php preg_replace usage -

Question

This is the very first time I am trying to use regular expressions, so please have patience.

I would like to use php preg_replace function to transform a string such as this:

"r1=r2+robby+ara"

into

"ar1b=ar2b+robby+ara"

so for every occurrence of "r" followed by an integer, I want to place an "a" before it and "b" after it.

I have figured out the search part, but not the replace part - the question-mark below indicates where I am having troubles. Is there a way to "remember" the integer found by [0-9] and use it in replace string?

$s="r1=r2+robby + ara";

$s1=preg_replace ( "/r[0-9]/","ar?b",$s);

Thank you

elixenide · Accepted Answer · 2014-02-21 06:31:08Z

1

You were close, but need parentheses:

$s="r1=r2+robby + ara";
$s1=preg_replace ( "/r([0-9])/","ar$1b", $s); // produces ar1b=ar2b+robby + ara

The ( and ) tell PHP to capture whatever is in between (the digit). The $1 in the replacement tells PHP to use that value (the digit) in that location.

Note: if your problem really is this simple, you don't need parentheses and can use $0, which means "the entire matching portion," like in xdazz's answer:

$s1=preg_replace ( "/r[0-9]/","a$0b", $s); // produces ar1b=ar2b+robby + ara

answered Feb 21, 2014 at 6:31

elixenide

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xdazz · Accepted Answer · 2014-02-21 06:31:20Z

1

// \0 means the matched part
$s1 = preg_replace('/r\d/', 'a\0b', $s);

answered Feb 21, 2014 at 6:31

xdazz

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Toto · Accepted Answer · 2014-02-21 08:50:01Z

1

I'd do:

$str_out = preg_replace('/\b(r\d)\b/', 'a$1b', $str_in);

\b is a word boundary, it prevents, in this case, the replacement of xr1y

answered Feb 21, 2014 at 8:50

Toto

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