convert this: $300
to this : 300
can't do it with intval() or (int)
typecasting
if the non-numerical character is
suffixed (300$), both works and
returns 300
if it is prefixed it returns 0
the non-numerical character can be anything other than the "$"(i.e. "askldjflksdjflsd")
Please help
EDIT : list items are not requirements, they are a list of activities and observations I have made. Sorry:(
preg_match('/\d+/', $num, $matches);
echo $matches[0];
You can get rid of all the non-digits in the input by doing:
$input = preg_replace('/\D/','',$input);
$number = filter_var($number, FILTER_SANITIZE_NUMBER_INT);
print (int) trim('$300', '$');
No need for a regex.
$number = (integer) str_replace('$', '', $number);
Related
i have input form currency with format like this 1,000.00
and i want to save in database in decimal format like 1000.00
how to change it in controller, when save it it will automatically change to decimal format?
$num = '3,500.20';
$formattedNum = number_format($num, 3, '', '');
echo $formattedNum;
the code above convert to 3000 not 3500.20
you can use str_replace of php to get you result:
<?php
$num = '3,500.20';
$formattedNum = str_replace(',', '', $num);
echo $formattedNum;
?>
str_replace funcation replace your ',' with '' a blank value,
To read more about str_replace you can read here
You can just remove comma with str_replace(',', '', $num) for example.
But better to get correct value right from your input.
Input format shouldn't affect actual input value which is 3500.20, that doesn't look as good practice.
Also number_format is not for converting string to numbers, on the contrary it makes pretty string from number(integer/float).
You can try str_replace for the desired result:-
$number = '3,500.20';
$number = str_replace(',', '', $number);
echo $number;
Output
3500.20
I'm trying to get all numeric before space/alpha in PHP string.
Example:
<?php
//string
$firstStr = '12 Car';
$secondStr = '412 8all';
$thirdStr = '100Pen';
//result I need
firstStr = 12
SecondStr = 412
thirdStr = 100
How do I can get all the number of a string just like example above?
I've an idea to get the position of first Alpha, then get all numeric before that position.
I've successfully get the position using
preg_match('~[a-z]~i', $value, $match, PREG_OFFSET_CAPTURE);
But I'm not done yet to get the numeric before the posisition.
How do I can do that, or anybody know how to fix my idea?
Anyhelp will be appreciated.
You don't need to use regex for strings like the examples you've shown, or any functions at all for that matter. You can just cast them to ints.
$number = (int) $firstStr; // etc.
The PHP rules for string conversion to number will handle it for you.
However, because of those rules, there are some other types of strings that this won't work for. For example, '-12 Car' or '412e2 8all'.
If you do use a regex, be sure to anchor it to the beginning of the string with ^ or it will match digits anywhere in the string as the other regex answers here do.
preg_match('/^\d+/', $string, $match);
$number = $match[0] ?? '';
Here's an extremely hackish approach that will work in most situations:
$s = "1001BigHairyCamels";
$n = intval($s);
$my_number = str_replace($n, '', $s);
$input = '100Pen';
if (preg_match('~(\d+)[ a-zA-Z]~', $input, $m)) {
echo $m[1];
}
This function will do the job!
<?php
function getInt($str){
preg_match_all('!\d+!', $str, $matches);
return $matches[0][0];
}
$firstStr = '12 Car';
$secondStr = '412 8all';
$thirdStr = '100Pen';
echo 'firstStr = '.getInt($firstStr).'<br>';
echo 'secondStr = '.getInt($secondStr).'<br>';
echo 'thirdStr = '.getInt($thirdStr);
?>
I am making application where I receive a string from user. The string is concatenated with - character between them. First part of string contains alphabetic data whereas later part contains integers or floating point numbers. For example: A string might be 3 Cups Tea-5.99.I want to get the later part of string 5.99 separated by - character. How to do that? I know about PHP substr() function but that takes fixed characters to retrieve substring from. But in this case the later part will not be fixed. For example: 2 Jeans-65.99. In this case I would need last 4 characters meaning that I can't use substr() function.
Anybody with solution?
I know I would need to apply regex but I am completely novice in Regex.
Waiting for your help.
Thanks!
Simply
$result = explode('-', $string)[1];
For PHP<5.4 you'll have to use temporary variable:
$data = explode('-', $string);
$result = $data[1];
Edit
As mentioned in comments, if there is more than 1 part, that will be:
$result = array_pop(explode('-', $string));
$bits = explode('-', $inputstring);
echo $bits[1];
You can use substr() with strpos():
$str = '3 Cups Tea-5.99';
echo substr($str, strpos($str, "-") + 1);
Output:
5.99
Demo!
If data will be like this: "1-Cup tea-2.99", then
$data = "1-Cup tea-2.99";
$data = explode('-', $string);
$result = $data[count($data)-1];
I have a small problem. I am tryng to convert a string like "1 234" to a number:1234
I cant't get there. The string is scraped fro a website. It is possible not to be a space there? Because I've tried methods like str_replace and preg_split for space and nothing. Also (int)$abc takes only the first digit(1).
If anyone has an ideea, I'd be greatefull! Thank you!
This is how I would handle it...
<?php
$string = "Here! is some text, and numbers 12 345, and symbols !£$%^&";
$new_string = preg_replace("/[^0-9]/", "", $string);
echo $new_string // Returns 12345
?>
intval(preg_replace('/[^0-9]/', '', $input))
Scraping websites always requires specific code, you know how you receive the input - and you write code that is required to make it usable.
That is why first answer is still str_replace.
$iInt = (int)str_replace(array(" ", ".", ","), "", $iInt);
$str = "1 234";
$int = intval(str_replace(' ', '', $str)); //1234
I've just came into the same issue, however the answer that was provided wasn't covering all the different cases I had...
So I made this function (the idea popped in my mind thanks to Dan) :
function customCastStringToNumber($stringContainingNumbers, $decimalSeparator = ".", $thousandsSeparator = " "){
$numericValues = $matches = $result = array();
$regExp = null;
$decimalSeparator = preg_quote($decimalSeparator);
$regExp = "/[^0-9$decimalSeparator]/";
preg_match_all("/[0-9]([0-9$thousandsSeparator]*)[0-9]($decimalSeparator)?([0-9]*)/", $stringContainingNumbers, $matches);
if(!empty($matches))
$matches = $matches[0];
foreach($matches as $match):
$numericValues[] = (float)str_replace(",", ".", preg_replace($regExp, "", $match));
endforeach;
$result = $numericValues;
if(count($numericValues) === 1)
$result = $numericValues[0];
return $result;
}
So, basically, this function extracts all the numbers contained inside of a string, no matter how many text there is, identifies the decimal separator and returns every extracted number as a float.
One can specify what decimal separator is used in one's country with the $decimalSeparator parameter.
Use this code for removing any other characters like .,:"'\/, !##$%^&*(), a-z, A-Z :
$string = "This string involves numbers like 12 3435 and 12.356 and other symbols like !## then the output will be just an integer number!";
$output = intval(preg_replace('/[^0-9]/', '', $string));
var_dump($output);
I have strings:
17s 283ms
48s 968ms
The string values are never the same and I want to extract the "second" value from it. In this case, the 17 and the 48.
I'm not very good with regex, so the workaround I did was this:
$str = "17s 283ms";
$split_str = explode(' ', $str);
foreach($split_str as $val){
if(strpos($val, 's') !== false) $sec = intval($val);
}
The problem is, the character 's' exists in both split_str[0] and split_str[1], so my $sec variable keeps obtaining 283, instead of 17.
Again, I'm not very good with regex, and I'm pretty sure regex is the way to go in this case. Please assist. Thanks.
You don't even need to use regex for this.
$seconds = substr($str, 0, strspn($str, '1234567890'));
The above solution will extract all the digits from the beginning of the string. Doesn't matter if the first non-digit character is "s", a space, or anything else.
But why bother?
You can even just cast $str to an int:
$seconds = (int)$str; // equivalent: intval($str)
See it in action.
Regular expressions are definite overkill for such a simple task. Don't use dynamite to drill holes in the wall.
You could do this like so:
preg_match('/(?<seconds>\d+)s\s*(?<milliseconds>\d+)ms/', $var, $matches);
print_r($matches);
If the string will always be formatted in this manner, you could simply use:
<?php
$timeString = '17s 283ms';
$seconds = substr($timeString, 0, strpos($timeString, 's'));
?>
Well, i guess that you can assume seconds always comes before milliseconds. No need for regexp if the format is consistent. This should do it:
$parts = explode(' ', $str);
$seconds = rtrim($parts[0], 's')
echo $seconds; // 17s
This will split the string by space and take the first part 17s. rtrim is then used to remove 's' and you're left with 17.
(\d+s) \d+ms
is the right regexp. Usage would be something like this:
$str = "17s 283ms";
$groups = array();
preg_match("/(\d+)s \d+ms/", $str, $groups);
Then, your number before ms would be $groups[1].