I have a string that has a number inside exactly one pair of quotes. How can I get this number using php?
I have tried iterating over the string and or using str_split but I'm not sure how to limit it to what is inside the quotation marks. The fact that the number can be any length makes it even harder.
EDIT: Here is an example
Hello this is my "100"th string!,
I would need to return 100;
echo intval('123string');
results in 123
preg_match() is a way to get the number from string. Example:
$str = 'Hello this is my "100"th string!,';
preg_match('/\d+/', $str, $match);
echo $match[0];
Demo
use preg_replace:
echo preg_replace("~[^\d+]~", '', 'Hello this is my "100"th string!,');
trim the quotes from around the string, and php will immediately see the number inside.
$str = "'1234'";
$num = trim($str, "'\"");
echo $num + 1;
// => 1235
or if you have text as well a string, replace all non-digits with spaces, then have
php automatically parse the string when it is used in an arithmetic expression.
$num = preg_replace('/\D/', ' ', $string) + 0;
Related
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 a question. I need to add a + before every word and see all between quotes as one word.
A have this code
preg_replace("/\w+/", '+\0', $string);
which results in this
+test +demo "+bla +bla2"
But I need
+test +demo +"bla bla2"
Can someone help me :)
And is it possible to not add a + if there is already one? So you don't get ++test
Thanks!
Maybe you can use this regex:
$string = '+test demo between "double quotes" and between \'single quotes\' test';
$result = preg_replace('/\b(?<!\+)\w+|["|\'].+?["|\']/', '+$0', $string);
var_dump($result);
// which will result in:
string '+test +demo +between +"double quotes" +and +between +'single quotes' +test' (length=74)
I've used a 'negative lookbehind' to check for the '+'.
Regex lookahead, lookbehind and atomic groups
I can't test this but could you try it and let me know how it goes?
First the regex: choose from either, a series of letters which may or may not be preceded by a '+', or, a quotation, followed by any number of letters or spaces, which may be preceded by a '+' followed by a quotation.
I would hope this matches all your examples.
We then get all the matches of the regex in your string, store them in the variable "$matches" which is an array. We then loop through this array testing if there is a '+' as the first character. If there is, do nothing, otherwise add one.
We then implode the array into a string, separating the elements by a space.
Note: I believe $matches in created when given as a parameter to preg_match.
$regex = '/[((\+)?[a-zA-z]+)(\"(\+)?[a-zA-Z ]+\")]/';
preg_match($regex, $string, $matches);
foreach($matches as $match)
{
if(substr($match, 0, 1) != "+") $match = "+" + $match;
}
$result = implode($matches, " ");
How can I get the value between the {} eg. 1 in "{1}" and use it?
if (preg_match('/\{(\d+)\}/', $str, $mtch))
echo $mtch[1];
where $str is '{1}'
If you just want to get rid of the brackets, you may use trim() function
$str = "{1}";
$str = trim($str, "{}");
echo $str; //output: 1
EDIT: I removed the comma - "{}" is enough as the secont parameter for trim() (was "{,}" before the edit)
You can use a preg_replace function to perform a regular expression.
What exactly do you need to do the string.
If you want to get a certain character of the string, $str = '{1}'; $str[1] will return the 2nd character of the string.
Depends on what you actually have .. in the example above this would be enough:
$number = $string{1};
But i guess you need more something like
preg_match('/{([0-9]+)}/', $string, $matches);
$number = $matches[1];