I’d question the wisdom, but perhaps: source.TakeWhile(x => x != value).Count(); (using EqualityComparer<T>.Default to emulate != if needed) – but you need to watch to return -1 if not found… so perhaps just do it the long way public static int IndexOf<T>(this IEnumerable<T> source, T value) { int index = 0; var comparer = EqualityComparer<T>.Default; … Read more
At least three ways: A regular expression: var result = /[^/]*$/.exec(“foo/bar/test.html”)[0]; …which says “grab the series of characters not containing a slash” ([^/]*) at the end of the string ($). Then it grabs the matched characters from the returned match object by indexing into it ([0]); in a match object, the first entry is the … Read more
You can use indexOf: var imageList = [100,200,300,400,500]; var index = imageList.indexOf(200); // 1 You will get -1 if it cannot find a value in the array.
There are a couple of ways to accomplish this using the Arrays utility class. If the array is not sorted and is not an array of primitives: java.util.Arrays.asList(theArray).indexOf(o) If the array is primitives and not sorted, one should use a solution offered by one of the other answers such as Kerem Baydoğan’s, Andrew McKinlay’s or … Read more
Instances of the String constructor have a .search() method which accepts a RegExp and returns the index of the first match. To start the search from a particular position (faking the second parameter of .indexOf()) you can slice off the first i characters: str.slice(i).search(/re/) But this will get the index in the shorter string (after … Read more
Versions of IE before IE9 don’t have an .indexOf() function for Array, to define the exact spec version, run this before trying to use it: if (!Array.prototype.indexOf) { Array.prototype.indexOf = function(elt /*, from*/) { var len = this.length >>> 0; var from = Number(arguments[1]) || 0; from = (from < 0) ? Math.ceil(from) : Math.floor(from); … Read more
Not sure what would perform faster but this would be more reliable when it comes to extension like .jpeg or .html x.replace(/\.[^/.]+$/, “”)