DriverManager.getConnection(“jdbc:odbc:thin:@localhost:1521:orcl”,”shaheena”,”shaheena”);error: cannot find symbol DriverManager, Statement st solved DriverManager.getConnection(“jdbc:odbc:thin:@localhost:1521:orcl”,”shaheena”,”shaheena”);error: cannot find symbol DriverManager, Statement st
Since IBS Integrator compiles to .class files, you should be able to write JUnit tests in Java against those classes, and run them however you’d normally run JUnit tests (kick off Ant or Maven, open Eclipse and run them from there, etc.). And I can’t think of any reason to use another technology (phpunit, rspec, … Read more
Translate to: for (int i = 0; i< n; i++) { int temp1 = arr[i]; int temp2 = temp1%k; int temp3 = arr[temp2]; arr[temp2] = temp3+k; } Edit: thanks for the correction @R Sahu 1 solved What does a[a[]] do in C?
Use URLEncoder to encode your URL params such as student_name[k] and roll_no[k]: final String queryString = “student_name=” + URLEncoder.encode(student_name[k], “UTF-8”) + “&roll_no=”+ URLEncoder.encode(roll_no[k], “UTF-8″); Use equals() and not == or != for comparing Strings: if(!””.equals(queryString)) 2 solved running application it’s showing this error in android
I’m not exactly sure what you are trying to accomplish…maybe this? StreamReader reader = new StreamReader(“C:/tut.txt”); String data = reader.ReadLine(); List<String> tok = new List<string>(); foreach (char s in data) { Console.WriteLine(s); tok.Add(s.ToString()); } 1 solved Problems With List [closed]
You should empty the stdin buffer before read something else. void emptyBuffer() { char c=”a”; while (c != ‘\n’ && c != EOF) { c = getchar(); } return; } This function empty the stdin buffer. printf(“Enter your name: “); scanf(“%s”, &string); emptyBuffer(); fprintf(fw, “%d\n%s\n”,p, string); printf(“Enter your telephone number: “); scanf(“%d”,&cislo); emptyBuffer(); fprintf(fw, “%d\n”,cislo); … Read more
It is simple, but you must provide the user way enter that value. For example by the GET parameter number$userInput = $_GET[‘number’]; Then you only need to create basic if statement. You can read about it at http://php.net/manual/en/control-structures.if.php $somma = $bxs+$bs+$bm+$bl+$bxl+$b2xl+$nxs+$ns+$nm+$nl+$nxl+$n2xl; if($userInput <= 30){ $somma *=$m1; }else if($userInput >30) { $somma *=$m2; } else if($userInput … Read more
Notice: Undefined variable: message in /Users/kobigo/Sites/yedikitamuhendislik.com/sendEmail.php on line 61 Look at what appears to be line 61: $message .= “Gönd.: ” . $name . “<br />”; You’re trying to append to $message, but you never defined it in the first place. You can append to it after it’s been defined, but on this first line … Read more
Check out: RandomStringUtils and pick the most appropriate method! I would suggest: RandomStringUtils.randomAlphanumeric(8) 1 solved Generate a alphanumeric random and unique string with size of 8 in java(uppercase letter required)
ID of an element must be unique, so need to use class to group similar elements, also in the click handler you need to find the title element which is a descendant of the clicked li jQuery(‘#restaurant-menu-content-tab’).on(‘click’, ‘#res-menu-filters li’, function(e) { var value = jQuery(this).find(‘.menu-extend-title’).text(); jQuery(‘#menu-title-layout-text’).text(value); return false; }); <script src=”https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js”></script> <div id=”restaurant-menu-content-tab”> <div class=”menu-title-layout-change”> … Read more
Your use of for does not behave as you expect. You’re using this for-comprehension: for (data <- category(i)) { if (data.startsWith(x)) true else false } This expression “desugars” into (i.e. is shorthand for): category(i).foreach(data => { if (data.startsWith(x)) true else false }) foreach returns Unit, and therefore the type of this expression (and the type … Read more
I do not use Java, but I will try to explain what is happening. The value 8188 is 0x1ff8 in big-endian hex. When your program is run the result is actually 8188 – 65536 = -57348. That is why you got the result you did. Because the input is big-endian binary, only the first bit … Read more
Extract the number from the string then increment it and put it back, e.g. using String.replaceXxx() and Integer.parseInt() etc. If you want to increment multiple independent numbers, try this: String input = “ABC999DEF999XYZ”; //find numbers Pattern p = Pattern.compile( “[0-9]+” ); Matcher m = p.matcher( input ); StringBuffer sb = new StringBuffer(); //loop through all … Read more
First you split the string into an array: String str = “1.2, 3.1, 5.3, 4.5”; String[] arrOfStr = str.split(“,”); Then you loop through the array and convert to floats: import java.util.ArrayList; ArrayList <Double> volts = new ArrayList<Double>(); for (int i = 0; i < arrOfStr.length; i++) { volts.add(Double.parseDouble(arrOfStr[i])); } System.out.println(volts); 4 solved How to convert … Read more
If what you want is to split the string at the first space, then you can use the string.split(s, maxsplit=n) where s is the string to split by and maxsplit=n is the number of splits to stop at. If you do not give any value for s that is only call the function as string.split(maxsplit=n) … Read more