Is it possible to write a regex which would take input like 'sqrt(2 * (2+2)) + sin(pi/6)' and transform it into '\sqrt{2 \cdot (2+2)} + \sin(\pi/6)'?
The problem is the 'sqrt' and parentheses in it. It is obvious I can't simply use something like this:
/sqrt\((.?)\)/ -> \\sqrt{$1}
because this code would create something like this '\sqrt{2 \cdot (2+2)) + \sin(\pi/6}'.
My solution: it simply go throw the string converted to char array and tests if a current substring starts with $latex, if it does second for-cycle go from this point in different direction and by parentheses decides where the function starts and ends. (startsWith function)
Code:
public static function formatFunction($function, $latex, $input) {
$input = preg_replace("/" . $function . "\(/", $latex . "{", $input);
$arr = str_split($input);
$inGap = false;
$gap = 0;
for ($i = count($arr) - 1; $i >= 0; $i--) {
if (startsWith(substr($input, $i), $latex)) {
for ($x = $i; $x < count($arr); $x++) {
if ($arr[$x] == "(" || $arr[$x] == "{") { $gap++; $inGap = true; }
else if ($arr[$x] == ")" || $arr[$x] == "}") { $gap--; }
if ($inGap && $gap == 0) {
$arr[$x] = "}";
$inGap = false;
break;
}
}
}
$gap = 0;
}
return implode($arr);
}
Use:
self::formatFunction("sqrt", "\\sqrt",
"sqrt(25 + sqrt(16 - sqrt(49)) + (7 + 1)) + sin(pi/2)");
Output:
\sqrt{25+\sqrt{16-\sqrt{49}}+(7+1)}+\sin (\pi/2)
Note: sin and pi aren't formated by this code, it's only str_replace function...
In general, no regular expression can effectively handle nested parentheses. Sorry to be the bearer of bad news! The MathJAX parser library can interpret LaTeX equations and you could probably add a custom output routine to do what you want.
For TeX questions, you can also try http://tex.stackexchange.com .
Some time ago i soved a similar problem in such way. Maybe it will be helpful for you
$str = 'sqrt((2 * (2+2)) + sin(pi/(6+7)))';
$from = []; // parentheses content
$to = []; // patterns for replace #<number>
$brackets = [['(', ')'], ['{', '}'], ['[', ']']]; // new parentheses for every level
$level = 0;
$count = 1; // count or replace made
while($count) {
$str = preg_replace_callback('(\(([^()]+)\))',
function ($m) use (&$to, &$from, $brackets, $level) {
array_unshift($to, $brackets[$level][0] . $m[1] . $brackets[$level][1]);
$i = '#' . (count($to)-1); // pattern for future replace.
// here it '#1', '#2'.
// Make it so they will be unique
array_unshift($from, $i);
return $i; }, $str, -1, $count);
$level++;
}
echo str_replace($from, $to, $str); // return content back
// sqrt[{2 * (2+2)} + sin{pi/(6+7)}]
I forgot all details, but it, seems, works
I have an array of numbers, these numbers are sometimes hyphenated, à la software version numbers. What I'm trying to do is echo "Missing!" or run a specific function when a number is missing.
For example:
$numbers = array('1', '2', '3', '5', '6', '8');
Prints:
1
2
3
Missing!
5
6
Missing!
8
I'm running into problems with the hyphens.
For example:
$numbers = array('1', '1-1', '1-3', '3-1-1', '3-1-3');
Prints:
1
1-1
Missing!
1-3
Missing!
3-1-1
Missing!
3-1-3
Plus my code seems awfully long/doing too many things for what -- seems to me -- should be a simple task. Is there a method or algorithm for this sort of thing?
Here's my code:
<?php
$numbers = array(
'1',
'1-1',
'1-3',
'3-1-1',
'3-1-3'
);
foreach ($numbers as $number) {
if (isset($prev_number)) {
$curr_number = explode('-', $number);
$prev_levels = explode('-', $prev_number);
if (preg_match('/-/', $number) and !preg_match('/-/', $prev_number)) {
if (current() - $prev_levels[0] >= 1) {
echo 'Missing!<br>' . PHP_EOL;
}
}
for ($missing = 1; ((count($curr_number) - count($prev_levels)) - $missing) >= 1; $missing++) {
echo 'Missing!<br>' . PHP_EOL;
}
foreach ($curr_number as $hyphen => $part) {
for ($missing = 1; ($part - $missing) - $prev_levels[$hyphen] >= 1; $missing++) {
echo 'Missing!<br>' . PHP_EOL;
}
}
} else {
if ($number != '1') {
echo 'Missing!<br>' . PHP_EOL;
foreach ($curr_number as $part) {
for ($missing = 1; $part > $missing; $missing++) {
echo 'Missing!<br>' . PHP_EOL;
}
}
}
}
echo $number . '<br>' . PHP_EOL;
$prev_number = $number;
}
?>
Only a partial answer.
Plus my code seems awfully long/doing too many things for what -- seems to me -- should be a simple task.
Right observation. If it's doing too many things, try to split up the task into pieces. Modularize.
For example: I see 6 explode calls in your code. You constantly struggling with transforming the input into a usable data format. Do a pre-process stage, convert the strings into arrays with explode once, and then work on that data format.
You can iterate through the list and for each pair you attempt to reach the second by applying a transformation on the first:
Increase the last value, e.g. "1" becomes "2" and "1-1" becomes "1-2".
Add a new sub value, e.g. "1" becomes "1-1" and "1-1" becomes "1-1-1".
#1 can be expanded by increasing from right to left.
If none of the transformations match the number is considered missing. In code:
$numbers = array('1', '1-1', '1-2', '1-3', '3-1-1', '3-1-3');
$first = array_shift($numbers);
echo "$first\n";
while (($second = array_shift($numbers)) !== null) {
if (find_next($first, $second) === false) {
echo "Missing\n";
}
echo "$second\n";
$first = $second;
}
// attempt to transform between current and next
function find_next($current, $next)
{
if (increment_last($current) == $next) {
return $next; // first transformation worked
} elseif (add_suffix($current) == $next) {
return $next; // second transformation worked
}
return false; // nothing worked
}
// transformation 1
function increment_last($value)
{
if (($pos = strpos($value, '-')) !== false) {
$last = substr($value, $pos + 1) + 1;
return substr_replace($value, $last, $pos + 1, strlen($last));
} else {
return $value + 1;
}
}
// transformation 2
function add_suffix($value)
{
return "$value-1";
}
This is not an algorithm, but a heuristic; it does a best effort at doing what you want but there's no formal proof that it works.
As designing a new platform we tried to integrate the IBAN numbers. We have to make sure that the IBAN is validated and the IBAN stored to the database is always correct. So what would be a proper way to validate the number?
As the logic was explained in my other question, I've created a function myself. Based on the logic explained in the Wikipedia article find a proper function below. Country specific validation.
Algorithm and character lengths per country at https://en.wikipedia.org/wiki/International_Bank_Account_Number#Validating_the_IBAN.
function checkIBAN($iban)
{
if(strlen($iban) < 5) return false;
$iban = strtolower(str_replace(' ','',$iban));
$Countries = array('al'=>28,'ad'=>24,'at'=>20,'az'=>28,'bh'=>22,'be'=>16,'ba'=>20,'br'=>29,'bg'=>22,'cr'=>21,'hr'=>21,'cy'=>28,'cz'=>24,'dk'=>18,'do'=>28,'ee'=>20,'fo'=>18,'fi'=>18,'fr'=>27,'ge'=>22,'de'=>22,'gi'=>23,'gr'=>27,'gl'=>18,'gt'=>28,'hu'=>28,'is'=>26,'ie'=>22,'il'=>23,'it'=>27,'jo'=>30,'kz'=>20,'kw'=>30,'lv'=>21,'lb'=>28,'li'=>21,'lt'=>20,'lu'=>20,'mk'=>19,'mt'=>31,'mr'=>27,'mu'=>30,'mc'=>27,'md'=>24,'me'=>22,'nl'=>18,'no'=>15,'pk'=>24,'ps'=>29,'pl'=>28,'pt'=>25,'qa'=>29,'ro'=>24,'sm'=>27,'sa'=>24,'rs'=>22,'sk'=>24,'si'=>19,'es'=>24,'se'=>24,'ch'=>21,'tn'=>24,'tr'=>26,'ae'=>23,'gb'=>22,'vg'=>24);
$Chars = array('a'=>10,'b'=>11,'c'=>12,'d'=>13,'e'=>14,'f'=>15,'g'=>16,'h'=>17,'i'=>18,'j'=>19,'k'=>20,'l'=>21,'m'=>22,'n'=>23,'o'=>24,'p'=>25,'q'=>26,'r'=>27,'s'=>28,'t'=>29,'u'=>30,'v'=>31,'w'=>32,'x'=>33,'y'=>34,'z'=>35);
if(array_key_exists(substr($iban,0,2), $Countries) && strlen($iban) == $Countries[substr($iban,0,2)]){
$MovedChar = substr($iban, 4).substr($iban,0,4);
$MovedCharArray = str_split($MovedChar);
$NewString = "";
foreach($MovedCharArray AS $key => $value){
if(!is_numeric($MovedCharArray[$key])){
if(!isset($Chars[$MovedCharArray[$key]])) return false;
$MovedCharArray[$key] = $Chars[$MovedCharArray[$key]];
}
$NewString .= $MovedCharArray[$key];
}
if(bcmod($NewString, '97') == 1)
{
return true;
}
}
return false;
}
Slight modification of #PeterFox answer including support for bcmod() when bcmath is not available,
<?php
function isValidIBAN ($iban) {
$iban = strtolower($iban);
$Countries = array(
'al'=>28,'ad'=>24,'at'=>20,'az'=>28,'bh'=>22,'be'=>16,'ba'=>20,'br'=>29,'bg'=>22,'cr'=>21,'hr'=>21,'cy'=>28,'cz'=>24,
'dk'=>18,'do'=>28,'ee'=>20,'fo'=>18,'fi'=>18,'fr'=>27,'ge'=>22,'de'=>22,'gi'=>23,'gr'=>27,'gl'=>18,'gt'=>28,'hu'=>28,
'is'=>26,'ie'=>22,'il'=>23,'it'=>27,'jo'=>30,'kz'=>20,'kw'=>30,'lv'=>21,'lb'=>28,'li'=>21,'lt'=>20,'lu'=>20,'mk'=>19,
'mt'=>31,'mr'=>27,'mu'=>30,'mc'=>27,'md'=>24,'me'=>22,'nl'=>18,'no'=>15,'pk'=>24,'ps'=>29,'pl'=>28,'pt'=>25,'qa'=>29,
'ro'=>24,'sm'=>27,'sa'=>24,'rs'=>22,'sk'=>24,'si'=>19,'es'=>24,'se'=>24,'ch'=>21,'tn'=>24,'tr'=>26,'ae'=>23,'gb'=>22,'vg'=>24
);
$Chars = array(
'a'=>10,'b'=>11,'c'=>12,'d'=>13,'e'=>14,'f'=>15,'g'=>16,'h'=>17,'i'=>18,'j'=>19,'k'=>20,'l'=>21,'m'=>22,
'n'=>23,'o'=>24,'p'=>25,'q'=>26,'r'=>27,'s'=>28,'t'=>29,'u'=>30,'v'=>31,'w'=>32,'x'=>33,'y'=>34,'z'=>35
);
if (strlen($iban) != $Countries[ substr($iban,0,2) ]) { return false; }
$MovedChar = substr($iban, 4) . substr($iban,0,4);
$MovedCharArray = str_split($MovedChar);
$NewString = "";
foreach ($MovedCharArray as $k => $v) {
if ( !is_numeric($MovedCharArray[$k]) ) {
$MovedCharArray[$k] = $Chars[$MovedCharArray[$k]];
}
$NewString .= $MovedCharArray[$k];
}
if (function_exists("bcmod")) { return bcmod($NewString, '97') == 1; }
// http://au2.php.net/manual/en/function.bcmod.php#38474
$x = $NewString; $y = "97";
$take = 5; $mod = "";
do {
$a = (int)$mod . substr($x, 0, $take);
$x = substr($x, $take);
$mod = $a % $y;
}
while (strlen($x));
return (int)$mod == 1;
}
The accepted answer is not the preferred way of validation. The specification dictates the following:
Check that the total IBAN length is correct as per the country. If not, the IBAN is invalid
Replace the two check digits by 00 (e.g. GB00 for the UK)
Move the four initial characters to the end of the string
Replace the letters in the string with digits, expanding the string as necessary, such that A or a = 10, B or b = 11, and Z or z = 35. Each alphabetic character is therefore replaced by 2 digits
Convert the string to an integer (i.e. ignore leading zeroes)
Calculate mod-97 of the new number, which results in the remainder
Subtract the remainder from 98, and use the result for the two check digits. If the result is a single digit number, pad it with a leading 0 to make a two-digit number
I've written a class that validates, formats and parses strings according to the spec. Hope this helps some save the time required to roll their own.
The code can be found on GitHub here.
top rated function does NOT work.
Just try a string with '%' in it...
I'm using this one :
function checkIBAN($iban) {
// Normalize input (remove spaces and make upcase)
$iban = strtoupper(str_replace(' ', '', $iban));
if (preg_match('/^[A-Z]{2}[0-9]{2}[A-Z0-9]{1,30}$/', $iban)) {
$country = substr($iban, 0, 2);
$check = intval(substr($iban, 2, 2));
$account = substr($iban, 4);
// To numeric representation
$search = range('A','Z');
foreach (range(10,35) as $tmp)
$replace[]=strval($tmp);
$numstr=str_replace($search, $replace, $account.$country.'00');
// Calculate checksum
$checksum = intval(substr($numstr, 0, 1));
for ($pos = 1; $pos < strlen($numstr); $pos++) {
$checksum *= 10;
$checksum += intval(substr($numstr, $pos,1));
$checksum %= 97;
}
return ((98-$checksum) == $check);
} else
return false;
}
I found this solution in cakephp 3.7 validation class. Plain beautiful php realization.
/**
* Check that the input value has a valid International Bank Account Number IBAN syntax
* Requirements are uppercase, no whitespaces, max length 34, country code and checksum exist at right spots,
* body matches against checksum via Mod97-10 algorithm
*
* #param string $check The value to check
*
* #return bool Success
*/
public static function iban($check)
{
if (!preg_match('/^[A-Z]{2}[0-9]{2}[A-Z0-9]{1,30}$/', $check)) {
return false;
}
$country = substr($check, 0, 2);
$checkInt = intval(substr($check, 2, 2));
$account = substr($check, 4);
$search = range('A', 'Z');
$replace = [];
foreach (range(10, 35) as $tmp) {
$replace[] = strval($tmp);
}
$numStr = str_replace($search, $replace, $account . $country . '00');
$checksum = intval(substr($numStr, 0, 1));
$numStrLength = strlen($numStr);
for ($pos = 1; $pos < $numStrLength; $pos++) {
$checksum *= 10;
$checksum += intval(substr($numStr, $pos, 1));
$checksum %= 97;
}
return ((98 - $checksum) === $checkInt);
}
This function check the IBAN and need GMP activate http://php.net/manual/en/book.gmp.php.
function checkIban($string){
$to_check = substr($string, 4).substr($string, 0,4);
$converted = '';
for ($i = 0; $i < strlen($to_check); $i++){
$char = strtoupper($to_check[$i]);
if(preg_match('/[0-9A-Z]/',$char)){
if(!preg_match('/\d/',$char)){
$char = ord($char)-55;
}
$converted .= $char;
}
}
// prevent: "gmp_mod() $num1 is not an integer string" error
$converted = ltrim($converted, '0');
return strlen($converted) && gmp_strval(gmp_mod($converted, '97')) == 1;
}
enjoy !
Does anybody know a PHP function for IMEI validation?
Short solution
You can use this (witchcraft!) solution, and simply check the string length:
function is_luhn($n) {
$str = '';
foreach (str_split(strrev((string) $n)) as $i => $d) {
$str .= $i %2 !== 0 ? $d * 2 : $d;
}
return array_sum(str_split($str)) % 10 === 0;
}
function is_imei($n){
return is_luhn($n) && strlen($n) == 15;
}
Detailed solution
Here's my original function that explains each step:
function is_imei($imei){
// Should be 15 digits
if(strlen($imei) != 15 || !ctype_digit($imei))
return false;
// Get digits
$digits = str_split($imei);
// Remove last digit, and store it
$imei_last = array_pop($digits);
// Create log
$log = array();
// Loop through digits
foreach($digits as $key => $n){
// If key is odd, then count is even
if($key & 1){
// Get double digits
$double = str_split($n * 2);
// Sum double digits
$n = array_sum($double);
}
// Append log
$log[] = $n;
}
// Sum log & multiply by 9
$sum = array_sum($log) * 9;
// Compare the last digit with $imei_last
return substr($sum, -1) == $imei_last;
}
Maybe can help you :
This IMEI number is something like this: ABCDEF-GH-IJKLMNO-X (without “-” characters)
For example: 350077523237513
In our example ABCDEF-GH-IJKLMNO-X:
AB is Reporting Body Identifier such as 35 = “British Approvals Board of Telecommunications (BABT)”
ABCDEF is Type Approval Code
GH is Final Assembly Code
IJKLMNO is Serial Number
X is Check Digit
Also this can help you : http://en.wikipedia.org/wiki/IMEI#Check_digit_computation
If i don't misunderstood, IMEI numbers using Luhn algorithm . So you can google this :) Or you can search IMEI algorithm
Maybe your good with the imei validator in the comments here:
http://www.php.net/manual/en/function.ctype-digit.php#77718
But I haven't tested it
Check this solution
<?php
function validate_imei($imei)
{
if (!preg_match('/^[0-9]{15}$/', $imei)) return false;
$sum = 0;
for ($i = 0; $i < 14; $i++)
{
$num = $imei[$i];
if (($i % 2) != 0)
{
$num = $imei[$i] * 2;
if ($num > 9)
{
$num = (string) $num;
$num = $num[0] + $num[1];
}
}
$sum += $num;
}
if ((($sum + $imei[14]) % 10) != 0) return false;
return true;
}
$imei = '868932036356090';
var_dump(validate_imei($imei));
?>
IMEI validation uses Luhn check algorithm. I found a link to a page where you can validate your IMEI. Furthermore, at the bottom of this page is a piece of code written in JavaScript to show how to calculate the 15th digit of IMEI and to valid IMEI. I might give you some ideas. You can check it out here http://imei.sms.eu.sk/index.html
Here is a jQuery solution which may be of use: https://github.com/madeinstefano/imei-validator
good fun from kasperhartwich
function validateImei($imei, $use_checksum = true) {
if (is_string($imei)) {
if (ereg('^[0-9]{15}$', $imei)) {
if (!$use_checksum) return true;
for ($i = 0, $sum = 0; $i < 14; $i++) {
$tmp = $imei[$i] * (($i%2) + 1 );
$sum += ($tmp%10) + intval($tmp/10);
}
return (((10 - ($sum%10)) %10) == $imei[14]);
}
}
return false;
}