I have a string formed up by numbers and sometimes by letters.
Example AF-1234 or 345ww.
I have to get the numeric part and increment it by one.
how can I do that? maybe with regex?
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5 Answers
You can use preg_replace_callback as:
function inc($matches) {
return ++$matches[1];
}
$input = preg_replace_callback("|(\d+)|", "inc", $input);
Basically you match the numeric part of the string using the regex \d+ and replace it with the value returned by the callback function which returns the incremented value.
Alternatively this can be done using preg_replace() with the e modifier as:
$input = preg_replace("|(\d+)|e", "$1+1", $input);
7 Comments
tampe125
what if i have string like "XXX-342"? it will be in $matches[0] or $matches[1]?
zerkms
@codaddict: Don't you think that callback is a "little" overhead there? Why don't use regular
+ 1?zerkms
@tampe125: It will be in both [0] and [1]
codaddict
@zerkms: You are right, this can be easily done using
preg_replace with e modifier. Please see my updated answer.
Jan Richter
According to PHP documentation
e modifier has been removed in PHP 7, see php.net/manual/en/function.preg-replace.php |
If the string ends with numeric characters it is this simple...
$str = 'AF-1234';
echo $str++; //AF-1235
That works the same way with '345ww' though the result may not be what you expect.
$str = '345ww';
echo $str++; //345wx
@tampe125
This example is probably the best method for your needs if incrementing string that end with numbers.
$str = 'XXX-342';
echo $str++; //XXX-343
Comments
Here is an example that worked for me by doing a pre increment on the value
$admNo = HF0001;
$newAdmNo = ++$admNo;
The above code will output HF0002
1 Comment
gre_gor
This also changes
$admNo and shows a warning. Also doesn't work with "345ww".If you are dealing with strings that have multiple number parts then it's not so easy to solve with regex, since you might have numbers overflowing from one numeric part to another.
For example if you have a number INV00-10-99 which should increment to INV00-11-00.
I ended up with the following:
for ($i = strlen($string) - 1; $i >= 0; $i--) {
if (is_numeric($string[$i])) {
$most_significant_number = $i;
if ($string[$i] < 9) {
$string[$i] = $string[$i] + 1;
break;
}
// The number was a 9, set it to zero and continue.
$string[$i] = 0;
}
}
// If the most significant number was set to a zero it has overflowed so we
// need to prefix it with a '1'.
if ($string[$most_significant_number] === '0') {
$string = substr_replace($string, '1', $most_significant_number, 0);
}
Comments
This is similar to the answer above, but contains the code inline and does a full check for the last character.
function replace_title($title) {
$pattern = '/(\d+)(?!.*\d)+/';
return preg_replace_callback($pattern, function($m) { return ++$m[0]; }, $title);
}
echo replace_title('test 123'); // test 124
echo replace_title('test 12 3'); // test 12 4
echo replace_title('test 123 - 2'); // test 123 - 3
echo replace_title('test 123 - 3 - 5'); // test 123 - 3 - 6
echo replace_title('123test'); // 124test
