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;
}
Related
I am running into a problem trying to do a replacement on a few strings. Essentially what I have is a bunch of prices on my page that look like
RMB148.00
What i am trying to do is run a replace on only the last 2 numbers so i can do something like
RMB14800
Preg replace works fine for the RMB part because it is always there.
My problem is the last two numbers can be anything it all depends on the price so I cant just remove and replace, I need to just wrap HTML <sup> tags around them.
$string = $product['price'];
$string = preg_replace('/[\x00-\x1F\x80-\xFF]/', '', $string);
echo preg_replace('/RMB/', '<sup class="currency-sym">RMB</sup>', $string, 1);
Assuming the last two characters are digits, you could just
$string=preg_replace('/(\d\d)$/', '<sup class="currency-sym">\1</sup>', $string);
If not,
$string=preg_replace('/(..)$/', '<sup class="currency-sym">\1</sup>', $string);
should do the trick.
Alternativly use
$string=substr($string,0,-2).'<sup class="currency-sym">'.substr($string,-2).'</sup>';
Here is a regex solution that looks for the final digit notation at the end of your string.
$string = 'RMB148.00';
$string = preg_replace('/(\d+)\.(\d{2})\z/','$1<sup>$2</sup>',$string);
echo $string;
You could use the following with the explode () function
$string = explode ('.', $product['price']);
$new_string = $string[0].'<sup>'. $string [1]. '</sup>';
And do the regex for the RMB the same way.
Code.
<?php
$string = '14842.00';
$string = substr($string, 0, strlen($string) - 2) . '<sup>' . substr($string, strlen($string) - 2, 2) . '</sup>';
echo $string;
Try online sandbox.
Explanation.
substr($s, $i, $l) gets $l symbols of $s, started from $i index (indexes starts from zero).
So first substr($string, 0, strlen($string) - 2) gets all string except last two symbols.
Second substr($string, strlen($string) - 2, 2) gets only last two symbols.
More about substr.
You should use a pattern matching regex. Note the $1 in the replacement argument matches (\d{2}) in the pattern argument. preg_replace() only replaces the matched pattern. This pattern matches . followed by any two digits. Since . is not included in the replacement argument it does not show up in your $string.
$string = preg_replace('/\.(\d{2})$/', '<sup>$1</sup>', $string);
Of course, you could use one preg_replace to do what you want:
$string = preg_replace('/^(RMB)(\d+)(\.(\d{2}))?$/', "<sup class='currency-sym'>$1</sup>$2<sup>$4</sup>", $string);
The second example may be a good idea if you want DOM integrity, otherwise it creates an empty <sup></sup> when there is no decimal.
I'm trying to find and replace all words that ends with 'ing'. How would I do that?
$text = "dreaming";
if (strlen($text) >= 6) {
if (0 === strpos($text, "ing"))
//replace the last 3 characters of $text <---not sure how to do this either
echo $text;
echo "true";
}
Result:
null
Want Result:
dream
true
You could also use substr
$text = "dreaming";
if (substr($text, (strlen($text) - 3), 3) === 'ing') {
$text = substr($text, 0, (strlen($text) - 3));
}
echo $text;
This should work for replacing ing at the end of words whilst ignoring stuff starting with Ing as well as words with ing in the middle of them.
$output = preg_replace('/(\w)ing([\W]+|$)/i', '$1$2', $input);
Updated to reflect change specified in comments.
You could use two regexs depending on what you are trying to accomplish your question is a bit ambiguous.
echo preg_replace('/([a-zA-Z]+)ing((:?[\s.,;!?]|$))/', '$1$2', $text);
or
echo preg_replace('/.{3}$/', '', $text);
The first regex looks for word characters before an ing and then punctuation marks, white spaces, or the end of the string. The second just takes off the last three characters of the string.
You can use regex and word boundaries.
$str = preg_replace('/\Bing\b/', "", $str);
\B (non word boundary) matches where word characters are sticking together.
Be aware it substitutes king to k. See demo at regex101
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'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'
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 */);