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Filter table data from first character of string in jquery/javascript

I have a table with some records and a textbox. I want to filter table data based on string entered in textbox on keyup event.
Currently I am using a code block which filter the table data but it search the record in table which exist anywhere in the string.
For example:- If I enter 'ab' in textbox it filter the table record with strings contains the keyword 'ab' like abcd, babd, cdab etc.
But my requirement is when I enter the keyword 'ab' in textbox it search only those string which starts from 'ab' like abcd, abdc etc.
Here is my current code:-
function Search_Gridview(strKey, strGV) {
var strData = strKey.value.toLowerCase().split(" ");
var tblData = document.getElementById(strGV);
var rowData;
for (var i = 1; i < tblData.rows.length; i++) {
rowData = tblData.rows[i].cells[3].innerHTML;
var styleDisplay = 'none';
for (var j = 0; j < strData.length; j++) {
if (rowData.toLowerCase().indexOf(strData[j]) >= 0)
styleDisplay = '';
else {
styleDisplay = 'none';
break;
}
}
tblData.rows[i].style.display = styleDisplay;
}
}
Please help guys......

You can filter with jQuery the columns that contain a string beginning with e.g. "ab" of this way:
var re = $("#TABLE_ID td").filter(function(i){ return this.innerHTML.startsWith("ab") })
//You can after, get the values of each td of the result of this way
re.map(function(i){return this.innerHTML})

You can use RegExp's test method.
var stringData = [
'aaa', 'aab', 'aac',
'aba', 'abb', 'abc'
];
var searchPrefix = 'ab';
var result = stringData.filter(function (str) {
// return true if str has prefix with searchPrefix.
return (new RegExp('^' + searchPrefix)).test(str);
});
console.log(result);
JavaScript Regexp Reference

This appears the most elegant solution.
To change search behavior from "exists anywhere in the data" into "data starts with ". You only need to change one single character, on one single line of your original code and nothing more.
Change this line from this..
if (rowData.toLowerCase().indexOf(strData[j]) >= 0)
into this...
if (rowData.toLowerCase().indexOf(strData[j]) == 0)
What it does is forces the indexOf() to address zero, instead of allowing mid-string matches.
Below is the whole (already modified) code for copy and paste into a project, such as a html table filter.
function Search_Gridview(strKey, strGV) {
var strData = strKey.value.toLowerCase().split(" ");
var tblData = document.getElementById(strGV);
var rowData;
for (var i = 1; i < tblData.rows.length; i++) {
rowData = tblData.rows[i].cells[3].innerHTML;
var styleDisplay = 'none';
for (var j = 0; j < strData.length; j++) {
if (rowData.toLowerCase().indexOf(strData[j]) == 0)
styleDisplay = '';
else {
styleDisplay = 'none';
break;
}
}
tblData.rows[i].style.display = styleDisplay;
}
}
Search_Gridview() = the function's name.
strKey = input search characters
strGV = ID of html table></table

Find smallest substring containing a given set of letters in a larger string

Say you have the following string:
FJKAUNOJDCUTCRHBYDLXKEODVBWTYPTSHASQQFCPRMLDXIJMYPVOHBDUGSMBLMVUMMZYHULSUIZIMZTICQORLNTOVKVAMQTKHVRIFMNTSLYGHEHFAHWWATLYAPEXTHEPKJUGDVWUDDPRQLUZMSZOJPSIKAIHLTONYXAULECXXKWFQOIKELWOHRVRUCXIAASKHMWTMAJEWGEESLWRTQKVHRRCDYXNT
LDSUPXMQTQDFAQAPYBGXPOLOCLFQNGNKPKOBHZWHRXAWAWJKMTJSLDLNHMUGVVOPSAMRUJEYUOBPFNEHPZZCLPNZKWMTCXERPZRFKSXVEZTYCXFRHRGEITWHRRYPWSVAYBUHCERJXDCYAVICPTNBGIODLYLMEYLISEYNXNMCDPJJRCTLYNFMJZQNCLAGHUDVLYIGASGXSZYPZKLAWQUDVNTWGFFY
FFSMQWUNUPZRJMTHACFELGHDZEJWFDWVPYOZEVEJKQWHQAHOCIYWGVLPSHFESCGEUCJGYLGDWPIWIDWZZXRUFXERABQJOXZALQOCSAYBRHXQQGUDADYSORTYZQPWGMBLNAQOFODSNXSZFURUNPMZGHTAJUJROIGMRKIZHSFUSKIZJJTLGOEEPBMIXISDHOAIFNFEKKSLEXSJLSGLCYYFEQBKIZZTQQ
XBQZAPXAAIFQEIXELQEZGFEPCKFPGXULLAHXTSRXDEMKFKABUTAABSLNQBNMXNEPODPGAORYJXCHCGKECLJVRBPRLHORREEIZOBSHDSCETTTNFTSMQPQIJBLKNZDMXOTRBNMTKHHCZQQMSLOAXJQKRHDGZVGITHYGVDXRTVBJEAHYBYRYKJAVXPOKHFFMEPHAGFOOPFNKQAUGYLVPWUJUPCUGGIXGR
AMELUTEPYILBIUOCKKUUBJROQFTXMZRLXBAMHSDTEKRRIKZUFNLGTQAEUINMBPYTWXULQNIIRXHHGQDPENXAJNWXULFBNKBRINUMTRBFWBYVNKNKDFR
I'm trying to find the smallest substring containing the letters ABCDA.
I tried a regex approach.
console.log(str.match(/[A].*?[B].*?[C].*?[D].*?[A]/gm).sort((a, b) => a.length - b.length)[0]);
This works, but it only find strings where ABCDA appear (in that order). Meaning it won't find substring where the letters appear in a order like this: BCDAA
I'm trying to change my regex to account for this. How would I do that without using | and type out all the different cases?

You can't.
Let's consider a special case: Assume the letters you are looking for are A, A, and B. At some point in your regexp there will certainly be a B. However, the parts to the left and to the right of the B are independent of each other, so you cannot refer from one to the other. How many As are matched in the subexpression to the right of the B depends on the number of As being already matched in the left part. This is not possible with regular expressions, so you will have to unfold all the different orders, which can be many!
Another popular example that illustrates the problem is to match opening brackets with closing brackets. It's not possible to write a regular expression asserting that in a given string a sequence of opening brackets is followed by a sequence of closing brackets of the same length. The reason for this is that to count the brackets you would need a stack machine in contrast to a finite state machine but regular expressions are limited to patterns that can be matched using FSMs.

This algorithm doesn't use a regex, but found both solutions as well.
var haystack = 'FJKAUNOJDCUTCRHBYDLXKEODVBWTYPTSHASQQFCPRMLDXIJMYPVOHBDUGSMBLMVUMMZYHULSUIZIMZTICQORLNTOVKVAMQTKHVRIFMNTSLYGHEHFAHWWATLYAPEXTHEPKJUGDVWUDDPRQLUZMSZOJPSIKAIHLTONYXAULECXXKWFQOIKELWOHRVRUCXIAASKHMWTMAJEWGEESLWRTQKVHRRCDYXNTLDSUPXMQTQDFAQAPYBGXPOLOCLFQNGNKPKOBHZWHRXAWAWJKMTJSLDLNHMUGVVOPSAMRUJEYUOBPFNEHPZZCLPNZKWMTCXERPZRFKSXVEZTYCXFRHRGEITWHRRYPWSVAYBUHCERJXDCYAVICPTNBGIODLYLMEYLISEYNXNMCDPJJRCTLYNFMJZQNCLAGHUDVLYIGASGXSZYPZKLAWQUDVNTWGFFYFFSMQWUNUPZRJMTHACFELGHDZEJWFDWVPYOZEVEJKQWHQAHOCIYWGVLPSHFESCGEUCJGYLGDWPIWIDWZZXRUFXERABQJOXZALQOCSAYBRHXQQGUDADYSORTYZQPWGMBLNAQOFODSNXSZFURUNPMZGHTAJUJROIGMRKIZHSFUSKIZJJTLGOEEPBMIXISDHOAIFNFEKKSLEXSJLSGLCYYFEQBKIZZTQQXBQZAPXAAIFQEIXELQEZGFEPCKFPGXULLAHXTSRXDEMKFKABUTAABSLNQBNMXNEPODPGAORYJXCHCGKECLJVRBPRLHORREEIZOBSHDSCETTTNFTSMQPQIJBLKNZDMXOTRBNMTKHHCZQQMSLOAXJQKRHDGZVGITHYGVDXRTVBJEAHYBYRYKJAVXPOKHFFMEPHAGFOOPFNKQAUGYLVPWUJUPCUGGIXGRAMELUTEPYILBIUOCKKUUBJROQFTXMZRLXBAMHSDTEKRRIKZUFNLGTQAEUINMBPYTWXULQNIIRXHHGQDPENXAJNWXULFBNKBRINUMTRBFWBYVNKNKDFR';
var needle = 'ABCDA'; // the order of letters doesn't matter
var letters = {};
needle.split('').forEach(function(ch) {
letters[ch] = letters[ch] || 0;
letters[ch]++;
});
var shortestSubstringLength = haystack.length;
var shortestSubstrings = []; // storage for found substrings
var startingPos = 0;
var length;
var currentPos;
var notFound;
var letterKeys = Object.keys(letters); // unique leters
do {
lettersLeft = JSON.parse(JSON.stringify(letters)); // copy letters count object
notFound = false;
posStart = haystack.length;
posEnd = 0;
letterKeys.forEach(function(ch) {
currentPos = startingPos;
while (!notFound && lettersLeft[ch] > 0) {
currentPos = haystack.indexOf(ch, currentPos);
if (currentPos >= 0) {
lettersLeft[ch]--;
posStart = Math.min(currentPos, posStart);
posEnd = Math.max(currentPos, posEnd);
currentPos++;
} else {
notFound = true;
}
}
});
if (!notFound) {
length = posEnd - posStart + 1;
startingPos = posStart + 1; // starting position for next iteration
}
if (!notFound && length === shortestSubstringLength) {
shortestSubstrings.push(haystack.substr(posStart, length));
}
if (!notFound && length < shortestSubstringLength) {
shortestSubstrings = [haystack.substr(posStart, length)];
shortestSubstringLength = length;
}
} while (!notFound);
console.log(shortestSubstrings);

Maybe not as clear as using regex could be (well, for me regex are never really clear :D ) you can use brute force (not so brute)
Create an index of "valid" points of your string (those with the letters you want) and iterate with a double loop over it getting substrings containing at least 5 of those points, checking that they are valid solutions. Maybe not the most efficient way, but easy to implement, to understand, and probably to optimize.
var haystack="UGDVWUDDPRQLUZMSZOJPSIKAIHLTONYXAULECXXKWFQOIKELWOHRVRUCXIAASKHMWTMAJEWGEESLWRTQKVHRRCDYXNTLDSUPXMQTQDFAQAPYBGXPOLOCLFQNGNKPKOBHZWHRXAWAWJKMTJSLDLNHMUGVVOPSAMRUJEYUOBPFNEHPZZCLPNZKWMTCXERPZRFKSXVEZTYCXFRHRGEITWHRRYPWSVAYBUHCERJXDCYAVICPTNBGIODLYLMEYLISEYNXNMCDPJJRCTLYNFMJZQNCLAGHUDVLYIGASGXSZYPZKLAWQUDVNTWGFFYFFSMQWUNUPZRJMTHACFELGHDZEJWFDWVPYOZEVEJKQWHQAHOCIYWGVLPSHFESCGEUCJGYLGDWPIWIDWZZXRUFXERABQJOXZALQOCSAYBRHXQQGUDADYSORTYZQPWGMBLNAQOFODSNXSZFURUNPMZGHTAJUJROIGMRKIZHSFUSKIZJJTLGOEEPBMIXISDHOAIFNFEKKSLEXSJLSGLCYYFEQBKIZZTQQXBQZAPXAAIFQEIXELQEZGFEPCKFPGXULLAHXTSRXDEMKFKABUTAABSLNQBNMXNEPODPGAORYJXCHCGKECLJVRBPRLHORREEIZOBSHDSCETTTNFTSMQPQIJBLKNZDMXOTRBNMTKHHCZQQMSLOAXJQKRHDGZVGITHYGVDXRTVBJEAHYBYRYKJAVXPOKHFFMEPHAGFOOPFNKQAUGYLVPWUJUPCUGGIXGR";
var needle="ABCD";
var size=haystack.length;
var candidate_substring="";
var minimal_length=size;
var solutions=new Array();
var points=Array();
for(var i=0;i<size;i++){
if(needle.indexOf(haystack[i])>-1) points.push(i);
}
var limit_i= points.length-4;
var limit_k= points.length;
for (var i=0;i<limit_i;i++){
for(var k=i;k<limit_k;k++){
if(points[k]-points[i]+1<=minimal_length){
candidate_substring=haystack.substr(points[i],points[k]-points[i]+1);
if(is_valid(candidate_substring)){
solutions.push(candidate_substring);
if(candidate_substring.length < minimal_length) minimal_length=candidate_substring.length;
}
}
}
}
document.write('<p>Solution length:'+minimal_length+'<p>');
for(var i=0;i<solutions.length;i++){
if(solutions[i].length<=minimal_length) document.write('<p>Solution:'+solutions[i]+'<p>');
}
function is_valid(candidate_substring){
//verify we've got all characters
for(var j=0;j<candidate_substring.length;j++){
if(candidate_substring.indexOf(needle.charAt(j))<0) return false;
}
//...and verify we have two "A"
if(candidate_substring.indexOf("A")==candidate_substring.lastIndexOf("A")) return false;
return true;
}

Just had this problem in an interview as a coding assignment and came up with another solution, (it's not as optimal as the one above but maybe it's easier to understand).
function MinWindowSubstring(strArr) {
const N = strArr[0];
const K = strArr[1];
const letters = {};
K.split('').forEach( (character) => {
letters[character] = letters[character] ? letters[character] + 1 : 1;
});
let possibleSequencesList = [];
const letterKeys = Object.keys(letters);
for(let i=0; i< N.length; i++) {
const char = N[i];
if (new String(letterKeys).indexOf(char) !== -1) {
// found a character in the string
// update all previus sequences
possibleSequencesList.forEach((seq) => {
if(!seq.sequenceComplete) {
seq[char] = seq[char]-1;
seq.lastIndex = i;
// check if sequence is complete
var sequenceComplete = true;
letterKeys.forEach( (letter) => {
if(seq[letter] > 0) {
sequenceComplete = false;
}
});
seq.sequenceComplete = sequenceComplete
}
})
// create a new sequence starting from it
const newSeq = {
startPoint: i,
lastIndex: i,
sequenceComplete: false,
...letters
}
newSeq[char] = newSeq[char]-1;
possibleSequencesList.push(newSeq);
}
}
// cleanup sequences
let sequencesList = possibleSequencesList.filter(sequence => sequence.sequenceComplete);
let output = [];
let minLength = N.length;
// find the smalles one
sequencesList.forEach( seq => {
if( (seq.lastIndex - seq.startPoint) < minLength) {
minLength = seq.lastIndex - seq.startPoint;
output = N.substring(seq.startPoint, seq.lastIndex + 1);
}
})
return output;
}

How to get numbers from string in Javascript?

I have a strings that can look like this:
left 10 top 50
How can i extract the numbers, while the numbers can range from 0 to 100 and words can be left/right top/bottom? Thanks

Try match()
var text = "top 50 right 100 left 33";
var arr = text.match(/[0-9]{1,3}/g);
console.log(arr); //Returns an array with "50", "100", "33"
You can also use [\d+] (digits) instead of [0-9]

Place this string in a var, if you know every number will be seperated by a space you can easely do the following:
var string = "top 50 left 100";
// split at the empty space
string.split(" ");
var numbers = new Array();
// run through the array
for(var i = 0; i < string.length; i++){
// check if the string is a number
if(parseInt(string[i], 10)){
// add the number to the results
numbers.push(string[i]);
}
}
Now you can wrap the whole bit in a function to run it at any time you want:
function extractNumbers(string){
var temp = string.split(" ");
var numbers = new Array();
for(var i = 0; i < temp.length; i++){
if(parseInt(temp[i], 10)){
numbers.push(temp[i]);
}
}
return numbers;
}
var myNumbers = extractNumbers("top 50 left 100");
Update
After reading #AmirPopovich s answer, it helped me to improve it a bit more:
if(!isNaN(Number(string[i]))){
numbers.push(Number(string[i]));
}
This will return any type of number, not just Integers. Then you could technically extend the string prototype to extract numbers from any string:
String.prototype.extractNumbers = function(){ /*The rest of the function body here, replacing the keyword 'string' with 'this' */ };
Now you can do var result = "top 50 right 100".extractNumbers();

Split and extract the 2nd and 4th tokens:
var arr = "left 10 top 50".split(" ");
var a = +arr[1];
var b = +arr[3];

var str = 'left 10 top 50';
var splitted = str.split(' ');
var arr = [];
for(var i = 0 ; i < splitted.length ; i++)
{
var num = Number(splitted[i]);
if(!isNaN(num) && num >= 0 && num <= 100){
arr.push(num);
}
}
console.log(arr);
JSFIDDLE

If you want it dynamically by different keywords try something like this:
var testString = "left 10 top 50";
var result = getNumber("top", testString);
function getNumber(keyword, testString) {
var tmpString = testString;
var tmpKeyword = keyword;
tmpString = tmpString.split(tmpKeyword + " ");
tmpString = tmpString[1].split(' ')[0];
return tmpString;
}

var myArray = "left 10 top 50".split(" ");
var numbers;
for ( var index = 0; index < myArray.length; index++ ) {
if ( !isNaN(myArray[index]))
numbers= myArray[index]
}
find working example on the link below
http://jsfiddle.net/shouvik1990/cnrbv485/

How to sum up distinct value using javascript

I want to try and sum up distinct value from a list.. currently i am able to do so if theres only 2 similar record. If theres more than 2 i am not able to do the checking. Following is the javascript code:
function validateData(){
var total = document.frm.size.value;
var msg="";
var tbxA;
var tbxB;
var tbxA2;
var tbxB2;
var tbxC;
var totalValue =0;
var repeatedValue= 0;
var row = 0;
var row2 = 0;
for(var i=0; i<parseInt(total); i++){
tbxA = document.getElementById('tbx_A'+i).value;
tbxB = document.getElementById('tbx_B'+i).value-0;
tbxC = document.getElementById('tbx_C'+i).value;
for(var j=i+1; j<parseInt(total); j++){
tbxA2 = document.getElementById('tbx_A'+j).value;
tbxB2 = document.getElementById('tbx_B'+j).value-0;
if (tbxA==tbxA2) {
totalValue = tbxB + tbxB2;
}
if (totalValue != tbxC) {
repeatedValue= 1;
row = i;
row2 = j;
msg+="*total value does not add up at row " +(row2+1);
break;
}
}
if(repeatedValue== 1){
break;
}
}
return msg;
}
For example A:type of fruit, B: total of each fruit, C: how many bought at a time
total of C should be equal to B. i.e Apple: 3+3+4 = 10. So if the total is not equals to 10 it should prompt me an error.
A B C
Apple 10 3
Orange 10 10
Apple - 3
Apple - 4
My code above will prompt error bt it doesnt go beyond 2nd occurence of Apple.
So yes, how should i go about to ensure it loop through the whole list to sum up all similar values?
Thanks in advance for any possible help!

Try this:
var total = +document.frm.size.value,
data = {};
for(var i=0; i<total; ++i) {
var key = document.getElementById('tbx_A'+i).value;
data[key] = data[key] || {B:0, C:0};
data[key].B += +document.getElementById('tbx_B'+i).value || 0;
data[key].C += +document.getElementById('tbx_C'+i).value || 0;
}
for(var i in data) {
if(data.hasOwnProperty(i) && data[i].B != data[i].C) {
return "total value does not add up";
}
}
return "";
Some comments:
parseInt (and parseFloat) is very slow. + operator before string converts it to a number much faster. But if you really want to make sure the numbers are integers, use Math.floor(), Math.round(), Math.ceil() or the faster but illegible |0.
In case you really want parseInt (e.g. you want to convert '123foobar' into 123), always use a radix. For example: parseInt('123', 10)
Avoid doing calculations at the condition of a loop, because they run at each iteration. Just do the calculation once before the loop and save the result in a variable.

Javascript: How do I replace text between two delimiters?

var record = "HENRY|5|58|L581"
How do I change the above to:
record now equals "HENRY|Five|58|L581"
I know how to retrieve the index of the first '|' and the second '|' .. I know how to retrieve the number '5' into a string.
But I have no idea how to actually replace that 5 with the word Five.
The part |5| could be any number from 1-50

Something like that ?
record = record.replace('|5|', '|FIVE|');
Following edit :
To replace any number by FIVE, you can do
record = record.replace(/\|\d+\|/, '|FIVE|');
If you want to replace with something depending of the number (maybe you want TEN when the number is 10), then you'll have to do some work :
record = record.replace(/\|\d+\|/, function(str) {
var number = parseInt(str,10);
return 'FIVE'; // here build a new string and return it
});

You can do this, for example:
var record = "HENRY|5|58|L581"
var recordArray = record.split("|");
for (var i = 0; i < recordArray.length; i++) {
if (recordArray[i] === "5") {
recordArray[i] = "FIVE";
}
}
record = recordArray.join("|"); // or record = recordArray.toString();
Is this what you want to achieve?
UPDATE
If you want any number, you can set it into a function:
function changeNumber(textVar, valueToChange, replaceText) {
var recordArray = textVar.split("|");
for (var i = 0; i < recordArray.length; i++) {
if (recordArray[i] === valueToChange) {
recordArray[i] = replaceText;
}
}
return recordArray.join("|"); // or recordArray.toString();
}
See demo.

I presume you don't want to replace any number with "five", you want to replace with the actual string representing number.
var repl = [0, 1, ....];
var to = ["zero", "one", ...];
var recordArray = record.split("|");
for (var i = 0; i < recordArray.length; i++) {
recordArray[i] = to[indexOf(recordArray[i], repl)];
}
finStr = recordArray.join("|");