i have the sentence
something about something WORD still something...
what is the most efficient metod to delete the word "WORD" from sentence in php?
thanks
You could replace it with nothing:
$sentence = str_replace('word', '', $sentence);
Although that would also ruin words like swordfish, turning them into sfish. So you could put spaces around the edges:
$sentence = str_replace(' word ', ' ', $sentence);
But then it won't match words at the end and beginning of sentences. So you might have to use a regex:
$sentence = preg_replace('/\bword\b/', '', $sentence);
The \b is a word boundary, which could be a space or a beginning of a string or anything like that.
Depends, str_replace might be what you're looking for. But note that it removes all occurrences.
Try this:
$fixed_string = str_replace(" WORD ", " ", $your_string);
Related
I have a load of labels which are camel case. Some examples are
whatData
whoData
deliveryDate
importantQuestions
What I am trying to do is this. Any label which has the word Data needs to have this word removed. At the point of the capital letter, I need to provide a space. Finally, everything should be uppercase. I have done the removal of Data and the uppercase by doing this ($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.
$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
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.
I am working with some code in PHP that grabs the referrer data from a search engine, giving me the query that the user entered.
I would then like to remove certain stop words from that string if they exist. However, the word may or may not have a space at either end.
For example, I have been using str_replace to remove a word as follows:
$keywords = str_replace("for", "", $keywords);
$keywords = str_replace("sale", "", $keywords);
but if the $keywords value is "baby formula" it will change it to "baby mula" - removing the "for" part.
Without having to create further str_replace's that account for " for" and "for " - is there a preg_replace type command I could use that would remove the given word if it is found with a space at either end?
My idea would be to put all of the stop words into an array and step through them that way and I suspect that a preg_replace is going to be quicker than stepping through multiple str_replace lines.
UPDATE:
Solved thanks to you guys using the following combination:
$keywords = "...";
$stopwords = array("for","each");
foreach($stopwords as $stopWord)
{
$keywords = preg_replace("/(\b)$stopWord(\b)/", "", $keywords);
}
$keywords = "...";
$stopWords = array("for","sale");
foreach($stopWords as $stopWord){
$keywords = preg_replace("/(\b)$stopWord(\b)/", "", $keywords);
}
Try it this way
$keywords = preg_replace( '/(?!\w)(for|sale)(?>!\w)/', '', $keywords );
You can use word boundaries for this
$keywords = preg_replace('/\bfor\b/', '', $keywords);
or with multiple words
$keywords = preg_replace('/\b(?:for|sale)\b/', '', $keywords);
While Armel's answer will work, it is not performing optimally. Yes, your desired output will require wordboundaries and probably case-insensitive matching, but:
Wordboundaries gain nothing from being wrapped in parentheses.
Performing iterated preg_match() calls for each element in the blacklist array is not efficient. Doing so will ask the regex engine to perform wave after wave of individual keyword checks on the full string.
I recommend building a single regex pattern that will check for all keywords during each step of traversing the string -- one time. To generate the single pattern dynamically, you only need to implode your blacklist array of elements with | (pipes) which represent the "OR" command in regex. By wrapping all of the pipe-delimited keywords in a non-capturing group ((?:...)), the wordboundaries (\b) serve their purpose for all keywords in the blacklist array.
Code: (Demo)
$string = "Each person wants peaches for themselves forever";
$blacklist = array("for", "each");
// if you might have non-letter characters that have special meaning to the regex engine
//$blacklist = array_map(function($v){return preg_quote($v, '/');}, $blacklist);
//print_r($blacklist);
echo "Without wordboundaries:\n";
var_export(preg_replace('/' . implode('|', $blacklist) . '/i', '', $string));
echo "\n\n---\n";
echo "With wordboundaries:\n";
var_export(preg_replace('/\b(?:' . implode('|', $blacklist) . ')\b/i', '', $string));
echo "\n\n---\n";
echo "With wordboundaries and consecutive space mop up:\n";
var_export(trim(preg_replace(array('/\b(?:' . implode('|', $blacklist) . ')\b/i', '/ \K +/'), '', $string)));
Output:
Without wordboundaries:
' person wants pes themselves ever'
---
With wordboundaries:
' person wants peaches themselves forever'
---
With wordboundaries and consecutive space mop up:
'person wants peaches themselves forever'
p.s. / \K +/ is the second pattern fed to preg_replace() which means the input string will be read a second time to search for 2 or more consecutive spaces. \K means "restart the fullstring match from here"; effectively it releases the previously matched space. Then one or more spaces to follow are matched and replaced with an empty string.
I'm need to insert space before new-line-symbol if it's not yet. For ex, in the text like this:
Some text
need space dfgfgfkgkf
insert space between 'text' and 'need'..
$pattern = "/[^\s]\n/i";
$replace = " \n";
but this regexp don't work
Are you sure that your newline character is \n? I would go for a more general solution and use
echo preg_replace('/(?<=\S)(?=\r?\n)/',' ', $str);
I use a positive lookbehind and a positive lookahead assertion to find the position at the end of the row, therefor only a space as replacement is needed.
Try this:
$pattern = "/\n/";
$replace = " \n";
I tried the code below and it worked:
$str = "line1\nline2";
echo preg_replace('/\n/',' \n', $str);