I need to delete the first 3 letters of a string and the last 3 letters of a string. I know I can use substr() to start at a certain character but if I need to strip both first and last characters i'm not sure if I can actually use this. Any suggestions?
Pass a negative value as the length argument (the 3rd argument) to substr(), like:
$result = substr($string, 3, -3);
So this:
<?php
$string = "Sean Bright";
$string = substr($string, 3, -3);
echo $string;
?>
Outputs:
n Bri
Use
substr($var,1,-1)
this will always get first and last without having to use strlen.
Example:
<?php
$input = ",a,b,d,e,f,";
$output = substr($input, 1, -1);
echo $output;
?>
Output:
a,b,d,e,f
As stated in other answers you can use one of the following functions to reach your goal:
substr($string, 3,
-3) removes 3 chars from start and end
trim($string, ",") removes all specific chars from start and end
ltrim($string, ".") removes all specific chars from start
rtrim($string, ";") removes all specific chars from end
It depends on the amount of chars you need to remove and if the removal needs to be specific. But finally substr() answers your question perfectly.
Maybe someone thinks about removing the first/last char through string dereferencing. Forget that, it will not work as null is a char as well:
<?php
$string = 'Stackoverflow';
var_dump($string);
$string[0] = null;
var_dump($string);
$string[0] = null;
var_dump($string);
echo ord($string[0]) . PHP_EOL;
$string[1] = '';
var_dump($string);
echo ord($string[1]) . PHP_EOL;
?>
returns:
string(13) "Stackoverflow"
string(13) "tackoverflow"
string(13) "tackoverflow"
0
string(13) "ackoverflow"
0
And it is not possible to use unset($string[0]) for strings:
Fatal error: Cannot unset string offsets in /usr/www/***.php on line **
substr($string, 3, strlen($string) - 6)
I don't know php, but can't you take the length of the string, start as position 3 and take length-6 characters using substr?
$myString='123456789';
$newString=substr($myString,3,-3);
Related
i have more + symbol in my string and i want to remove last one and any character after it
ex
Giza+badrashen+test
You can explode your string on '+' and then join it ignoring the last element of the split (with array_slice with negative index), like this (assuming $str is your string)
$result = join('', array_slice(explode('+', $str), -1));
In case you suspect your string may not contain a '+', you can check for its presence first with strpos
if(strpos($str, '+') !== false) {
$result = join('', array_slice(explode('+', $str), -1));
}
A simple regex solution:
Assuming you have Giza+badrashen+test and want Giza+badrashen as result.
echo preg_replace("/\+[^\+]*$/", "", "Giza+badrashen+test");
Tests:
var_dump(preg_replace("/\+[^\+]*$/", "", "Giza+badrashen+test"));
var_dump(preg_replace("/\+[^\+]*$/", "", "Giza+badrashen+test+"));
var_dump(preg_replace("/\+[^\+]*$/", "", "Giza"));
Output:
string(14) "Giza+badrashen"
string(19) "Giza+badrashen+test"
string(4) "Giza"
$string = "abc1234+12+3455+xzyabc";
$string = substr($string, 0, strrpos($string,"+"));
echo $string;
> abc1234+12+3455
EDIT: and gives an empty string if there is no + but it doesn't crash/fail
EDIT2: I slightly mis-read the question the first time, my edited answer is even simpler
Regex with a negative lookahead might be the most compact solution:
$myString="foo-bar+foo+foobar";
$result = preg_split("/\+(?!.*\+)/", $myString);
echo $result[0];
//result: foo-bar+foo
No need of additional check, cause in case no + is found it just gives back the original string.
It's just worth pointing that the + must be escaped having special meaning in all flowers of regex...
I have a string that begins with an empty space and a + sign :
$s = ' +This is a string[...]';
I can't figure out how to remove the first + sign using PHP. I've tried ltrim, preg_replace with several patterns and with trying to escape the + sign, I've also tried substr and str_replace. None of them is removing the plus sign at the beginning of the string. Either it doesn't replace it or it remplace/remove the totality of the string. Any help will be highly appreciated!
Edit : After further investigation, it seems that it's not really a plus sign, it looks 100% like a + sign but I think it's not. Any ideas for how to decode/convert it?
Edit 2 : There's one white space before the + sign. I'm using get_the_excerpt Wordpress function to get the string.
Edit 3 : After successfully removing the empty space and the + with substr($s, 2);, Here's what I get now :
$s == '#43;This is a string[...]'
Wiki : I had to remove 6 characters, I've tried substr($s, 6); and it's working well now. Thanks for your help guys.
ltrim has second parameter
$s = ltrim($s,'+');
edit:
if it is not working it means that there is sth else at the beginning of that string, eg. white spaces. You can check it by using var_dump($s); which shows you exactly what you have there.
You can use explode like this:
$result = explode('+', $s)[0];
What this function actually does is, it removes the delimeter you specify as a first argument and breaks the string into smaller strings whenever that delimeter is found and places those strings in an array.
It's mostly used with multiple ocurrences of a certain delimeter but it will work in your case too.
For example:
$string = "This,is,a,string";
$results = explode(',', $string);
var_dump($results); //prints ['This', 'is', 'a', 'string' ]
So in your case since the plus sign appears ony once the result is in the zero index of the returned array (that contains only one element, your string obviously)
Here's a couple of different ways I can think of
str_replace
$string = str_replace('+', '', $string);
preg_replace
$string = preg_replace('/^\+/', '', $string);
ltrim
$string = ltrim($string, '+');
substr
$string = substr($string, 1);
try this
<?php
$s = '+This is a string';
echo ltrim($s,'+');
?>
You can use ltrim() or substr().
For example :
$output = ltrim($string, '+');
or you can use
$output = substr($string, 1);
You can remove multiple characters with trim. Perhaps you were not re-assigning the outcome of your trim function.
<?php
$s = ' +This is a string[...]';
$s = ltrim($s, '+ ');
print $s;
Outputs:
This is a string[...]
ltrim in the above example removes all spaces and addition characters from the left hand side of the original string.
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.
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 have strings with folowing structure:
7_string_12
7_string2_122
7_string3_1223
How I can get string before second "_" ?
I want my final result to be :
7_string
7_string2
7_string3
I am using explode('_', $string) and combine first two values, but my script was very slow!
$str = '7_string_12';
echo substr($str,0,strrpos($str,'_'));
echoes
7_string
no matter what's at the begining of the string
If it always starts with 7_ you can try this:
$string = substr($text, 0, strpos($text, '_', 2));
The strpos() searches for the first _ starting from character 3 (= s from string). Then you use substr() to select the whole string starting from the first character to the character returned by strpos().
$s1 = '7_string_12';
echo substr($s1, 0, strpos($s1, '_', 2));