[ad_1] Well, there are at least two ways to achieve this result. You can pick one of them depending on browsers compatibility. 1) You can choose a pure-css solution using a background image for your div and setting: background-size: cover; You cannot use this approach if IE8 is one of your target browsers. 2) You … Read more
[ad_1] Your Question is Simple.. If and else are used, When you make a true or false statements. Example: If ( 3>2) { printf (“3”);} else { printf ( “2”);} But in case of if and else if Example: if(3>2) {printf (“3”);} else if (2>3) {printf (“false”);} IN THIS WAY .. In else if. you … Read more
[ad_1] This is what I was looking for avgDelay <- function(carrier, dest) { TempTotal = 0 TempCount = 0 for(i in 1:dim(delays)[1]) { if(delays$CARRIER[i] == carrier & delays$ARR_DELAY[i] >0 & is.na(delays$ARR_DELAY[i]) == FALSE & delays$DEST[i] == dest) { TempTotal <-TempTotal + delays$ARR_DELAY[i] TempCount <-TempCount + 1 # keeps count of the number of delays } … Read more
[ad_1] Instead of forEach, use map. The map() method creates a new array with the results of calling a provided function on every element in this array. return array.map(logArrayElements); 0 [ad_2] solved Produce an ordered HTML list using JavaScript
[ad_1] All you have to do in this case is remove the int() call from the main(). Your string() function is expecting a string – if you send it an int, it’ll never work. Additionally, you can use the elif keyword: def string(x): if x==”1″: word = “one” elif x==”2″: word = “two” elif x==”3″: … Read more
[ad_1] Given that 98% of the time is spent in that function, I rewrote your “get a number” function: int Ten_To_One_Million_Ten(void) { unsigned Number = (((unsigned)rand() << 5) + (unsigned)rand()%32) % 1000000 + 10; assert(Number >= 10 && Number <= 1000010); return Number; } Now, on my machine, using clang++ (Version 3.7 from about 4 … Read more
[ad_1] First change the method signature to return a value: public static boolean checkUSPASS(String a,String b) Then return a value from within the method: return true; or: return false; Note that all code paths must return some value. So you have this in your try block: if (rs.next()) { return true; } else { return … Read more
[ad_1] just add a test : String firstWords = “”; if(splitVideoTitle.length>1){ firstWords = splitVideoTitle[0] + ” ” + splitVideoTitle[1]; } else firstWords=splitVideoTitle[0]; 0 [ad_2] solved Splitting a string caused ArrayIndexOutOfBoundsException [duplicate]
[ad_1] Let’s consider two ways to solve it that are two slow, and then merge them into one solution, that will be guaranteed to finish in milliseconds. Approach 1 (slow) Allocate an array v of size 2^16. Every time you add an element, do the following: void add(int s) { for (int i = 0; … Read more
[ad_1] claim is not an array, but another object. If you want it to be an array, it should be something like this: “claim”: [ “date”: “2015-08-06”, “comments”: “Some comments” ] (notice the [ ] instead of { }) If it’s meant to be an object, then the code should read: JSONObject claims = c.getJSONObject(TAG_CLAIM); … Read more
[ad_1] Not Possible Two problems: Where does the array end? In C, arrays are simply blocks of memory and do not have a length property. It is possible to read unpredictable numbers off the end of the array, from other portions of memory. In the posted code, sizeof(*arr) apparently is being used to get the … Read more
[ad_1] Change your code from $query = “SELECT COUNT( id ) FROM scores WHERE id = $id;”; $result = mysql_query($query) or die(‘Query failed: ‘ . mysql_error()); to $query = “SELECT COUNT( id ) as `total_ids` FROM scores WHERE id = $id”; $result = mysql_query($query) or die(‘Query failed: ‘ . mysql_error()); after that you need $count … Read more
[ad_1] You can use a function based on this formula: The basic formula for the frequencies of the notes of the equal tempered scale is given by fn = f0 * (a)n where f0 = the frequency of one fixed note which must be defined. A common choice is setting the A above middle C … Read more
[ad_1] I could not post it to comment section. Please see below: List<Integer> integers = Arrays.asList(1, 2, 3, 4); int i = 0; StringBuilder sb = new StringBuilder(“[“); for (int value : integers) { sb.append((i == 0) ? “” : (i % 2 == 0 ? “,” : “:”)).append(value); i++; } sb.append(“]”); System.out.println(sb.toString()); Output is: … Read more
[ad_1] I would go by a string.split(” “).lastOption. But keep in mind that other blank chars may be used instead of space [ad_2] solved Scala. How get last word in String [closed]