I have a strings like this.
$str = "-=!#?Bob-Green_Smith";
$str = "-_#!?1241482";
How can I explode them at the first alphanumeric match.
eg:
$str = "-=!#?Bob-Green_Smith";
becomes:
$val[0] = "-=!#?";
$val[1] = "Bob-Green_Smith";
Quick thought some times the string won't contain the initial string of characters,
so I'd need to check if the first character is alphanumeric or not.. otherwise Bob-Green_Smith would get split when he shouldn't.
Thanks
You can use preg_match.
This will match "non word characters" zero or more as first group.
Then the rest as the second.
The output will have three items, the first is the full string, so I use array_shift to remove it.
$str = "-=!#?Bob-Green_Smith";
Preg_match("/(\W*)(.*)/", $str, $val);
Array_shift($val); // remove first item
Var_dump($val);
https://3v4l.org/m2MCg
You can do this like :
$str = "-=!#?1Bob-Green_Smith";
preg_match('~[a-z0-9]~i', $str, $match, PREG_OFFSET_CAPTURE);
echo $bubString = substr($str, $match[0][1]);
Related
I have an array of words and a string and want to add a hashtag to the words in the string that they have a match inside the array. I use this loop to find and replace the words:
foreach($testArray as $tag){
$str = preg_replace("~\b".$tag."~i","#\$0",$str);
}
Problem: lets say I have the word "is" and "isolate" in my array. I will get ##isolate at the output. this means that the word "isolate" is found once for "is" and once for "isolate". And the pattern ignores the fact that "#isoldated" is not starting with "is" anymore and it starts with "#".
I bring an example BUT this is only an example and I don't want to just solve this one but every other possiblity:
$str = "this is isolated is an example of this and that";
$testArray = array('is','isolated','somethingElse');
Output will be:
this #is ##isolated #is an example of this and that
You may build a regex with an alternation group enclosed with word boundaries on both ends and replace all the matches in one pass:
$str = "this is isolated is an example of this and that";
$testArray = array('is','isolated','somethingElse');
echo preg_replace('~\b(?:' . implode('|', $testArray) . ')\b~i', '#$0', $str);
// => this #is #isolated #is an example of this and that
See the PHP demo.
The regex will look like
~\b(?:is|isolated|somethingElse)\b~
See its online demo.
If you want to make your approach work, you might add a negative lookbehind after \b: "~\b(?<!#)".$tag."~i","#\$0". The lookbehind will fail all matches that are preceded with #. See this PHP demo.
A way to do that is to split your string by words and to build a associative array with your original array of words (to avoid the use of in_array):
$str = "this is isolated is an example of this and that";
$testArray = array('is','isolated','somethingElse');
$hash = array_flip(array_map('strtolower', $testArray));
$parts = preg_split('~\b~', $str);
for ($i=1; $i<count($parts); $i+=2) {
$low = strtolower($parts[$i]);
if (isset($hash[$low])) $parts[$i-1] .= '#';
}
$result = implode('', $parts);
echo $result;
This way, your string is processed only once, whatever the number of words in your array.
How can i remove part of string from example:
##lang_eng_begin##test##lang_eng_end##
##lang_fr_begin##school##lang_fr_end##
##lang_esp_begin##test33##lang_esp_end##
I always want to pull middle of string: test, school, test33. from this string.
I Read about ltrim, substr and other but I had no good ideas how to do this. Becouse each of strings can have other length for example :
'eng', 'fr'
I just want have string from middle between ## and ##. to Maye someone can help me? I tried:
foreach ($article as $art) {
$title = $art->titl = str_replace("##lang_eng_begin##", "", $art->title);
$art->cleanTitle = str_replace("##lang_eng_end##", "", $title);
}
But there
##lang_eng_end##
can be changed to
##lang_ger_end##
in next row so i ahvent idea how to fix that
If your strings are always in this format, an explode way looks easy:
$str = "##lang_eng_begin##test##lang_eng_end## ";
$res = explode("##", $str)[2];
echo $res;
You may use a regex and extract the value in between the non-starting ## and next ##:
$re = "/(?!^)##(.*?)##/";
$str = "##lang_eng_begin##test##lang_eng_end## ";
preg_match($re, $str, $match);
print_r($match[1]);
See the PHP demo. Here, the regex matches a ## that is not at the string start ((?!^)##), then captures into Group 1 any 0+ chars other than newline as few as possible ((.*?)) up to the first ## substring.
Or, replace all ##...## substrings with `preg_replace:
$re = "/##.*?##/";
$str = "##lang_eng_begin##test##lang_eng_end## ";
echo preg_replace($re, "", $str);
See another demo. Here, we just remove all non-overlapping substrings beginning with ##, then having any 0+ chars other than a newline up to the first ##.
I have a question. I need to add a + before every word and see all between quotes as one word.
A have this code
preg_replace("/\w+/", '+\0', $string);
which results in this
+test +demo "+bla +bla2"
But I need
+test +demo +"bla bla2"
Can someone help me :)
And is it possible to not add a + if there is already one? So you don't get ++test
Thanks!
Maybe you can use this regex:
$string = '+test demo between "double quotes" and between \'single quotes\' test';
$result = preg_replace('/\b(?<!\+)\w+|["|\'].+?["|\']/', '+$0', $string);
var_dump($result);
// which will result in:
string '+test +demo +between +"double quotes" +and +between +'single quotes' +test' (length=74)
I've used a 'negative lookbehind' to check for the '+'.
Regex lookahead, lookbehind and atomic groups
I can't test this but could you try it and let me know how it goes?
First the regex: choose from either, a series of letters which may or may not be preceded by a '+', or, a quotation, followed by any number of letters or spaces, which may be preceded by a '+' followed by a quotation.
I would hope this matches all your examples.
We then get all the matches of the regex in your string, store them in the variable "$matches" which is an array. We then loop through this array testing if there is a '+' as the first character. If there is, do nothing, otherwise add one.
We then implode the array into a string, separating the elements by a space.
Note: I believe $matches in created when given as a parameter to preg_match.
$regex = '/[((\+)?[a-zA-z]+)(\"(\+)?[a-zA-Z ]+\")]/';
preg_match($regex, $string, $matches);
foreach($matches as $match)
{
if(substr($match, 0, 1) != "+") $match = "+" + $match;
}
$result = implode($matches, " ");
I have a string that contains many underscores followed by words ex: "Field_4_txtbox" I need to find the last underscore in the string and remove everything following it(including the "_"), so it would return to me "Field_4" but I need this to work for different length ending strings. So I can't just trim a fixed length.
I know I can do an If statement that checks for certain endings like
if(strstr($key,'chkbox')) {
$string= rtrim($key, '_chkbox');
}
but I would like to do this in one go with a regex pattern, how can I accomplish this?
The matching regex would be:
/_[^_]*$/
Just replace that with '':
preg_replace( '/_[^_]*$/', '', your_string );
There is no need to use an extremly costly regex, a simple strrpos() would do the job:
$string=substr($key,0,strrpos($key,"_"));
strrpos — Find the position of the last occurrence of a substring in a string
You can also just use explode():
$string = 'Field_4_txtbox';
$temp = explode('_', strrev($string), 2);
$string = strrev($temp[1]);
echo $string;
As of PHP 5.4+
$string = 'Field_4_txtbox';
$string = strrev(explode('_', strrev($string), 2)[1]);
echo $string;
I have an alpha numeric string say for example,
abc123bcd , bdfnd567, dfd89ds.
I want to trim all the characters before the first appearance of any integer in the string.
My result should look like,
abc , bdfnd, dfd.
I am thinking of using substr. But not sure how to check for a string before first appearance of an integer.
You can easily remove the characters you don't want with preg_replace [docs] and a regular expression:
$str = preg_replace('#\d.*$#', '', $str);
\d matches a digit and .*$ matches any character until the end of the string.
Learn more about regular expressions: http://www.regular-expressions.info/.
DEMO
A possible non-Regex solution would be:
strcspn — Find length of initial segment not matching mask
substr — Return part of a string
Example:
$string = 'foo1bar';
echo substr($string, 0, strcspn($string, '1234567890')); // gives foo
$string = 'abc123bcd';
preg_replace("/[0-9]/", "", $string);
or
trim($string, '0123456789');
I believe you are looking for this?
$matches = array();
preg_match("/^[a-z]+/", "dfd89ds", $matches);
echo $matches[0]; // returns dfd
You can use a regex for this:
$string = 'abc123bcd';
preg_match('/^[a-zA-Z]*/i', $string, $matches);
var_dump($matches[0]);
will produce:
abc
To remove the +/- sign, you can simply use:
abs($number)
and get the absolute value.
e.g
$abs = abs($signed_integer);