How to validate infix notation in javascript? - javascript

I have an infix expression: ((attribute1*attribute2)/attribute3+attribute4)
It may vary according to the user input. I want to check whether the expression is valid.
Valid example: ((attribute1*attribute2)/attribute3+attribute4)
Invalid example: (attrribute1*attribute2+*(attribute3)
The second one has no closing parenthesis; also the * operator is not needed. How can I perform this sort of validation in javascript?
Now this is my regex:
/ *\+? *\-? *[a-zA-Z0-9]+ *( *[\+\-\*\/\=\<\>\!\&\|\%] *\+? *\-? *[a-zA-Z0-9]+ *)*/
I need a regex for comparison operators like <= , >= , != , == etc. How can I implement this?

You could try something like this:
function validateInfix(infix) {
var balance = 0;
// remove white spaces to simplify regex
infix = infix.replace(/\s/g, '');
// if it has empty parenthesis then is not valid
if (/\(\)/.test(infix)) {
return false;
}
// valid values: integers and identifiers
var value = '(\\d+|[a-zA-Z_]\\w*)';
// the unary '+' and '-'
var unaryOper = '[\\+\\-]?';
// the arithmetic operators
var arithOper = '[\\+\\-\\*\\/]';
// the comparison operators
var compOper = '(\\<\\=?|\\>\\=?|\\=\\=|\\!\\=)';
// if it has more than one comparison operator then is not valid
if (infix.match(new RegExp(compOper, 'g')).length > 1) {
return false;
}
// the combined final regex: /[\+\-]?(\d+|[a-zA-Z_]\w*)(([\+\-\*\/]|(\<\=?|\>\=?|\=\=|\!\=))[\+\-]?(\d+|[a-zA-Z_]\w*))*/
var regex = new RegExp(unaryOper + value + '((' + arithOper + '|' + compOper + ')' + unaryOper + value + ')*');
// validate parenthesis balance
for (var i = 0; i < infix.length; i++) {
if (infix[i] == '(') {
balance++;
}
else if (infix[i] == ')') {
balance--;
}
if (balance < 0) {
return false;
}
}
if (balance > 0) {
return false;
}
// remove all the parenthesis
infix = infix.replace(/[\(\)]/g, '');
return regex.test(infix);
}
The idea is to check first the parenthesis balance, then remove them all given that we only want to validate and not evaluate, and then match the remaining expression to a regex (which may not be perfect, I'm not a regex expert). And... just in case: infix argument must be a string.
Edit
I noticed a couple of details and changed the code a bit:
Added the operators you needed the regex to match too.
Removed white spaces to get rid of regex junk.
Checked if the expression had empty parenthesis.
Checked if the expression had more than one comparison operators.
Changed this \+?\-? by this [\+\-]?.
Changed string match method by regex test method where possible.
Changed this [a-zA-Z0-9] by this (\d+|[a-zA-Z_]\w*) since the first one matches wrong identifiers like 53abc.
For better understanding and clarity, extracted pieces of regex into separate variables and built the final one from these.
Hope this is ok for you now :)

Related

javascript indexof regex A-Za-z0-9 always returns false

I have created a JS fiddle https://jsfiddle.net/95r110s9/#&togetherjs=Emdw6ORNpc
HTML
<input id="landlordstreetaddress2" class="landlordinputs" onfocusout="validateinputentries()" />
JS
validateinputentries(){
landlordstreetaddress2 = document.getElementById('landlordstreetaddress2').value;
goodcharacters = "/^[a-zA-Z0-9#.,;:'\s]+$/gi";
for (var i = 0; i < landlordstreetaddress2.length; i++){
if (goodcharacters.indexOf(landlordstreetaddress2.charAt(i)) != -1){
console.log('Character is valid');
}
}
}
Its pulling the value from an input and running an indexOf regex expression with A-Z a-z and 0-9 with a few additional characters as well.
The problem is that it works with the entry of BCDEFG...etc and 12345...etc, but when I type "A" or "Z" or "0" or "1", it returns incorrectly.
I need it to return the same with 0123456789, ABCDEF...XYZ and abcdef...xyz
I should point out that the below does work as intended:
var badcharacters = "*|,\":<>[]`\';#?=+/\\";
badcharacter = false;
//firstname
for (var i = 0; i < landlordfirstname.value.length; i++){
if (badcharacters.indexOf(landlordfirstname.value.charAt(i)) != -1){
badcharacter = true;
break;
}
if(landlordfirstname.value.charAt(0) == " "){
badcharacter = true;
break;
}
}
String.prototype.indexOf()
The indexOf() method returns the index within the calling String object of the first occurrence of the specified value, starting the search at fromIndex. Returns -1 if the value is not found.
So, you're trying to search this value "/^[a-zA-Z0-9#.,;:'\s]+$/gi" which "never" will be found in the entered string.
You actually want to test that regexp against the entered value.
/^[a-zA-Z0-9#.,;:'\s]+$/gi.test(landlordstreetaddress2)
function validateinputentries() {
var landlordstreetaddress2 = document.getElementById('landlordstreetaddress2').value;
if (/^[a-zA-Z0-9#.,;:'\s]+$/gi.test(landlordstreetaddress2)) {
console.log('Characters are valid');
} else {
console.log('Characters are invalid');
}
}
<input id="landlordstreetaddress2" class="landlordinputs" onfocusout="validateinputentries()" />
You're trying to combine two different methods of testing a string -- one way is with a regex; the other way is by checking each character against a list of allowed characters. What you've wound up with is checking each character against a list of what would have been a regex, if you hadn't declared it as a string.
Those methods conflict with each other; you need to pick one or the other.
Check each character:
This is closest to what you were attempting. You can't use character ranges here (like a-zA-Z) as you would in a regex; you have to spell out each allowed character individually:
var validateinputentries = function() {
var address = document.getElementById('landlordstreetaddress2').value;
var goodcharacters = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789#.,;:' ";
var badcharactersfound = false;
for (var i = 0; i < address.length; i++) {
if (goodcharacters.indexOf(address.charAt(i)) == -1) {
badcharactersfound = true;
console.log("not allowed: ", address.charAt(i));
}
}
if (badcharactersfound) {
// Show validation error here
}
}
<input id="landlordstreetaddress2" class="landlordinputs" onfocusout="validateinputentries()" />
Regular Expressions
The regex version is much simpler, because the regular expression is doing most of the work. You don't need to step through the string, just test the whole string against the regex and see what comes out. In this case you're looking to see if the input contains any characters that aren't allowed, so you want to use the character exception rule: [^abc] will match any character that is not a, b, or c. You don't want to anchor the match to the beginning or the end of the string, as you were doing with the initial ^ and the trailing $; and you can leave out the + because you don't care if there are sequential bad characters, you just care if they exist at all.
var validateinputentries = function() {
var address = document.getElementById('landlordstreetaddress2').value;
var regex = new RegExp("[^a-zA-Z0-9#.,;:'\\s]","g")
var badcharactersfound = address.match(regex);
// or the above two lines could also have been written like this:
// var bad = address.match(/[^a-zA-Z0-9#.,;:'\s]/g)
// In either case the "g" operator could be omitted; then it would only return the first bad character.
if (badcharactersfound) {
console.log("Not allowed: ", badcharactersfound);
}
}
<input id="landlordstreetaddress2" class="landlordinputs" onfocusout="validateinputentries()" />

Regular Expression Returns Undefined

I'm attempting one of the beginner coderByte challenges, Simple Symbols. Challenge summary below.
"Using the JavaScript language, have the function SimpleSymbols(str) take the str parameter being passed and determine if it is an acceptable sequence by either returning the string true or false. The str parameter will be composed of + and = symbols with several letters between them (ie. ++d+===+c++==a) and for the string to be true each letter must be surrounded by a + symbol. So the string to the left would be false. The string will not be empty and will have at least one letter."
function SimpleSymbols(str){
var RegExp = /\+\w\+/gi;
var regexp1 = /^\w/gi;
var regexp2 = /\w$/g;
if(regexp1.test(str) == true){
return false
} else if(regexp2.test(str) == true){
return false
} else if(RegExp == true){
return true
}
}
console.log(SimpleSymbols('+d+=3=+s+'));
console.log(SimpleSymbols('f++d+'));
The first regular expression I'm testing, /^\w/gi, comes back undefined, and I can't figure out why?
https://regex101.com/ is a great tool I've used before, and my expression does identify f as the first character in the string, but when I test it in codepen, it comes back undefined in the console.
Code
See regex in use here
^[+=\d]*\+(?:[a-z]\+[+=\d]*)+$
Alternatively, using the opposite logic (catching invalid strings instead of valid ones), you can use (?:^|[^+])[a-z]|[a-z](?:[^+]|$)
Usage
Please note the valid/invalid strings below have been created according to the OP's explanation of valid and invalid strings: That each letter must be surrounded by a + symbol. and that the plus sign + may be shared between characters such that +a+a+ is valid (specified in comments below the question).
var a = [
// valid
"++d+===+c++==+a++",
"+a+a+a+",
"+a++a+",
"+a+",
// invalid
"++d+===+c++==a",
"+=d+",
"+dd+",
"+d=+",
"+d+d",
"d+d+"
];
var r = /^[+=\d]*\+(?:[a-z]\+[+=\d]*)+$/mi;
a.forEach(function(s){
console.log(r.test(s));
});
Explanation
^ Assert position at the start of the line
[+=\d]* Match any number of characters in the set (+, =, or digit)
\+ Match a literal plus sign +
(?:[a-z]\+[+=\d]*)+ Match one or more of the following
[a-z] Match a lowercase ASCII letter
\+ Match a literal plus sign +
[+=\d]* Match any number of characters in the set (+, =, or digit)
$ Assert position at the end of the line
It's returning undefined because your expression does not meet any of the criteria. Since you have no else {} defined, than nothing gets returned. Thus you get undefined. Try this:
function SimpleSymbols(str){
var RegExp = /\+\w\+/gi;
var regexp1 = /^\w/gi;
var regexp2 = /\w$/g;
if(regexp1.test(str) == true){
return false
} else if(regexp2.test(str) == true){
return false
} else if(RegExp == true){
return true
} else {
return "catch all here";
}
}
console.log(SimpleSymbols('+d+=3=+s+'));
console.log(SimpleSymbols('f++d+'));
You can use a single regex and the string.test() method (which returns just true/false).
Below are 2 different ways (regex) to do it .
First requires a separate + between word chars. Example +a++b+ (true)
^
(?: [+=]* \+ \w \+ [+=]* )+
$
Second can take a common + between word chars. Example +a+b+ (true)
^
(?:
[+=]* \+ \w
(?= \+ )
)+
[+=]*
$
var patt1 = new RegExp("^(?:[+=]*\\+\\w\\+[+=]*)+$");
function SimpleSymbols_1(str){
return patt1.test(str);
}
var patt2 = new RegExp("^(?:[+=]*\\+\\w(?=\\+))+[+=]*$");
function SimpleSymbols_2(str){
return patt2.test(str);
}
console.log(SimpleSymbols_1('+d+=3=+s+'));
console.log(SimpleSymbols_1('f++d+'));
console.log(SimpleSymbols_1('+a+b+c+'));
console.log(SimpleSymbols_2('+a+b+c+'));
console.log(SimpleSymbols_2('+a+=+c+'));
console.log(SimpleSymbols_2('+a++c+'));
Thank you all for throwing some support/comments my way. Again, I am new to JavaScript and Regular Expressions are fairly foreign to me, though I am gaining some traction in understanding them. Here is the updated solution I posted. It's quite convoluted and perhaps a more inelegant and non-simple way to come to the right answer, but it worked.
function SimpleSymbols(str){
var RegExp = /\+[a-z]\+/gi;
var regexp1 = /^[a-z]/gi;
var regexp2 = /[a-z]$/g;
var regexp3 = /[a-z]\=/gi;
var regexp4 = /\=[a-z]/gi;
if(regexp1.test(str) === true){
return false
} else if(regexp2.test(str) === true){
return false
} else if(regexp3.test(str) === true){
return false
} else if(regexp4.test(str) === true){
return false
} else {
return true
}
}
console.log(SimpleSymbols('+d+=3=+s+'));
console.log(SimpleSymbols('f++d+'));
console.log(SimpleSymbols('+d===+a+'));
console.log(SimpleSymbols('+a='));
console.log(SimpleSymbols('2+a+a+'));
console.log(SimpleSymbols('==a+'));
I was sure there had to be a way to use only one regular expression, but again, I'm still very much a novice.
Thanks again everyone.

Regular Expressions required format

I want to validate following text using regular expressions
integer(1..any)/'fs' or 'sf'/ + or - /integer(1..any)/(h) or (m) or (d)
samples :
1) 8fs+60h
2) 10sf-30m
3) 2fs+3h
3) 15sf-20m
i tried with this
function checkRegx(str,id){
var arr = strSplit(str);
var regx_FS =/\wFS\w|\d{0,9}\d[hmd]/gi;
for (var i in arr){
var str_ = arr[i];
console.log(str_);
var is_ok = str_.match(regx_FS);
var err_pos = str_.search(regx_FS);
if(is_ok){
console.log(' ID from ok ' + id);
$('#'+id).text('Format Error');
break;
}else{
console.log(' ID from fail ' + id);
$('#'+id).text('');
}
}
}
but it is not working
please can any one help me to make this correct
This should do it:
/^[1-9]\d*(?:fs|sf)[-+][1-9]\d*[hmd]$/i
You were close, but you seem to be missing some basic regex comprehension.
First of all, the ^ and $ just make sure you're matching the entire string. Otherwise any junk before or after will count as valid.
The formation [1-9]\d* allows for any integer from 1 upwards (and any number of digits long).
(?:fs|sf) is an alternation (the ?: is to make the group non-capturing) to allow for both options.
[-+] and [hmd] are character classes allowing to match any one of the characters in there.
That final i allows the letters to be lowercase or uppercase.
I don't see how the expression you tried relates anyhow to the description you gave us. What you want is
/\d+(fs|sf)[+-]\d+[hmd]/
Since you seem to know a bit about regular expressions I won't give a step-by-step explanation :-)
If you need exclude zero from the "integer" matches, use [1-9]\d* instead. Not sure whether by "(1..any)" you meant the number of digits or the number itself.
Looking on the code, you
should not use for in enumerations on arrays
will need string start and end anchors to check whether _str exactly matches the regex (instead of only some part)
don't need the global flag on the regex
rather might use the RegExp test method than match - you don't need a result string but only whether it did match or not
are not using the err_pos variable anywhere, and it hardly will work with search
function checkRegx(str, id) {
var arr = strSplit(str);
var regx_FS = /^\d+(fs|sf)[+-]\d+[hmd]$/i;
for (var i=0; i<arr.length; i++) {
var str = arr[i];
console.log(str);
if (regx_FS.test(str) {
console.log(' ID from ok ' + id);
$('#'+id).text('Format Error');
break;
} else {
console.log(' ID from fail ' + id);
$('#'+id).text('');
}
}
}
Btw, it would be better to separate the validation (regex, array split, iteration) from the output (id, jQuery, logs) into two functions.
Try something like this:
/^\d+(?:fs|sf)[-+]\d+[hmd]$/i

Javascript IndexOf with integers in string not working

Can anyone tell me why does this not work for integers but works for characters? I really hate reg expressions since they are cryptic but will if I have too. Also I want to include the "-()" as well in the valid characters.
String.prototype.Contains = function (str) {
return this.indexOf(str) != -1;
};
var validChars = '0123456789';
var str = $("#textbox1").val().toString();
if (str.Contains(validChars)) {
alert("found");
} else {
alert("not found");
}
Review
String.prototype.Contains = function (str) {
return this.indexOf(str) != -1;
};
This String "method" returns true if str is contained within itself, e.g. 'hello world'.indexOf('world') != -1would returntrue`.
var validChars = '0123456789';
var str = $("#textbox1").val().toString();
The value of $('#textbox1').val() is already a string, so the .toString() isn't necessary here.
if (str.Contains(validChars)) {
alert("found");
} else {
alert("not found");
}
This is where it goes wrong; effectively, this executes '1234'.indexOf('0123456789') != -1; it will almost always return false unless you have a huge number like 10123456789.
What you could have done is test each character in str whether they're contained inside '0123456789', e.g. '0123456789'.indexOf(c) != -1 where c is a character in str. It can be done a lot easier though.
Solution
I know you don't like regular expressions, but they're pretty useful in these cases:
if ($("#textbox1").val().match(/^[0-9()]+$/)) {
alert("valid");
} else {
alert("not valid");
}
Explanation
[0-9()] is a character class, comprising the range 0-9 which is short for 0123456789 and the parentheses ().
[0-9()]+ matches at least one character that matches the above character class.
^[0-9()]+$ matches strings for which ALL characters match the character class; ^ and $ match the beginning and end of the string, respectively.
In the end, the whole expression is padded on both sides with /, which is the regular expression delimiter. It's short for new RegExp('^[0-9()]+$').
Assuming you are looking for a function to validate your input, considering a validChars parameter:
String.prototype.validate = function (validChars) {
var mychar;
for(var i=0; i < this.length; i++) {
if(validChars.indexOf(this[i]) == -1) { // Loop through all characters of your string.
return false; // Return false if the current character is not found in 'validChars' string.
}
}
return true;
};
var validChars = '0123456789';
var str = $("#textbox1").val().toString();
if (str.validate(validChars)) {
alert("Only valid characters were found! String validates!");
} else {
alert("Invalid Char found! String doesn't validate.");
}
However, This is quite a load of code for a string validation. I'd recommend looking into regexes, instead. (Jack's got a nice answer up here)
You are passing the entire list of validChars to indexOf(). You need to loop through the characters and check them one-by-one.
Demo
String.prototype.Contains = function (str) {
var mychar;
for(var i=0; i<str.length; i++)
{
mychar = this.substr(i, 1);
if(str.indexOf(mychar) == -1)
{
return false;
}
}
return this.length > 0;
};
To use this on integers, you can convert the integer to a string with String(), like this:
var myint = 33; // define integer
var strTest = String(myint); // convert to string
console.log(strTest.Contains("0123456789")); // validate against chars
I'm only guessing, but it looks like you are trying to check a phone number. One of the simple ways to change your function is to check string value with RegExp.
String.prototype.Contains = function(str) {
var reg = new RegExp("^[" + str +"]+$");
return reg.test(this);
};
But it does not check the sequence of symbols in string.
Checking phone number is more complicated, so RegExp is a good way to do this (even if you do not like it). It can look like:
String.prototype.ContainsPhone = function() {
var reg = new RegExp("^\\([0-9]{3}\\)[0-9]{3}-[0-9]{2}-[0-9]{2}$");
return reg.test(this);
};
This variant will check phones like "(123)456-78-90". It not only checks for a list of characters, but also checks their sequence in string.
Thank you all for your answers! Looks like I'll use regular expressions. I've tried all those solutions but really wanted to be able to pass in a string of validChars but instead I'll pass in a regex..
This works for words, letters, but not integers. I wanted to know why it doesn't work for integers. I wanted to be able to mimic the FilteredTextBoxExtender from the ajax control toolkit in MVC by using a custom Attribute on a textBox

Javascript search string for numbers regex maybe?

I have a string like
:21::22::24::99:
And I want to find say if :22: is in said string. But is there a means of searching a string like above for one like I want to match it to with javascript, and if there is, does it involve regex magic or is there something else? Either way not sure how to do it, more so if regex is involved.
You can build the regular expression you need:
function findNumberInString(num, s) {
var re = new RegExp(':' + num + ':');
return re.test(s);
}
var s = ':21::22::24::99';
var n = '22';
findNumberInString(n, s); // true
or just use match (though test is cleaner to me)
!!s.match(':' + n + ':'); // true
Edit
Both the above use regular expressions, so a decimal ponit (.) will come to represent any character, so "4.1" will match "461" or even "4z1", so better to use a method based on String.prototype.indexOf just in case (unless you want "." to represent any character), so per Blender's comment:
function findNumberInString(num, s) {
return s.indexOf(':' + num + ':') != -1;
}
like this:
aStr = ':21::22::24::99:';
if(aStr.indexOf(':22:') != -1){
//':22:' exists in aStr
}
else{
//it doesn't
}

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