If you’re able to use Underscore.js in your project, the _.filter() array function makes this a snap: // find all strings in array containing ‘thi’ var matches = _.filter( [ ‘item 1’, ‘thing’, ‘id-3-text’, ‘class’ ], function( s ) { return s.indexOf( ‘thi’ ) !== -1; } ); The iterator function can do whatever you … Read more
You need to use return value of replaceAll() method. replaceAll() does not replace the characters in the current string, it returns a new string with replacement. String objects are immutable, their values cannot be changed after they are created. You may use replace() instead of replaceAll() if you don’t need regex. String str = “abcd=0; … Read more
There’s a builtin method find on string objects. s = “Happy Birthday” s2 = “py” print(s.find(s2)) Python is a “batteries included language” there’s code written to do most of what you want already (whatever you want).. unless this is homework 🙂 find returns -1 if the string cannot be found.
UPDATE: I liked this question so much, I just blogged it. See Strings, immutability and persistence The short answer is: O(n) is O(1) if n does not grow large. Most people extract tiny substrings from tiny strings, so how the complexity grows asymptotically is completely irrelevant. The long answer is: An immutable data structure built … Read more
match returns an array. The default string representation of an array in JavaScript is the elements of the array separated by commas. In this case the desired result is in the second element of the array: var tesst = “afskfsd33j” var test = tesst.match(/a(.*)j/); alert (test[1]);
Using a combination of SUBSTR, INSTR, and NVL (for strings without an underscore) will return what you want: SELECT NVL(SUBSTR(‘ABC_blah’, 0, INSTR(‘ABC_blah’, ‘_’)-1), ‘ABC_blah’) AS output FROM DUAL Result: output —— ABC Use: SELECT NVL(SUBSTR(t.column, 0, INSTR(t.column, ‘_’)-1), t.column) AS output FROM YOUR_TABLE t Reference: SUBSTR INSTR Addendum If using Oracle10g+, you can use regex … Read more
According to the Java API doc, substring throws an error when the start index is greater than the Length of the String. IndexOutOfBoundsException – if beginIndex is negative or larger than the length of this String object. In fact, they give an example much like yours: “emptiness”.substring(9) returns “” (an empty string) I guess this … Read more
Why the c++’s implemented string::substr() doesn’t use the KMP algorithm (and doesn’t run in O(N + M)) and runs in O(N * M)? I assume you mean find(), rather than substr() which doesn’t need to search and should run in linear time (and only because it has to copy the result into a new string). … Read more
0: U 1: n 2: i 3: v 4: e 5: r 6: s 7: i 8: t 9: y Start index is inclusive End index is exclusive Javadoc link
Not sure what would perform faster but this would be more reliable when it comes to extension like .jpeg or .html x.replace(/\.[^/.]+$/, “”)