A service provider interface (or SPI) for short is a concrete implementation of an API formalized as a set of java interfaces. So the short answer is to create a set of classes which implement the API interfaces for which you want to create a concrete implementation. Here is an article which goes into more … Read more
For the desktop application I would look at using jdbc to connect to say a MySql database ( should work on most desktop OS not just windows) . Then you can investigate how to import the sqlite sb file into MySql a google search will throw up quite a few linksfor this 0 solved Windows … Read more
From the UIWebView docs This is the method you need to load your html: – (void)loadHTMLString:(NSString *)string baseURL:(NSURL *)baseURL 3 solved How to make the HTML file [closed]
You don’t need the % character, since the function Contains already implied that. So : select Subject from [dbo].[Email] where Contains(Subject,’t’) The % are for LIKE function select Subject from [dbo].[Email] where Subject LIKE ‘%t%’ 1 solved Search box to work like google search [closed]
I think you have a mistake inside your script. Check this: jsfiddle.net/Frv8G/1/ I changed “s-o” to “g-o”. $(“#containergames”).mouseleave(function () { $(“.g-o”).animate({ opacity: 0 }, function () { $(“.g-o”).hide(); }); }); 0 solved Link still clickable after .hide() [closed]
A += 1 means A = A + 1. So A += A = a would be A = A + A = a (which obviously doesnt work). The error you seem to experience is not caused by this operator. Perhaps you mixed spaces with tabs, or you simply did not indent a certain line … Read more
You’re on the right track with what you have. You are currently only trying to get the numeric characters, which is why that’s all you are getting. Also, after some testing with different strings, it turned out that your particular regex seems to only works when all the numbers are next to each other. I … Read more
It is not clear why you expect this code to behave differently. Look at these lines: var ExchangeRate = c.GrabBTCAmount(); var AmountInBTC = (c.USDtoBTC(15000, ExchangeRate)); var AmountAtMarket = (AmountInBTC * ExchangeRate); If I inline USDtoBTC and inline everything to calculate AmountAtMarket the formula would be AmountAtMarket = (15000 / ExchangeRate) * ExchangeRate So you should … Read more
Introduction Math is a subject that can be difficult to understand and master. It requires a lot of practice and dedication to get it right. Unfortunately, even with all the practice and dedication, sometimes math can still be confusing and mistakes can be made. If you are having trouble understanding why your math isn’t right, … Read more
i resolve my above problem by applying this code <?php echo $_REQUEST[‘n’];?> solved What is required to work PHP post form method?
global variables ——-> data static variables ——-> data constant data types —–> code and/or data. Consider string literals for a situation when a constant itself would be stored in the data segment, and references to it would be embedded in the code local variables(declared and defined in functions) ——–> stack variables declared and defined in … Read more
You could achieve this as follows: import java.nio.file.Paths val path = “/home/gmc/exists.csv” val fileName = Paths.get(path).getFileName // Convert the path string to a Path object and get the “base name” from that path. val extension = fileName.toString.split(“\\.”).last // Split the “base name” on a . and take the last element – which is the extension. … Read more
You can use the LINQ All method. @if(!Model.PaymentList.All(f=>f.Status==”Paid”)) { <button>click me</button> } Or the Any method @if(Model.PaymentList.All(f=>f.Status!=”Paid”)) { <button>click me</button> } 1 solved Hiding button if all status is “Paid” using LINQ [closed]
Introduction This article will discuss how to use LINQ to hide a button if all the status values are “Paid”. LINQ is a powerful language-integrated query language that can be used to query and manipulate data in a variety of ways. We will look at how to use LINQ to check if all the status … Read more
The compact way: struct Node { int x; Node *next; } nodes[] = { {1, nodes + 1}, {2, nodes + 2}, {3, nodes + 3}, {4, nodes + 4}, {5, nodes + 5}, {6, nodes + 6}, {7, nodes + 7}, {8, nodes + 8}, {9, nullptr} }; Node* root = nodes; solved How … Read more