php regex - replace on "\${1}"

found this regex:

insert " " every 10 characters:  开发者_StackOverflow社区
$text =  preg_replace("|(.{10})|u", "\${1}"." ", $text);

can you, please, explain what \${1} means. Why using \ and what curly brackets means?

Quoting some portions of the manual page of preg_replace :

replacement may contain references of the form \\n or $n, with the latter form being the preferred one.

You are, apparently, in the second case : $n

And, later :

When working with a replacement pattern where a backreference is immediately followed by another number (i.e.: placing a literal number immediately after a matched pattern), you cannot use the familiar \\1 notation for your backreference.

\\11, for example, would confuse preg_replace() since it does not know whether you want the \\1 backreference followed by a literal 1, or the \\11 backreference followed by nothing.

In this case the solution is to use \${1}1. This creates an isolated $1 backreference, leaving the 1 as a literal.

Here, you don't have anything after what would be $1 -- but I suppose it cannot hurt to use the \${1} notation : I find it makes code easier to read, having those {} ; and it makes sure you won't forget to add them the day they are needed.

The first curly bracket is responsible for the counting of the characters. .{10} means: 10 times any character.
The \${1} represents everything that is matched in the first pair of parantheses. So, to paraphrase it: "Substitute the ten characters (.{10}) with the same 10 characters (\${1}) plus a space.