Currently I have an array something like this
[0] => IS-001 開花した才能「篠ノ之 箒」
From this, I would like to extract only the IS-001 part and leave the Japanese character behind to something like this
[0] => 開花した才能「篠ノ之 箒」
Normal preg_split I am using currently only for white space but it seems like having some issue on the 箒」 character to fall into next array. So I decided if only I can split those non Japanese characters out?
Try this
echo preg_replace('/^[a-zA-Z0-9\-_]+/u','','IS-001 開花した才能「篠ノ之 箒」');
^ assert position at start of the string
[a-zA-Z0-9\-_] match a single character present in the list
+ Between one and unlimited times, as many times as possible, giving back as needed
u modifier unicode: Pattern strings are treated as UTF-16.
A solution to this is by using multibyte string functions.
So $char = substr($str, $i, 1); will become $char = mb_substr($str, $i, 1, 'UTF-8'); and strlen($str) will become mb_strlen($str, 'UTF-8').
$str="IS-001 開花した才能「篠ノ之 箒」";
$japanese = preg_replace(array('/[^\p{Han}？]/u', '/(\s)+/'), array('', '$1'), $str);
echo $japanese;
(or)
Remove latin letters and digits from string
$res = preg_replace('/[a-zA-Z0-9-]+/', '', $str);
echo $res;
If your string is the same in all your cases, you can use explode with limit parameter :
$string = 'IS-001 開花した才能「篠ノ之 箒」';
$array = explode(' ', $string, 2);
echo $array[1];
Related
I need to know how to do this reliably with the least amount of calls.
I have a string and I need to remove the 8th character from it. It doesn't matter what the char is, I just need to remove ONLY the 8th char.
I came up with this but a little too unwieldy for me.
// 12345678901234567890
$str = '5NMSG3AB1AH353158';
// after removing char, result should be:
// 5NMSG3A1AH353158
$r = str_split($str);
unset($r[7]);
echo join('', $r);
Possibly Regex?
Here are some solutions:
$str = substr($str, 0, 7) . substr($str, 8);
$str = substr_replace($str, '', 7, 1);
$str = preg_replace('/(.{7})./', '$1', $str, 1);
$str = preg_replace('/.{7}\K./', '', $str, 1);
I'd go for substr() or better substr_replace() as this will certainly be fastest and most easy to read.
substr_replace makes this very simple.
$string = substr_replace($string, '', 7, 1);
It can also take an array of strings as its first parameter, and do the same replacement in all of them, which can be pretty handy.
Like this
$str = '5NMSG3AB1AH353158';
echo preg_replace('/^(.{7})./', '\1', $str);
Output:
5NMSG3A1AH353158
Sandbox
Explanation
^ start of string
(...) capture
. match any
{7} seven times
. match any one time
Then the replacement
\1 first capture group
Basically capture the first 7 chars, then match the 8th, and replace that with the captured 7 chars. Which effectively removes the 8th.
UPDATE
here is another way I like (I haven't used that function sense college, I think):
$s = '5NMSG3A1AH353158';
echo substr_replace($s,'',7,1); //strings are 0 based
sandbox
substr_replace() replaces a copy of string delimited by the start and (optionally) length parameters with the string given in replacement.
mixed substr_replace( mixed $string, mixed $replacement, mixed $start [, mixed $length ] )
http://php.net/manual/en/function.substr-replace.php
Simple.
I have a string:
3 pk. Ready-Dough White Loaves Included $3.99 - 47500 - 00892, 48101
I want to keep only groups of digits longer than 5 characters, and if possible, any dashes or commas between them.
e.g.
47500-00892,48101
My first step was to strip out groups of digits < 4:
preg_replace('/\d{1,4}/', '', $string);
My thinking was "replace any block of digits from 1 to 4 with nothing", but that doesn't do exactly what I thought. Maybe I'm just missing an operator?
Then I was going to strip out all letters and all punctuation except , and -. In my example I would've been left with a starting - because of it being in a string, but a trim() would've been fine to clean that up.
Any help is appreciated!
Had I been patient for 5 more minutes, I would've found the answer: \b
For some reason, working with digits didn't trigger that I needed to use 'word boundaries'.
$string = preg_replace('/\b\d{1,4}\b/', '', $string);
$string = preg_replace('/[^0-9-,]/', '', $string);
$string = trim($string, ',-');
Since there's no reason to perform a replacement, you can use preg_match_all to take what you want and reduce the result array:
$re = '/\d{5,}(?:(?=\s*([-,])\s*\d{5}))?/';
$str = '3 pk. Ready-Dough White Loaves Included $3.99 - 47500 - 00892, 48101';
if ( preg_match_all($re, $str, $matches, PREG_SET_ORDER) ) {
$result = array_reduce($matches, function ($c,$i) { return $c . implode('', $i); });
echo $result;
}
I am trying to something like this.
Hiding users except for first 3 characters.
EX)
apple -> app**
google -> goo***
abc12345 ->abc*****
I am currently using php like this:
$string = "abcd1234";
$regex = '/(?<=^(.{3}))(.*)$/';
$replacement = '*';
$changed = preg_replace($regex,$replacement,$string);
echo $changed;
and the result be like:
abc*
But I want to make a replacement to every single character except for first 3 - like:
abc*****
How should I do?
Don't use regex, use substr_replace:
$var = "abcdef";
$charToKeep = 3;
echo strlen($var) > $charToKeep ? substr_replace($var, str_repeat ( '*' , strlen($var) - $charToKeep), $charToKeep) : $var;
Keep in mind that regex are good for matching patterns in string, but there is a lot of functions already designed for string manipulation.
Will output:
abc***
Try this function. You can specify how much chars should be visible and which character should be used as mask:
$string = "abcd1234";
echo hideCharacters($string, 3, "*");
function hideCharacters($string, $visibleCharactersCount, $mask)
{
if(strlen($string) < $visibleCharactersCount)
return $string;
$part = substr($string, 0, $visibleCharactersCount);
return str_pad($part, strlen($string), $mask, STR_PAD_RIGHT);
}
Output:
abc*****
Your regex matches all symbols after the first 3, thus, you replace them with a one hard-coded *.
You can use
'~(^.{3}|(?!^)\G)\K.~'
And replace with *. See the regex demo
This regex matches the first 3 characters (with ^.{3}) or the end of the previous successful match or start of the string (with (?!^)\G), and then omits the characters matched from the match value (with \K) and matches any character but a newline with ..
See IDEONE demo
$re = '~(^.{3}|(?!^)\G)\K.~';
$strs = array("aa","apple", "google", "abc12345", "asdddd");
foreach ($strs as $s) {
$result = preg_replace($re, "*", $s);
echo $result . PHP_EOL;
}
Another possible solution is to concatenate the first three characters with a string of * repeated the correct number of times:
$text = substr($string, 0, 3).str_repeat('*', max(0, strlen($string) - 3));
The usage of max() is needed to avoid str_repeat() issue a warning when it receives a negative argument. This situation happens when the length of $string is less than 3.
I'm having trouble knowing how many times that the - character is in front of my string.
Some examples:
$string = "-Lorem Ipsum"; // 1
$string = "--Lorem Ipsum"; // 2
$string = "---Lorem Ipsum"; // 3
$string = "--Lorem-Ipsum"; // 2
But how can I find this? I know you can search the number of occurrences of a character in a string. But I want the number of - characters before an alphabet letter. Not all the sequences (see last example).
How should I approach this?
You can use the old school trick of using a string as an array here as such:
$search="-";
$i=0;
while($string[$i]==$search)
{
$i++;
}
echo "Found $i instances at the start of the string.";
What about using ltrim() and strlen()
echo strlen($string) - strlen(ltrim($string, "-"));
See example at eval.in
it would also work -
preg_match('/(?!-)/', $string, $match, PREG_OFFSET_CAPTURE);
$match - the position of any character but - which is indeed the count of -.
I have a string Trade Card Catalogue 1988 Edition I wish to remove everything apart from 1988.
I could have an array of all letters and do a str_replace and trim, but I wondered if this was a better solution?
$string = 'Trade Card Catalogue 1988 Edition';
$letters = array('a','b','c'....'x','y','z');
$string = str_to_lower($string);
$string = str_replace($letters, '', $string);
$string = trim($string);
Thanks in advance
Regular expression?
So assuming you want the number (and not the 4th word or something like that):
$str = preg_replace('#\D#', '', $str);
\D means every character that is not a digit. The same as [^0-9].
If there could be more numbers but you only want to get a four digit number (a year), this will also work (but obviously fails if you there are several four digit numbers and you want to get a specific one) :
$str = preg_replace('#.*?(\d{4,4}).*#', '\1', $str);
You can actually just pass the entire set of characters to be trimmed as a parameter to trim:
$string = trim($string, 'abc...zABC...Z ' /* don't forget the space */);