在php中查找条件并将其替换为条件

Question

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.

Answer 1

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.