#indexof

Using indexOf is a very common operation in Javascript. Though a little known feature is that indexOf can start searching from any point in a string. At first, this seems like a feature purely for performance optimization. But it useful when doing more complex cases of string parsing.

The following replaces all content between 2 tags (containing anything) within a string with another string:

function replaceTaggedContent(contentString, repeaterTag, replaceWith) { let newString = contentString; let index = 0; while (index < contentString.length) { const startIndex = newString.indexOf(repeaterTag, index); // No point in continuing, there is nothing more to replace. if (startIndex < 0) { break; } // No point in continuing, there is no second string to match. const endIndex = newString.indexOf(repeaterTag, startIndex + repeaterTag.length); if (endIndex < 0) { break; } const endPoint = endIndex + repeaterTag.length; newString = newString.substring(0, startIndex) + replaceWith + newString.substring(endPoint); index = startIndex + replaceWith.length; } return newString; } const result = replaceTaggedContent( 'BEFORE-@should replace@-MIDDLE-@should-replace@-AFTER' , '@', 'REPLACED'); console.log('After :', result); // BEFORE-REPLACED-MIDDLE-REPLACED-AFTER

Arguably, this can be done with regex though at that point the tags would need to be escaped and the expression would look quite complicated. This is also relatively easy to upgrade to other scenarios like supporting replacing a list of words corresponding to the index of the current replacement.

Dica para localizar um item dentro de um combobox.

Para localizar um item dentro de um combobox faça o seguinte: cboNomes.ItemIndex := cboNomes.Items.IndexOf('Maria'); Um abraço, Fabiano Catrinck

Searching for Elements in Array

IndexOf():It can search for any element and get the index. If not found returns -1

LastIndexOf():It can search for element as last occurance.

find():It can search using a condition and return first occurance

filter():It can search using condition and return all occurances.

Syntax:

                arrayName.indexOf(“Value”);

Sorting Array Elements

sort():        It arranges…

JS Searching for Elements in Array

IndexOf()                        It can search for any element and

                                get the index. If not found returns -1

LastIndexOf()    It can search for element as last

                                occurance.

find()                     It can search using a condition

                                and return first occurance

filter()                   It…

Escaping Characters in String:

A string literal can contain special characters.- Few special characters are ignored by compiler and they escape printing.- Any escaping character can be defined with “\”. It will prevents escaping and prints the non-printable characters

\\ → Prints \

Ex:<script>

function f1(){

var path= “\”C:\\Images\\Cars\\Audi.jpg\””;…

Cómo procesar muchos datos para controlar tu robot por Bluetooth

En estos momentos ya sabremos hacer un montón de cosas con nuestras aplicaciones. Mandar un texto, mandar un número, definir instrucciones con un conjunto de datos. En fin, se podría decir que estamos empezando a crear un protocolo de comunicaciones personalizado. En el post anterior nos quedamos con la forma de enviar datos desde nuestro robot a nuestras aplicaciones por Bluetooth. Ahora vamos a…

Return true if the string in the first element of the array contains all of the letters of the string in the second element of the array.

For example, ["hello", "Hello"], should return true because all of the letters in the second string are present in the first, ignoring case.

The arguments ["hello", "hey"] should return false because the string "hello" does not contain a "y".

Lastly, ["Alien", "line"], should return true because all of the letters in "line" are present in "Alien".

function mutation(arr) {

 var secondWord = arr[0].toLowerCase();

 var firstWord = arr[1].toLowerCase();

 for (i = 0; i < firstWord.length; i++) {    if (secondWord.indexOf(firstWord[i]) === -1) {       return false; }

   }  return true; }

mutation(["Mary", "Army"]);

use indexOf to solve this