i have the bellow string
$LINE = TCNU1573105 HDPE HTA108 155 155 000893520918 PAL990 25.2750 MT 28.9750 MT
and i want extract the PAL990 from the above string. actually extract PAL990 string or any string that has PAL followed by some digits Like PAL222 or PAL123
i tried many ways and could not get the result. i used,
substr ( $LINE, 77, 3)
but when the value in different position i get the wrong value.
You may use
$LINE = "TCNU1573105 HDPE HTA108 155 155 000893520918 PAL990 25.2750 MT 28.9750 MT";
if (preg_match('~\bPAL\d+\b~', $LINE, $res)) {
echo $res[0]; // => PAL990
}
See the PHP demo and this regex demo.
Details
\b - a word boundary
PAL - a PAL substring
\d+ - 1+ digits
\b - a word boundary.
The preg_match function will return the first match.
Note that in case your string contains similar strings in between hyphens/whitespace you will no longer be able to rely on word boundaries, use custom whitespace boundaries then, i.e.:
'~(?<!\S)PAL\d+(?!\S)~'
See this regex demo
EDIT
If you may have an optional whitespace between PAL and digits, you may use
preg_replace('~.*\b(PAL)\s?(\d+)\b.*~s', '$1$2', $LINE)
See this PHP demo and this regex demo.
Or, match the string you need with spaces, and then remove them:
if (preg_match('~\bPAL ?\d+\b~', $LINE, $res)) {
echo str_replace(" ", "", $res[0]);
}
See yet another PHP demo
Note that ? makes the preceding pattern optional (1 or 0 occurrences are matched).
$string = "123ABC1234 *$%^&abc.";
$newstr = preg_replace('/[^a-zA-Z\']/','',$string);
echo $newstr;
Output:ABCabc
Related
I'm using the below code to display only the first letter of each word in my string. For example, "Hello World!" would be displayed as "H W". However, I want to also include punctuation like this: "H W!"
How can I modify my code so punctuation is preserved?
$editversetext = preg_split("/[\s,_-]+/", $editversetext);
$initials = "";
foreach ($editversetext as $w) {
$initials .= $w[0];
}
$initials = implode(' ',str_split($initials));
echo $initials . ".";
You may use the following regex to match what you need:
'~\b(\p{L})\p{L}*(\p{P}?)~u'
See the regex demo.
Details
\b - a word boundary
(\p{L}) - Capturing group #1: a letter
\p{L}* - 0+ letters
(\p{P}?) - Capturing group #2: an optional punctuation (NOTE: if you also want to match symbols, replace \p{P} with [\p{P}\p{S}])
u - the "Unicode" modifier that enables PCRE_UTF and PCRE_UCP verbs to fully enable Unicode support.
Depending on the input you have, you may either use a replacing approach, or you may collect the matches and then combine them into the result you need in a similar way you are doing it now.
See the PHP demo:
$str = 'Hello World!';
// Replacing approach (if all words are matches):
echo preg_replace('~\b(\p{L})\p{L}*(\p{P}?)~u', '$1$2', $str) . "\n"; // => H W!
// Collecting/post-processing (if there are non-matching sequences)
$res = [];
preg_replace_callback('~\b(\p{L})\p{L}*(\p{P}?)~u', function($m) use (&$res) {
$res[] = $m[1].$m[2];
return '';
}, $str);
print_r(implode(" ", $res)); // => H W!
To match and remove all word characters that are not the first one, use \B a non word boundary.
$str = preg_replace('/\B\w+/', "", $str);
See regex demo or PHP demo
Be aware that digitis belong to \w. Use [A-Za-z] or unicode \pL with u flag instead if desired.
I have a string:
access":"YOU HAVE 0 BALANCE","machine
How can I extract the string in between the double quotes and have only the text (without the double quotes):
YOU HAVE 0 BALANCE
I have tried
if(preg_match("'access":"(.*?)","machine'", $tok2, $matches)){
but with no luck :( .
You may use
'/access":"(.*?)","machine/'
See the regex demo. The value you need is in Group 1.
Details
access":" - a literal substring
(.*?) - Group 1: any 0+ chars other than line break chars, as few as possible, since *? is a lazy quantifier
","machine - a literal substring
See the PHP online demo:
$re = '/access":"(.*?)","machine/';
$str = 'access":"YOU HAVE 0 BALANCE","machine';
if (preg_match($re, $str, $matches)) {
print_r($matches[1]);
}
// => YOU HAVE 0 BALANCE
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 as
This is a sample text. This text will be used as a dummy for "various" RegEx "operations" using PHP.
I want to select and replace all the first alphabet of each word (in the example : T,i,a,s,t,T,t,w,b,u,a,d,f,",R,",u,P). How do I do it?
I tried /\b.{1}\w+\b/. I read the expression as "select any character that has length of 1 followed by word of any length" but didn't work.
You may try this regex as well:
(?<=\s|^)([a-zA-Z"])
Demo
Your regex - /\b.{1}\w+\b/ - matches any string that is not enclosed in word characters, starts with any symbol that is in a position after a word boundary (thus, it can even be whitespace if there is a letter/digit/underscore in front of it), followed with 1 or more alphanumeric symbols (\w) up to the word boundary.
That \b. is the culprit here.
If you plan to match any non-whitespace preceded with a whitespace, you can just use
/(?<!\S)\S/
Or
/(?<=^|\s)\S/
See demo
Then, replace with any symbol you need.
You may try to use the following regex:
(.)[^\s]*\s?
Using the preg_match_all and implode the output result group 1
<?php
$string = 'This is a sample text. This text will be used as a dummy for'
. '"various" RegEx "operations" using PHP.';
$pattern = '/(.)[^\s]*\s?/';
$matches;
preg_match_all($pattern, $string, $matches);
$output = implode('', $matches[1]);
echo $output; //Output is TiastTtwbuaadf"R"uP
For replace use something like preg_replace_callback like:
$pattern = '/(.)([^\s]*\s?)/';
$output2 = preg_replace_callback($pattern,
function($match) { return '_' . $match[2]; }, $string);
//result: _his _s _ _ample _ext. _his _ext _ill _e _sed _s _ _ummy _or _various" _egEx _operations" _sing _HP.
I'm trying to remove all words of less than 3 characters from a string, specifically with RegEx.
The following doesn't work because it is looking for double spaces. I suppose I could convert all spaces to double spaces beforehand and then convert them back after, but that doesn't seem very efficient. Any ideas?
$text='an of and then some an ee halved or or whenever';
$text=preg_replace('# [a-z]{1,2} #',' ',' '.$text.' ');
echo trim($text);
Removing the Short Words
You can use this:
$replaced = preg_replace('~\b[a-z]{1,2}\b\~', '', $yourstring);
In the demo, see the substitutions at the bottom.
Explanation
\b is a word boundary that matches a position where one side is a letter, and the other side is not a letter (for instance a space character, or the beginning of the string)
[a-z]{1,2} matches one or two letters
\b another word boundary
Replace with the empty string.
Option 2: Also Remove Trailing Spaces
If you also want to remove the spaces after the words, we can add \s* at the end of the regex:
$replaced = preg_replace('~\b[a-z]{1,2}\b\s*~', '', $yourstring);
Reference
Word Boundaries
You can use the word boundary tag: \b:
Replace: \b[a-z]{1,2}\b with ''
Use this
preg_replace('/(\b.{1,2}\s)/','',$your_string);
As some solutions worked here, they had a problem with my language's "multichar characters", such as "ch". A simple explode and implode worked for me.
$maxWordLength = 3;
$string = "my super string";
$exploded = explode(" ", $string);
foreach($exploded as $key => $word) {
if(mb_strlen($word) < $maxWordLength) unset($exploded[$key]);
}
$string = implode(" ", $exploded);
echo $string;
// outputs "super string"
To me, it seems that this hack works fine with most PHP versions:
$string2 = preg_replace("/~\b[a-zA-Z0-9]{1,2}\b\~/i", "", trim($string1));
Where [a-zA-Z0-9] are the accepted Char/Number range.