I have some plain text used to generate html, this is the text:
lots of stuff
<a onclick="javascript:do_things('http://somelink.to.something.com', 'string'with'bad'quotes'');">
lots of stuff
The structure of the text is always the same because that text is in turn generated, but the last string used as an argument to the javascript function can change, it may have any number of single quotes or not at all. I want to replace those quotes with \' so that the result is:
lots of stuff
<a onclick="javascript:do_things('http://somelink.to.something.com', 'string\'with\'bad\'quotes\'');">
lots of stuff
I got this far:
onclick="javascript:do_things\('.*', '(.*)'\)
which gives me this match:
string'with'bad'quotes'
But I haven't been able to match the quotes inside, I mean, I could match a quote with .*'.*, but how do I match any number of quotes in any position?
Thanks
How about this?
$string = 'lots of stuff
<a onclick="javascript:do_things(\'http://somelink.to.something.com\', \'string\'with\'bad\'quotes\'\');">
lots of stuff';
echo preg_replace_callback('~(<a\h*onclick="javascript:do_things\(\'.*?\',\h*\')(.*)(\'\);">)~', function($match){
return $match[1] . str_replace("'", "\'", $match[2]) . $match[3];}, $string);
Output:
lots of stuff
<a onclick="javascript:do_things('http://somelink.to.something.com', 'string\'with\'bad\'quotes\'');">
lots of stuff
Regex101 Demo: https://regex101.com/r/rM5mM3/3
We capture the second part of the function then execute a replacement on all single quotes in the found string.
Related
I have a multi-line string shown below -
?a="text1
?bc="text23
I need to identify a pattern like using below regex
'/[?][a-z^A-Z]+[=]["]/'
and replace my string by just remove the double quote (") in it, expected output is shown below
?a=text1
?b=text23
Please help in solving the above issue using php.
Capture everything except the quote in a capture group () and replace:
$string = preg_replace('/([?][a-z^A-Z]+[=])["]/', '$1', $string);
But you really don't need all those character classes []:
/(\?[a-z^A-Z]+=)"/
I will give another solution because i see the php tag also. So let's say you have these:
$a='"text1';
$b='"text2';
if i echo them i get
"text1
"text2
in order to get rid of double quote there is a function trim in php that you can use like that:
echo trim($a,'"');
echo trim($b,'"');
the results will be
text1
text2
I dont think you need regex in this occasion as long as you use php. Php can take care of those small things without bother with complex regex expressions.
I would like to replace extra spaces (instances of consecutive whitespace characters) with one space, as long as those extra spaces are not in double or single quotes (or any other enclosures I may want to include).
I saw some similar questions, but I could not find a direct response to my needs above. Thank you!
Hope you're still looking, or come back to check! This seems to work for me:
'/\s+((["\']).*?(?=\2)\2)|\s\s+/'
...and replace with $1
EDIT
Also, if you need to allow for escaped quotes like \" or \', you could use this expression:
'/\s+((["\'])(\\\\\2|(?!\2).)*?(?=\2)\2)|\s\s+/'
It gets a bit stickier if you want to add support for "balanced" quotes like brackets (e.g. () or {})
END EDIT
Let me know if you find problems or would like some explanation!
HOPEFULLY FINAL EDIT AND WARNINGS
Potential problem: If a quoted string starts at the beginning of the string variable (or file), it will either not count as a quoted string (and have any whitespace reduced) or it will throw off the whole thing, making anything NOT in quotes get treated as though it was in quotes and vice versa -
A potential change that might remedy this is to use the following match expression
/(?:^|\s+)((["\'])(\\\\\2|(?!\2).)*?(?=\2)\2)|\s\s+/
this replaces \s+ with (?:^|\s+) at the beginning of the expression
this will add a space at the beginning of the variable if the string starts with a quote - just trim() or remove that whitespace to continue
I seem to have used the "line by line" approach (like sed, if I'm not mistaken) to reach my original results - if you use the "whole file" or "whole string" setting or approach, carriage-return-line-feed seems to count as two whitespace characters (can't imagine why...), thus turning any newlines into single spaces (unless they are inside quotes and "dot-matches-newline" is used, of course)
this could be resolved by replacing the . and \s shorthand character classes with the specific characters you want to match, like the following:
/(?:^|[ \t]+)((["\'])(\\\\\2|(?!\2)[\s\S])*?(?=\2)\2)|[ \t]{2,}/
this does not require the dot-matches-newline switch and only replaces multiple spaces or tabs - not newlines - with a single space (and of course, only if they are not quoted)
EXAMPLE
This link shows an example of the first expression and last expression in use on sample text on http://codepad.viper-7.com
You could do it in several steps. Consider the following example:
$str = 'This is a string with "Bunch of extra spaces". Leave them "untouched !".';
$id = 0;
$buffer = array();
$str = preg_replace_callback('|".*?"|', function($m) use (&$id, &$buffer) {
$buffer[] = $m[0];
return '__' . $id++;
}, $str);
$str = preg_replace('|\s+|', ' ', $str);
$str = preg_replace_callback('|__(\d+)|', function($m) use ($buffer) {
return $buffer[$m[1]];
}, $str);
echo $str;
This will output the string:
This is a string with "Bunch of extra spaces". Leave them "untouched !".
Although this is is not the prettiest solution.
I have a PHP page which gets text from an outside source wrapped in quotation marks. How do I strip them off?
For example:
input: "This is a text"
output: This is a text
Please answer with full PHP coding rather than just the regex...
This will work quite nicely unless you have strings with multiple quotes like """hello""" as input and you want to preserve all but the outermost "'s:
$output = trim($input, '"');
trim strips all of certain characters from the beginning and end of a string in the charlist that is passed in as a second argument (in this case just "). If you don't pass in a second argument it trims whitespace.
If the situation of multiple leading and ending quotes is an issue you can use:
$output = preg_replace('/^"|"$/', '', $input);
Which replaces only one leading or trailing quote with the empty string, such that:
""This is a text"" becomes "This is a text"
$output = str_replace('"', '', $input);
Of course, this will remove all quotation marks, even from inside the strings. Is this what you want? How many strings like this are there?
The question was on how to do it with a regex (maybe for curiosity/learning purposes).
This is how you would do that in php:
$result = preg_replace('/(")(.*?)(")/i', '$2', $subject);
Hope this helps,
Buckley
I have a string as below
$str = '"Mark Zuckerberg" facebook "A social utility connecting friends" profile';
I want it to be manipulated as follows
$output = '"Mark Zuckerberg" OR facebook OR "A social utility connecting friends" OR profile';
What I am trying to have in output is all the units combined with OR in between them. Here a unit is wither a single word when its not in double quotes or the complete string that falls within the single quotes.
I wanted to try with preg_replace. But am unable to get a correct regular expression to match. Kindly help!
$result = preg_replace('/ (?=(?:[^"]*"[^"]*")*[^"]*$)/', ' OR ', $subject);
works if you don't have any escaped quotes in your string. It replaces spaces only if they are followed by an even number of double quotes.
Any reason why preg_replace rather than a simple
$output = '"'.implode('" OR "',str_getcsv($str,' ')).'"';
I currently have this regex:
$text = preg_replace("#<sup>(?:(?!</?sup).)*$key(?:(?!</?sup).)*<\/sup>#is", '<sup>'.$val.'</sup>', $text);
The objective of the regex is to take <sup>[stuff here]$key[stuff here]</sup> and remove the stuff within the [stuff here] locations.
What I actually would like to do, is not remove $key[stuff here]</sup>, but simply move the stuff to $key</sup>[stuff here]
I've tried using $1-$4 and \\1-\\4 and I can't seem to get the text to be added after </sup>
Try this;
$text = preg_replace(
'#<sup>((?:(?!</?sup).)*)'.$key.'((?:(?!</?sup).)*)</sup>#is',
'<sup>'.$val.'</sup>\1\2',
$text
);
The (?:...)* bit isn't actually a sub-pattern, and is therefor not available using backreferences. Also, if you use ' rather than " for string literals, you will only need to escape \ and '
// Cheers, Morten
You have to combine preg_match(); and preg_replace();
You match the desired stuff with preg_match() and store in to the variable.
You replace with the same regex to empty string.
Append the variable you store to at the end.