Preg_replace exact strings only - php

$string = "recruitment offer human resource IT for before of";
$string = str_replace(array('it', 'for', 'of'), '-', $string);
I want to remove some unnecessary words from the string (in this example - I want to replace it, for, and of with -) but I don't want others words to be affected (the above example will also affect the words recuITment, OFfer and beFORe
Result : recruitment offer human resource - - before -
Note : I need a solution that does not limit only to these words / string.

Use preg_replace() with \b, the word boundary assertion:
$string = preg_replace( '#\b(it|for|of)\b#i', '-', $string);

It's better to use lookarounds instead of word boundaries. Because \b(it|for|of)\b would match it in :it: string. I think this is not you want.
$string = preg_replace( '#(?<=\s|^)(?:it|for|of)(?=\s|$)#i', '-', $string);
DEMO

Related

Adding space to string based on condition

I have a load of labels which are camel case. Some examples are
whatData
whoData
deliveryDate
importantQuestions
What I am trying to do is this. Any label which has the word Data needs to have this word removed. At the point of the capital letter, I need to provide a space. Finally, everything should be uppercase. I have done the removal of Data and the uppercase by doing this ($data->key is the label)
strtoupper(str_replace('Data', '', $data->key))
The part I am struggling with is adding the spaces between words. So basically the above words should end up like this
WHAT
WHO
DELIVERY DATE
IMPORTANT QUESTIONS
How can I factor in the last part of this?
Thanks
It will add spaces before every capital letters. Try this:
$String = 'whatData';
$Words = preg_replace('/(?<!\ )[A-Z]/', ' $0', $String);
Problem
Your regex '~^[A-Z]~' will match only the first capital letter. Check out Meta Characters in the Pattern Syntax for more information.
Your replacement is a newline character '\n' and not a space.
Solution
Use preg_replace(). Try below code.
$string = "whatData";
echo preg_replace('/(?<!\ )[A-Z]/', ' $0', $string);
Output
what Data
Try following:
$string = 'importantQuestions';
$string = strtoupper(ltrim(preg_replace('/[A-Z]/', ' $0', $string)));
echo $string;
This will give you output as:
IMPORTANT QUESTIONS
Try this:
preg_split: split on camel case
array_map: UPPER case all the element
implode: Implode the array
str_replace: Replace the `DATE` with empty
trim: trim the white spaces.
Do this simple things:
echo trim(str_replace("DATE", "", implode(" ", array_map("strtoupper", preg_split('/(?=[A-Z])/', 'deliveryDate', -1, PREG_SPLIT_NO_EMPTY))))); // DELIVERY
This is result exactly what you want.

PHP trim and return a string from its right

I'm trying to take a string that's output from MySql like this (MySql outputs X characters):
$str = 'Buddy you're a boy make a big noise Playin in the stre';
and trying to start from the right side, trim whatever is there up till the first space. Sounded simple when I got down to it, but now, it has my brain and fingers in knots.
The output I'm tying to achieve is simple:
$str = 'Buddy you're a boy make a big noise Playin in the';
Notice, that characters starting from the right, till the first space, are removed.
Can you help?
My Fiddle
$str = 'Buddy you\'re a boy make a big noise Playin in the stre';
//echo rtrim($str,' ');
It's a useful idiom to remember on its own: to remove all the characters preceding a specific one from the right side of the string (including that special character), use the following:
$trimmed = substr($str, 0, strrpos($str, ' '));
... where ' ' is that special character.
Demo
If you don't know, however, whether or not the character is present, you'd check the result of sttrrpos first:
$last_space_index = strrpos($str, ' ');
$trimmed = $last_space_index !== false
? substr($str, 0, $last_space_index)
: $str;
And if there can be more than one character that you need to trim, like in 'hello there test' line, just rtrim the result:
$trimmed = rtrim(substr($str, 0, strrpos($str, ' ')), ' ');
In this case, however, a regex-based solution looks more appropriate:
$trimmed = preg_replace('/ +[^ ]*$/', '', $str);
I think your best option would be a regex replace:
preg_replace('/\s+\S*$/', '', $str);
which outputs Buddy you're a boy make a big noise Playin in the
And the Fiddle
it's probably easier to do it with regex, but I'm sooo bad with that! You shoud try this:
// Get all the words in an array
$strArray = explode(" ", $str);
// Remove the last word.
array_pop($strArray);
// Get it back into a sentence
$newString = implode(" ", $strArray);
There's a hundred ways to do this, here are some options:
array_pop'ing the last word off an array we create from explode:
$arr = explode(" ", $str);
$fixed_arr = array_pop($arr);
$result = implode(" ", $arr);
Using regular expressions:
$result = preg_replace('/\s+\S*$/', '', $str);
and using strrpos and substr:
$spacePos = strrpos($str, ' ');
$result = substr($str, 0, $spacePos);
In mysql use
left(field,length)
to output only the strlen first digits
right(field,length) having opposite effects
otherwise use substr($string,0,$length) or regex in php
As a matter of regex performance comparison, the regex engine can move faster through the string when it can perform greedy matching with minimal backtracking.
/ +[^ ]*$/ uses 68 steps. (#raina77ow)
/(?:[^ ]+\K )+.*/ uses 56 steps. (#mickmackusa)
/(?:\K [^ ]*)+/ uses 48 steps. (#mickmackusa)
\s+\S*$ uses 34 steps. (#ChrisBornhoft and #RyanKempt)
/.*\K .*/ uses just 15 steps. (#mickmackusa)
Based on these comparisons, I recommend greedily matching any characters, then restarting the fullstring match before matching the last occurring space, then matching zero or more characters until the end of the string.
Code: (Demo)
$string = "Buddy you're a boy make a big noise Playin in the stre";
var_export(
preg_replace('/.*\K .*/', '', $string)
);
Output:
'Buddy you\'re a boy make a big noise Playin in the'

PHP Regex: Remove words less than 3 characters

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.

regexp add space after comma but not when comma is thousands separator?

Using php regexp in a simple way, is it possible to modify a string to add a space after commas and periods that follow words but not after a comma or period that is preceded and followed by a number such as 1,000.00?
String,looks like this with an amount of 1,000.00
Needs to be changed to...
String, looks like this with an amount of 1,000.00
This should allow for multiple instances of course... Here is what I am using now but it is causing numbers to return as 1, 000. 00
$punctuation = ',.;:';
$string = preg_replace('/(['.$punctuation.'])[\s]*/', '\1 ', $string);
You could replace '/(?<!\d),|,(?!\d{3})/' with ', '.
Something like:
$str = preg_replace('/(?<!\d),|,(?!\d{3})/', ', ', $str);
I was searching for this regex.
This post really help me, and I improve solution proposed by Qtax.
Here is mine:
$ponctuations = array(','=>', ','\.'=>'. ',';'=>'; ',':'=>': ');
foreach($ponctuations as $ponctuation => $replace){
$string = preg_replace('/(?<!\d)'.$ponctuation.'(?!\s)|'.$ponctuation.'(?!(\d|\s))/', $replace, $string);
}
With this solution, "sentence like: this" will not be changed to "sentence like: this" (whith 2 blanck spaces)
That's all.
Although this is quite old, I was looking for this same question, and after understanding solutions given I have a different answer.
Instead of checking for character before the comma this regex checks the character after the comma and so it can be limited to alphabetic ones. Also this will not create a string with two spaces after a comma.
$punctuation = ',.;:';
$string = preg_replace("/([$punctuation])([a-z])/i",'\1 \2', $string);
Test script can be checked here.

Replace all characters in string apart from PHP

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 */);

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