I am trying to write a regular expression such that if a number have decimal point then the zeros (0) at the end must be removed.
Example:
$value = 234.8076000
After Regexp Replace it should become
234.8076
I am trying the following regexp [0]+$ in preg_replace but the problem is that if the value does not have decimal point and it contain zero at the end then that zero is also removed.
Example:
$value = 2340
It becomes 234 but it should remain 2340
Any idea? Is there any in-built function in php that can do this?
Yes, you really can do it with regex:
$pattern = '/(\.\d*[^0])0+$/';
echo preg_replace($pattern, '$1', '2340'); // 2340
echo preg_replace($pattern, '$1', '2340.0'); // 2340.0
echo preg_replace($pattern, '$1', '2340.07600'); // 2340.076
... but the simplest way is just convert a string value into a float value.
echo (float)'2340'; // 2340
echo (float)'2340.0'; // 2340
echo (float)'2340.07600'; // 2340.076
Echoing floats that are really integer values drops the decimal part apparently - but it seems from your comments it's actually what you want.
You can do it without regexpes:
php > echo ((float)"21.40200")."\n";
21.402
/(\.\d*?)0+$/, works except /\d+\.0{n}/ cases
Related
I have a question about a String i want to count all the characters in the string. Like if i have a string
"Hello world & good morning. The date is 18.05.2016"
You can use explode() to convert string into array and then use count() function to count length of array.
echo count(explode(' ', "Hello world & good morning. The date is 18.05.2016"))
You can try this code.
<?php
$file = "C:\Users\singh\Desktop\Talkative Challenge\base_example.txt";
$document = file_get_contents($file);
$return_array = preg_split("/[\s,]+/",$document);
echo count($return_array);
echo $document;
?>
Hopefully it will be working fine.
The 3rd parameter of str_word_count allows you to set additional characters to be counted as words:
str_word_count($document, 0, '&.0..9');
&.0..9 means it will consider &, ., and range from 0 to 9.
You can count the spaces with substr_count and add one.
echo substr_count($str, " ")+1;
// 9
https://3v4l.org/oJJkt
I have a STRING $special which is formatted like £130.00 and is also an ex TAX(VAT) price.
I need to strip the first char so i can run some simple addition.
$str= substr($special, 1, 0); // Strip first char '£'
echo $str ; // Echo Value to check its worked
$endPrice = (0.20*$str)+$str ; // Work out VAT
I don't receive any value when i echo on the second line ? Also would i then need to convert the string to an integer in order to run the addition ?
Thanks
Matt
+++ UPDATE
Thanks for your help with this, I took your code and added some of my own, There are more than likely nicer ways to do this but it works :) I found out that if the price was below 1000 would look like £130.00 if the price was a larger value it would include a break. ie £1,400.22.
$str = str_replace('£', '', $price);
$str2 = str_replace(',', '', $str);
$vatprice = (0.2 * $str2) + $str2;
$display_vat_price = sprintf('%0.2f', $vatprice);
echo "£";
echo $display_vat_price ;
echo " (Inc VAT)";
Thanks again, Matt
You cannot use substr the way you are using it currently. This is because you are trying to remove the £ char, which is a two-byte unicode character, but substr() isn't unicode safe. You can either use $str = substr($string, 2), or, better, str_replace() like this:
$string = '£130.00';
$str = str_replace('£', '', $string);
echo (0.2 * $str) + $str; // 156
Original answer
I'll keep this version as it still can give some insight. The answer would be OK if £ wouldn't be a 2byte unicode character. Knowing this, you can still use it but you need to start the sub-string at offset 2 instead of 1.
Your usage of substr is wrong. It should be:
$str = substr($special, 1);
Check the documentation the third param would be the length of the sub-string. You passed 0, therefore you got an empty string. If you omit the third param it will return the sub-string starting from the index given in the first param until the end of the original string.
$value = preg_replace("/[^0-9]+/", '', $value);
How could I edit this regex to get rid of everything after the decimal point? There may or may not be a decimal point.
Currently "100.1" becomes 1001 but it should be 100.
Complete function:
function intfix($value)
{
$value = preg_replace("/[^0-9]+/", '', $value);
$value = trim($value);
return $value + 0;
}
It is used to format user input for numbers as well as servers output to format numbers for the DB. The functions deals with very large numbers, so I can't use intval or similar. Any extra comments to improve this function are welcome.
You could just change the regex to /[^0-9].*/s.
.* matches zero or more characters, so the first character that is not a digit, and the digits that immediately follow, would be deleted.
You need to have a pattern that starts the search with a decimal place. At the moment you're only deleting the . not the numbers after it... So you could do '/\.[\d]+/'
$text = "1201.21 12 .12 12.21";
$text = preg_replace('/\.[\d]+/', '' ,$text);
The above code would result in $text = "1201 12 12"
Why not $value = round($value, 0);? This can handle large values and is meant to get rid of the following decimals mathematically (I'd rather work with numbers as numbers not as strings). You can pass PHP_ROUND_HALF_DOWN as a third parameter if you want to just get rid of the decimals 10.7 -> 10. Or floor($value); could work too.
assuming i have these texts 'x34' , '150px' , '650dpi' , 'e3r4t5' ... how can i get only numbers ? i mean i want 34 , 150 , 650 , 345 without any other character . i mean get the numbers this string has into one variable .
$str = "e3r4t5";
$str_numbers_only = preg_replace("/[^\d]/", "", $str);
// $number = (int) $str;
Sorry for joining the bandwagon late, rather than using Regex, I would suggest you use PHP's built in functions, which may be faster than Regex.
filter_var
flags for the filters
e.g. to get just numbers from the given string
<?php
$a = '!a-b.c3#j+dk9.0$3e8`~]\]2';
$number = str_replace(['+', '-'], '', filter_var($a, FILTER_SANITIZE_NUMBER_INT));
// Output is 390382
?>
To adhere to more strict standards for your question, I have updated my answer to give a better result.
I have added str_replace, as FILTER_SANITIZE_NUMBER_FLOAT or INT flag will not strip + and - chars from the string, because they are part of PHP's exception rule.
Though it has made the filter bit long, but it's now has less chance of failing or giving you unexpected results, and this will be faster than REGEX.
Edit:
1: Realized that with FILTER_SANITIZE_NUMBER_FLOAT, PHP won't strip these characters optionally .,eE, hence to get just pure numbers kindly use FILTER_SANITIZE_NUMBER_INT
2: If you have a PHP version less than 5.4, then kindly use array('+', '-') instead of the short array syntax ['+', '-'].
You can use a regular expression to remove any character that is not a digit:
preg_replace('/\D/', '', $str)
Here the pattern \D describes any character that is not a digit (complement to \d).
Use PHP FILTER functions if you are using PHP 5.2.X, 5.3.x,5.4 . Its highly recommended
$mixed_input = "e3r4t5";
$only_numbers = filter_var($mixed_input, FILTER_SANITIZE_NUMBER_INT);
Please Go through with this link to know more
Replace everything that isn't a number and use that value.
$str = "foo1bar2baz3";
$num = intval(preg_replace("/[^0-9]/", "", $str));
You could use the following function:
function extract_numbers($string) {
preg_match_all('/([\d]+)/', $string, $match);
return $match;
}
$vari = "testing 245";
$numb = 0..9;
$numb_pos = strpos($vari,$numb);
echo substr($vari,0,$numb_pos);
The $numb is numbers from 0 to 9
Where am I wrong here, all I need to echo is testing
You want to cut out the numbers from a string?
$string = preg_replace('/(\d+)/', '', 'String with 1234 numbers');
Use a regular expression to strip numeric characters from your string.
or, use a regular expression to find the first instance of one either way...
Your code won't work as-is, as it'll fail if the number if the first character in the string. (You need to check $numb_pos !== false prior to the substr.)
Irrespective, if you just want to check for the existance of a number in a string, something like the following would probably be more efficient.
$digitMatched = preg_match('/\\d/im', $vari);