JAVASCRIPT

substring

//substring() is similar to slice().
//The difference is that start and end values less than 0 are treated as 0 in substring()

let str = "Apple, Banana, Kiwi";
let part = str.substring(7, 13);

// >> Banana

//if you find this answer is useful ,
//upvote ⇑⇑ , so can the others benefit also . @mohammad alshraideh ( ͡~ ͜ʖ ͡°)

substring

const str = 'Mozilla';

console.log(str.substring(1, 3));
// expected output: "oz"

console.log(str.substring(2));
// expected output: "zilla"

substring

var str = "Some String";

public string Substring(int startIndex);

var valueOfSubstring = str.Substring(5);

public string Substring(int startIndex, int length);

var valueOfSubstring = str.Substring(5, 3);

string substring

string a = "String";
string b = a.Replace("i", "o"); // Strong
       b = a.Insert(0, "My ");  // My String
       b = a.Remove(0, 3);      // ing
       b = a.Substring(0, 3);   // Str
       b = a.ToUpper();         // STRING
int    i = a.Length;            // 6

substring

# To find the index of a substring in another string
# Take for instance, below take two strings: str and substr. 

str="hello world"
substr="world"
 
prefix=${str%%$substr*}
index=${#prefix}
 
if [[ index -eq ${#str} ]];
then
    echo "Substring is not present in string."
else
    echo "Index of substring in string : $index"
fi

substring

var str = "The MEAN stack is MongoDB, Express.js, AngularJS, and Node.js";
str.indexOf('MongoDB') !== -1 // true
str.indexOf('Java') !== -1 //false
str.indexOf('Node', 5) !== -1 //true

Substring

/*Delimiter variables, first and second position*/
DECLARE @dfp AS CHAR(1);
DECLARE @dsp AS CHAR(1);
DECLARE @text VARCHAR(MAX);

SET @dfp = ';';
SET @dsp = '@';
SET @text = 'I want you to ;Extract this@ substring for me please.';

SELECT SUBSTRING(@text, (CHARINDEX(@dfp, @text) + 1), (CHARINDEX(@dsp, @text) - 2) - CHARINDEX(@dfp, @text) + Len(@dsp))

String Substring

let text = "Microsoft Google";
text.substr(2, 10);

Substring Of String

s  ="abcdefghijklmnopqrstuvwxyz"
print(s[0:len(s)])
#prints the entire string from 0 to len(s)-1

substrings

s = '   hello   '
s = s[3:8]  # no crash if s[3:20]
# 'hello'