I'm doing str_replace on a very long string and my $search is an array.
$search = array(
" tag_name_item ",
" tag_name_item_category "
);
$replace = array(
" tag_name_item{$suffix} ",
" tag_name_item_category{$suffix} "
);
echo str_replace($search, $replace, $my_really_long_string);
The reason why I added spaces on both $search and $replace is because I want to only match whole words. As you would have guessed from my code above, if I removed the spaces and my really long string is:
...
tag_name_item ...
tag_name_item_category ...
...
Then I would get something like
...
tag_name_item_sfx ...
tag_name_item_sfx_category ...
...
This is wrong because I want the following result:
...
tag_name_item_sfx ...
tag_name_item_category_sfx ...
...
So what's wrong?
Nothing really, it works. But I don't like it. Looks dirty, not well coded, inefficient.
I realized I can do something like this using regular expressions using the \b modifier but I'm not good with regex and so I don't know how to preg_replace.
A possible approach using regular expressions would/could look like this:
$result = preg_replace(
'/\b(tag_name_item(_category)?)\b/',
'$1' . $suffix,
$string
);
How it works:
\b: As you say are word boundaries, this is to ensure we're only matching words, not word parts
(: We want to use part of our match in the replacement string (tag_name_index has to be replaced with itself + a suffix). That's why we use a match group, so we can refer back to the match in the replacement string
tag_name_index is a literal match for that string.
(_category)?: Another literal match, grouped and made optional through use of the ? operator. This ensures that we're matching both tag_name_item and tag_name_item_category
): end of the first group (the optional _category match is the second group). This group, essentially, holds the entire match we're going to replace
\b: word boundary again
These matches are replaced with '$1' . $suffix. The $1 is a reference to the first match group (everything inside the outer brackets in the expression). You could refer to the second group using $2, but we're not interested in that group right now.
That's all there is to it really
More generic:
So, you're trying to suffix all strings starting with tag_name, which judging by your example, can be followed by any number of snake_cased words. A more generic regex for that would look something like this:
$result = preg_replace(
'/\b(tag_name[a-z_]*)\b/',
'$1' . $suffix,
$string
);
Like before, the use of \b, () and the tag_name literal remains the same. what changed is this:
[a-z_]*: This is a character class. It matches characters a-z (a to z), and underscores zero or more times (*). It matches _item and _item_category, just as it would match _foo_bar_zar_fefe.
These regex's are case-sensitive, if you want to match things like tag_name_XYZ, you'll probably want to use the i flag (case-insensitive): /\b(tag_name[a-z_]*)\b/i
Like before, the entire match is grouped, and used in the replacement string, to which we add $suffix, whatever that might be
To avoid the problem, you can use strtr that parses the string only once and chooses the longest match:
$pairs = [ " tag_name_item " => " tag_name_item{$suffix} ",
" tag_name_item_category " => " tag_name_item_category{$suffix} " ];
$result = strtr($str, $pairs);
This function replaces the entire whole word but not the substring with an array element which matches the word
<?PHP
function removePrepositions($text){
$propositions=array('/\b,\b/i','/\bthe\b/i','/\bor\b/i');
if( count($propositions) > 0 ) {
foreach($propositions as $exceptionPhrase) {
$text = preg_replace($exceptionPhrase, '', trim($text));
}
$retval = trim($text);
}
return $retval;
}
?>
See the entire example
Related
thanks by your help.
my target is use preg_replace + pattern for remove very sample strings.
then only using preg_replace in this string or others, I need remove ANY content into <tag and next symbol >, the pattern is so simple, then:
$x = '#<\w+(\s+[^>]*)>#is';
$s = 'DATA<td class="td1">111</td><td class="td2">222</td>DATA';
preg_match_all($x, $s, $Q);
print_r($Q[1]);
[1] => Array
(
[0] => class="td1"
[1] => class="td2"
)
work greath!
now I try remove strings using the same pattern:
$new_string = '';
$Q = preg_replace($x, "\\1$new_string", $s);
print_r($Q);
result is completely different.
what is bad in my use of preg_replace?
using only preg_replace() how I can remove this strings?
(we can use foreach(...) for remove each string, but where is the error in my code?)
my result expected when I intro this value:
$s = 'DATA<td class="td1">111</td><td class="td2">222</td>DATA';
is this output:
$Q = 'DATA<td>111</td><td>222</td>DATA';
Let's break down your RegEx, #<\w+(\s+[^>]*)>#is, and see if that helps.
# // Start delimiter
< // Literal `<` character
\w+ // One or more word-characters, a-z, A-Z, 0-9 or _
( // Start capturing group
\s+ // One or more spaces
[^>]* // Zero or more characters that are not the literal `>`
) // End capturing group
> // Literal `>` character
# // End delimiter
is // Ignore case and `.` matches all characters including newline
Given the input DATA<td class="td1">DATA this matches <td class="td1"> and captures class="td1". The difference between match and capture is very important.
When you use preg_match you'll see the entire match at index 0, and any subsequent captures at incrementing indexes.
When you use preg_replace the entire match will be replaced. You can use the captures, if you so choose, but you are replacing the match.
I'm going to say that again: whatever you pass as the replacement string will replace the entirety of the found match. If you say $1 or \\=1, you are saying replace the entire match with just the capture.
Going back to the sample after the breakdown, using $1 is the equivalent of calling:
str_replace('<td class="td1">', ' class="td1"', $string);
which you can see here: https://3v4l.org/ZkPFb
To your question "how to change [0] by $new_string", you are doing it correctly, it is your RegEx itself that is wrong. To do what you are trying to do, your pattern must capture the tag itself so that you can say "replace the HTML tag with all of the attributes with just the tag".
As one of my comments noted, this is where you'd invert the capturing. You aren't interesting in capturing the attributes, you are throwing those away. Instead, you are interested in capturing the tag itself:
$string = 'DATA<td class="td1">DATA';
$pattern = '#<(\w+)\s+[^>]*>#is';
echo preg_replace($pattern, '<$1>', $string);
Demo: https://3v4l.org/oIW7d
I am newbie in PHP. I want to replace certain characters in a string. My code is in below:
$str="this 'is' a new 'string and i wanna' replace \"in\" \"it here\"";
$find = [
'\'',
'"'
];
$replace = [
['^', '*']
['#', '#']
];
$result = null;
$odd = true;
for ($i=0; $i < strlen($str); $i++) {
if (in_array($str[$i], $find)) {
$key = array_search($str[$i], $find);
$result .= $odd ? $replace[$key][0] : $replace[$key][1];
$odd = !$odd;
} else {
$result .= $str[$i];
}
}
echo $result;
the output of the above code is:
this ^is* a new ^string and i wanna* replace #in# #it here#.
but I want the output to be:
this ^is* a new 'string and i wanna' replace #in# "it here".
That means character will replace for both quotation(left quotation and right quotation- condition is for ' and "). for single quotation, string will not be replaced either if have left or right quotation. it will be replaced for left and right quotation.
Ok, I don't know what all that code is trying to accomplish.
But anyway here is my go at it
$str = "this 'is' a new 'string and i wanna' replace \"in\" \"it here\"";
$str = preg_replace(["/'([^']+)'/",'/"([^"]+)"/'], ["^$1*", "#$1#"], $str, 1);
print_r($str);
You can test it here
Ouptput
this ^is* a new 'string and i wanna' replace #in# "it here"
Using preg_replace and a fairly simple Regular expression, we can replace the quotes. Now the trick here is the fourth parameter of preg_replace is $count And is defined as this:
count If specified, this variable will be filled with the number of replacements done.
Therefore, setting this to 1 limits it to the first match only. In other words it will do $count replacements, or 1 in this case. Now because it's an array of patterns, each pattern is treated separately. So each one is basically treated as a separate operation, and thus each is allowed $count matches, or each get 1 match/replacement.
Now rather or not this fits every use case you have I cannot say, but it's the most straight forward way to do it for the example you provided.
As for the match itself /'([^']+)'/
/ opening and closing "delimiters" for the Expression (its a required thing, although it doesn't have to be /)
' literal match, matches ' one time (the opening quote)
( ... ) capture group (group1) so we can use it in the replacement, as $1
[^']+ character set with a [^ not modifier, match anything not in the set, so anything that is not a ' one or more times, greedy
' literal match, matches ' one time (the ending quote)
The replacement "^$1*"
^ literal, adds this char in
$1 use the contents of the capture group (group1)
* literal, adds the char in
Hope that helps understand how it works.
UPDATE
Ok I think I finally deciphered what you want:
string will be replaced for if any word have left and right quotation. example..'word'..here string will be changed..but 'word...in this case not change or word' also not be changed.
This seems like you are trying to say only "whole" words with no spaces.
So in that case we have to adjust our regular expression like this:
$str = preg_replace(["/'([-\w]+)'/",'/"([-\w]+)"/'], ["^$1*", "#$1#"], $str);
So we removed the limit $count and we changed what is in the character group to be more strict:
[-\w]+ the \w means the working set, or in other words a-zA-Z0-9_ then the - is a literal (it has to/should go first in this case)
What we are saying with this is to match only strings that start and end with a quote(single|double) and only if the string within them match the working set plus the hyphen. This does not include the space. This way in the first case, your example, it produces the same result, but if you were to flip it to
//[ORIGINAL] this 'is' a new 'string and i wanna' replace \"in\" \"it here\"
this a new 'string and i wanna' replace 'is' \"it here\" \"in\"
You would get his output
this a new 'string and i wanna' replace ^is* \"it here\" #in#
Before this change you would have gotten
this a new ^string and i wanna* replace 'is' #it here# "in"
In other words it would have only replaced the first occurrence, now it will replace anything between the quotes if and only if it's a whole word.
As a final note you can be even more strict if you only want alpha characters by changing the character set to this [a-zA-Z]+, then it will match only a to z, upper or lower case. Whereas the example above will match 0 to 9 (or any combination of them) the - hyphen, the _ underline and the previously mentioned alpha sets.
Hope that is what you need.
Maybe it can not be solved this issue as I want, but maybe you can help me guys.
I have a lot of malformed words in the name of my products.
Some of them has leading ( and trailing ) or maybe one of these, it is same for / and " signs.
What I do is that I am explode the name of the product by spaces, and examines these words.
So I want to replace them to nothing. But, a hard drive could be 40GB ATA 3.5" hard drive. I need to process all the word, but I can not use the same method for 3.5" as for () or // because this 3.5" is valid.
So I only need to replace the quotes, when it is at the start of the string AND at end of the string.
$cases = [
'(testone)',
'(testtwo',
'testthree)',
'/otherone/',
'/othertwo',
'otherthree/',
'"anotherone',
'anothertwo"',
'"anotherthree"',
];
$patterns = [
'/^\(/',
'/\)$/',
'~^/~',
'~/$~',
//Here is what I can not imagine, how to add the rule for `"`
];
$result = preg_replace($patterns, '', $cases);
This is works well, but can it be done in one regex_replace()? If yes, somebody can help me out the pattern(s) for the quotes?
Result for quotes should be this:
'"anotherone', //no quote at end leave the leading
'anothertwo"', //no quote at start leave the trailin
'anotherthree', //there are quotes on start and end so remove them.
You may use another approach: rather than define an array of patterns, use one single alternation based regex:
preg_replace('~^[(/]|[/)]$|^"(.*)"$~s', '$1', $s)
See the regex demo
Details:
^[(/] - a literal ( or / at the start of the string
| - or
[/)]$ - a literal ) or / at the end of the string
| - or
^"(.*)"$ - a " at the start of the string, then any 0+ characters (due to /s option, the . matches a linebreak sequence, too) that are captured into Group 1, and " at the end of the string.
The replacement pattern is $1 that is empty when the first 2 alternatives are matched, and contains Group 1 value if the 3rd alternative is matched.
Note: In case you need to replace until no match is found, use a preg_match with preg_replace together (see demo):
$s = '"/some text/"';
$re = '~^[(/]|[/)]$|^"(.*)"$~s';
$tmp = '';
while (preg_match($re, $s) && $tmp != $s) {
$tmp = $s;
$s = preg_replace($re, '$1', $s);
}
echo $s;
This works
preg_replace([[/(]?(.+)[/)]?|/\"(.+)\"/], '$1', $string)
I want to manipulate a string like "...4+3(4-2)-...." to become "...4+3*(4-2)-....", but of course it should recognize any number, d, followed by a '(' and change it to 'd*('. And I also want to change ')(' to ')*(' at the same time if possible. Would nice if there is a possibility to add support for constants like pi or e too.
For now, I just do it this stupid way:
private function make_implicit_multiplication_explicit($string)
{
$i=1;
if(strlen($string)>1)
{
while(($i=strpos($string,"(",$i))!==false)
{
if(strpos("0123456789",substr($string,$i-1,1)))
{
$string=substr_replace($string,"*(",$i,1);
$i++;
}
$i++;
}
$string=str_replace(")(",")*(",$string);
}
return $string;
}
But I Believe this could be done much nicer with preg_replace or some other regex function? But those manuals are really cumbersome to grasp, I think.
Let's start by what you are looking for:
either of the following: ((a|b) will match either a or b)
any number, \d
the character ): \)
followed by (: \(
Which creates this pattern: (\d|\))\(. But since you want to modify the string and keep both parts, you can group the \( which results in (\() making it worse to read but better to handle.
Now everything left is to tell how to rearrange, which is simple: \\1*\\2, leaving you with code like this
$regex = "/(\d|\))(\()/";
$replace = "\\1*\\2";
$new = preg_replace($regex, $replace, $test);
To see that the pattern actually matches all cases, see this example.
To recognize any number followed by a ( OR a combination of a )( and place an asterisk in between them, you can use a combination of lookaround assertions.
echo preg_replace("/
(?<=[0-9)]) # look behind to see if there is: '0' to '9', ')'
(?=\() # look ahead to see if there is: '('
/x", '*', '(4+3(4-2)-3)(2+3)');
The Positive Lookbehind asserts that what precedes is either a number or right parentheses. While the Positive Lookahead asserts that the preceding characters are followed by a left parentheses.
Another option is to use the \K escape sequence in replace of the Lookbehind. \K resets the starting point of the reported match. Any previously consumed characters are no longer included ( throws away everything that it has matched up to that point. )
echo preg_replace("/
[0-9)] # any character of: '0' to '9', ')'
\K # resets the starting point of the reported match
(?=\() # look ahead to see if there is: '('
/x", '*', '(4+3(4-2)-3)(2+3)');
Your php code should be,
<?php
$mystring = "4+3(4-2)-(5)(3)";
$regex = '~\d+\K\(~';
$replacement = "*(";
$str = preg_replace($regex, $replacement, $mystring);
$regex1 = '~\)\K\(~';
$replacement1 = "*(";
echo preg_replace($regex1, $replacement1, $str);
?> //=> 4+3*(4-2)-(5)*(3)
Explanation:
~\d+\K\(~ this would match the one or more numbers followed by a (. Because of \K it excludes the \d+
Again it replaces the matched part with *( which in turn produces 3*( and the result was stored in another variable.
\)\K\( Matches )( and excludes the first ). This would be replaced by *( which in turn produces )*(
DEMO 1
DEMO 2
Silly method :^ )
$value = '4+3(4-2)(1+2)';
$search = ['1(', '2(', '3(', '4(', '5(', '6(', '7(', '8(', '9(', '0(', ')('];
$replace = ['1*(', '2*(', '3*(', '4*(', '5*(', '6*(', '7*(', '8*(', '9*(', '0*(', ')*('];
echo str_replace($search, $replace, $value);
I have a string that has the following structure:
ABC_ABC_PQR_XYZ
Where PQR has the structure:
ABC+JKL
and
ABC itself is a string that can contain alphanumeric characters and a few other characters like "_", "-", "+", "." and follows no set structure:
eg.qWe_rtY-asdf or pkl123
so, in effect, the string can look like this:
qWe_rtY-asdf_qWe_rtY-asdf_qWe_rtY-asdf+JKL_XYZ
My goal is to find out what string constitutes ABC.
I was initially just using
$arrString = explode("_",$string);
to return $arrString[0] before I was made aware that ABC ($arrString[0]) itself can contain underscores, thus rendering it incorrect.
My next attempt was exlpoding it on "_" anyway and then comparing each of the exploded string parts with the first string part until I get a semblance of a pattern:
function getPatternABC($string)
{
$count = 0;
$pattern ="";
$arrString = explode("_", $string);
foreach($arrString as $expString)
{
if(strcmp($expString,$arrString[0])!==0 || $count==0)
{
$pattern = $pattern ."_". $arrString[$count];
$count++;
}
else break;
}
return substr($pattern,1);
}
This works great - but I wanted to know if there was a more elegant way of doing this using regular expressions?
Here is the regex solution:
'^([a-zA-Z0-9_+-]+)_\1_\1\+'
What this does is match (starting from the beginning of the string) the longest possible sequence consisting of the characters inside the square brackets (edit that per your spec). The sequence must appear exactly twice, each time followed by an underscore, and then must appear once more followed by a plus sign (this is actually the first half of PQR with the delimiter before JKL). The rest of the input is ignored.
You will find ABC captured as capture group 1.
So:
$input = 'qWe_rtY-asdf_qWe_rtY-asdf_qWe_rtY-asdf+JKL_XYZ';
$result = preg_match('/^([a-zA-Z0-9_+-]+)_\1_\1\+/', $input, $matches);
if ($result) {
echo $matches[2];
}
See it in action.
Sure, just make a regular expression that matches your pattern. In this case, something like this:
preg_match('/^([a-zA-Z0-9_+.-]+)_\1_\1\+JKL_XYZ$/', $string, $match);
Your ABC is in $match[1].
If the presence of underscores in these strings has a low frequency, it may be worth checking to see if a simple explode() will do it before bothering with regex.
<?php
$str = 'ABC_ABC_PQR_XYZ';
if(substr_count($str, '_') == 3)
$abc = reset(explode('_', $str));
else
$abc = regexy_function($str);
?>