Replace all the _ except one before last one - php

I have the following url:
http://distilleryimage3_s3_amazonaws_com/8af11cdcf11e286b022000ae90285_7_jpg
I want to replace the _ with .. However, the _7 at the end should be kept - it is not a dot.
So, basically it should look like:
http://distilleryimage3.s3.amazonaws.com/8af11cdcf11e286b022000ae90285_7.jpg
If I use str_replace it will replace all the _ but I need to keep that one there. How can I do this?

Use this (corrected now!)
<?php
$subject = 'http://distilleryimage3_s3_amazonaws_com/8af11cdcf11e286b022000ae90285_7_jpg';
$pattern = '/(_)(?!\d_jpg)/';
var_dump(preg_replace($pattern, '.', $subject));
This outputs
http://distilleryimage3.s3.amazonaws.com/8af11cdcf11e286b022000ae90285_7.jpg

Here's a solution that doesn't use regex (works for all numbers, so this won't fail if the number is different from 7; it doesn't have to be a number -- a string works, too):
<?php
$haystack = 'http://distilleryimage3_s3_amazonaws_com/8af11cdcf11e286b022000ae90285_7_jpg';
//replacing all '_' with '.'
$haystack = str_replace('_', '.', $haystack);
//finding second last occurence of '.'
$n = strrpos($haystack, '.', strrpos($haystack, '.') - strlen($haystack) - 1);
//replacing the nth character to '_'
$haystack[$n] = '_';
echo $haystack;
Output:
http://distilleryimage3.s3.amazonaws.com/8af11cdcf11e286b022000ae90285_7.jpg
Demo!

You can use this negative lookahead based regex code:
$s='http://distilleryimage3_s3_amazonaws_com/8af11cdcf11e286b022000ae90285_7_jpg';
$repl = preg_replace('/_(?![^_]*_[^_]*$)/', '.', $s);
//=> http://distilleryimage3.s3.amazonaws.com/8af11cdcf11e286b022000ae90285_7.jpg

The final _ in a string that is not itself.
_(?=[^_]+$)
Edit: my assumption is that this question needs specific knowledge of file extensions to answered correctly.

Related

php regex replace each character with asterisk

I am trying to something like this.
Hiding users except for first 3 characters.
EX)
apple -> app**
google -> goo***
abc12345 ->abc*****
I am currently using php like this:
$string = "abcd1234";
$regex = '/(?<=^(.{3}))(.*)$/';
$replacement = '*';
$changed = preg_replace($regex,$replacement,$string);
echo $changed;
and the result be like:
abc*
But I want to make a replacement to every single character except for first 3 - like:
abc*****
How should I do?
Don't use regex, use substr_replace:
$var = "abcdef";
$charToKeep = 3;
echo strlen($var) > $charToKeep ? substr_replace($var, str_repeat ( '*' , strlen($var) - $charToKeep), $charToKeep) : $var;
Keep in mind that regex are good for matching patterns in string, but there is a lot of functions already designed for string manipulation.
Will output:
abc***
Try this function. You can specify how much chars should be visible and which character should be used as mask:
$string = "abcd1234";
echo hideCharacters($string, 3, "*");
function hideCharacters($string, $visibleCharactersCount, $mask)
{
if(strlen($string) < $visibleCharactersCount)
return $string;
$part = substr($string, 0, $visibleCharactersCount);
return str_pad($part, strlen($string), $mask, STR_PAD_RIGHT);
}
Output:
abc*****
Your regex matches all symbols after the first 3, thus, you replace them with a one hard-coded *.
You can use
'~(^.{3}|(?!^)\G)\K.~'
And replace with *. See the regex demo
This regex matches the first 3 characters (with ^.{3}) or the end of the previous successful match or start of the string (with (?!^)\G), and then omits the characters matched from the match value (with \K) and matches any character but a newline with ..
See IDEONE demo
$re = '~(^.{3}|(?!^)\G)\K.~';
$strs = array("aa","apple", "google", "abc12345", "asdddd");
foreach ($strs as $s) {
$result = preg_replace($re, "*", $s);
echo $result . PHP_EOL;
}
Another possible solution is to concatenate the first three characters with a string of * repeated the correct number of times:
$text = substr($string, 0, 3).str_repeat('*', max(0, strlen($string) - 3));
The usage of max() is needed to avoid str_repeat() issue a warning when it receives a negative argument. This situation happens when the length of $string is less than 3.

PHP wrapping last two letters of string with HTML

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.

PHP get first few charactors and remove from string

Does anyone know how to remove the first few characters from a string and remove them in PHP.
Like in the string "str_filename" I need to remove the "str_" and save the "filename".
But it has to remove as many charactors as it takes to get to the "_".
In other words, i need to remove all the characters up until and including the first "_" in the string.
You can do this:
if (strpos($string, '_') !== false)
$string = substr($string, strpos($string, '_') + 1);
It works as you can see here: http://codepad.org/g12ENLGY
Note: The if is useful because your string could not have the '_' char.
Could you use something like:
$files = explode("_", $filename);
echo $files[1];
So this would split the string on the _ character, and then echo the second part (after the underscore).
This assumes that there is only one underscore though.
Please see http://www.php.net/manual/en/function.explode.php
An additional solution than using explode and substr as have been mentioned you can solve it using regex:
^(?:[^_]*_)(.*)$
Here's an example using it:
$str = "str_filename";
$pattern = "/^(?:[^_]*_)(.*)$/";
preg_match($pattern, $str, $matches);
echo $matches[1]; //prints "filename"

php remove everything in the string, keep the last digits

I have values like below:
$var1 = car-123-244343
$var2 = boat-2-1
$var3 = plane-311-23
I need to remove everything and keep the last digit/ditgits after the second hyphen
Expecting values:
244343
1
23
This is what I've got
$stripped = preg_replace('^[a-z]+[-]$', '', 'car-123-244343');
I got a big red error No ending delimiter '^' found
Without regex:
$var1 = substr($var1, strrpos($var1, '-') + 1);
What this does is the same as:
$pos = strrpos($var1, '-') + 1 takes the last postion of '-' and adds 1 for starting at the next character
substr($var, $pos) takes the $var and returns the substring starting in $pos.
I think is less expensive than using regex.
Edit:
As pointed below by konforce, if you are not sure which all the strings have that format, you have to verify it.
this function will work:
function foo($value)
{
$split = explode('-', $value);
return $split[count($split)-1];
}
Here is a fun version with explode:
list($vehicle, $no1, $no2) = explode('-', $data);
First, that error means your regex needs to be enclosed in delimiters (below I use the classic /).
Second, I would rewrite your regex to this:
$stripped = preg_replace('/.+?(\d+)$/', '$1', 'car-123-244343');
If you can operate on the assumption that what comes after the last - is always a number, the other solutions also work.
With regex:
$endnumber = preg_replace('/.*[^0-9]/', '', $input);
Remove everything up till, and including, the last non-digit.

How to replace the Last "s" with "" in PHP

I need to know how I can replace the last "s" from a string with ""
Let's say I have a string like testers and the output should be tester.
It should just replace the last "s" and not every "s" in a string
how can I do that in PHP?
if (substr($str, -1) == 's')
{
$str = substr($str, 0, -1);
}
Update: Ok it is also possible without regular expressions using strrpos ans substr_replace:
$str = "A sentence with 'Testers' in it";
echo substr_replace($str,'', strrpos($str, 's'), 1);
// Ouputs: A sentence with 'Tester' in it
strrpos returns the index of the last occurrence of a string and substr_replace replaces a string starting from a certain position.
(Which is the same as Gordon proposed as I just noticed.)
All answers so far remove the last character of a word. However if you really want to replace the last occurrence of a character, you can use preg_replace with a negative lookahead:
$s = "A sentence with 'Testers' in it";
echo preg_replace("%s(?!.*s.*)%", "", $string );
// Ouputs: A sentence with 'Tester' in it
$result = rtrim($str, 's');
$result = str_pad($result, strlen($str) - 1, 's');
See rtrim()
Your question is somewhat unclear whether you want to remove the s from the end of the string or the last occurence of s in the string. It's a difference. If you want the first, use the solution offered by zerkms.
This function removes the last occurence of $char from $string, regardless of it's position in the string or returns the whole string, when $char does not occur in the string.
function removeLastOccurenceOfChar($char, $string)
{
if( ($pos = strrpos($string, $char)) !== FALSE) {
return substr_replace($string, '', $pos, 1);
}
return $string;
}
echo removeLastOccurenceOfChar('s', "the world's greatest");
// gives "the world's greatet"
If your intention is to inflect, e.g singularize/pluralize words, then have a look at this simple inflector class to know which route to take.
$str = preg_replace("/s$/i","",rtrim($str));
The very simplest solution is using rtrim()
That is exactly what that function is intended to be used for:
Strip whitespace (or other characters) from the end of a string.
Nothing simpler than that, I am not sure why, and would not follow the suggestions in this thread going from regex to "if/else" blocks.
This is your code:
$string = "Testers";
$stripped = rtrim( $string, 's' );
The output will be:
Tester

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