC limit limit 50; In PHP, how am i setting $ query $ query = "SELECT count (*) agents @ agentcount; SELECT user_agent_parsed, User_agent_original , COUNT (user_agent_parsed) ascount, COUNT (*) / (@AgentCount) as a percentage by user_agent_parced by URI date DESC limit 50 "; This exact query will work properly if I put it in MySQL directly through a command line session. Do I need to release two separate php calls from mysql_query () and store the results first? I get the following error: There is an error in your SQL syntax; Instead of using a sub-selection and instead of using the 'SELECT user_agent_parsed, user_agent_original, ne...
Read more

url rewriting - How to implement the returnurl like SO in PHP? -

-
This URL should work even after the rewrite. Is this possible? Otherwise, I can only hardcoded everything If you are trying to create a login system with redirection You need to use the HTTP_REFERER header to find out where the user came from so that you can send it back there. The best way to do this is to use something like this in your login form: // - Login php & Lt; Input type = "password" name = "password" /> & Lt; Input type = "hidden" name = "returnrelale" value = "& lt ;? php echo $ redirectURL ;? & gt;" / & Gt; & Lt; / Form & gt; Please note that some ISP, browser or proxy will delete the referrer header, so you can not rely on each user to work fully. If you use a statement, to check whether the header exists or not, you can send those users back to the homepage or whatever. Edit If you are looking to use $ _SERVER ['REQUEST_URI'], if you are running Apache It sho...
Read more

kernel - How to create a "Block device" in Windows -

-
For people familiar with Linux, it is possible to create a module and register it as a block device. This allows the user to mount it as a regular disk (whereas all block I / O is controlled by the module, for example USB-MastStorage). Is there a way to do this in windows? (Windows needs to create a volume. Raw data will come from the interface with justification.) Thanks The easiest way (and this is not easy!) Is to write one, you can do it also with KMDF. The latest issue of the "NT Insider" of the OSR is an article in how to do this, but it not is going to be easy anyway You will learn a ton, though, So if you are interested in kernel development, then this is a good way to start!
Read more