REGEX: chain substitution

[carlos1234]

Hi everybody,

I fill a variable with html, once has been charged I need to add a "<br />" at the begining and at the ending of each link found in the variable. As an example "<a ......>.....</a>" to "<br /><a ......>.....</a><br />"

I use the regex expression "%(<a.*?>)%is" to find the initial link tag but I don't know how to tell the preg_replace to do the substitution with "<br /><a.*?>". The link closing tag is easier to do so....:).

Thanks.

PD: I don't know what I'm puting into the variable, that's why I have to do it afterwoods, once the variable is charged. Is like to be reading a text with html without knowing its content.

[lucy24]

I need to add a "<br />" at the begining and at the ending of each link

Approaching it purely in RegEx terms, without reference to php:

pattern:

(<a[^>]+>[^<>]+</a>)

substitution:
<br>$1<br>

(Or, if you insist,

<br/>

)

Here it's much easier to treat the whole address as a package. Can we safely assume there will be no html markup within the anchor text? Otherwise the pattern becomes a good bit more complicated.

But what if your text says something like

check out this terrific site at example.com/pagename.

i.e.

check out this terrific site at <a href = "http://www.example.com/pagename">pagename</a>.

Applying the replacement would leave the final punctuation stranded on a new line, even if it happened to be the end of a paragraph. Is this OK?

Will your text ever contain other kinds of anchors, like

<a name = "something" id = "something">

? Those would need to be excluded, again making for a more complicated pattern, but still perfectly doable.

[penders]

Do you need to use a regex...?

$result = str_ireplace(array('<a','a>'), array('<br><a','a><br>'), $source);

[lucy24]

Spoilsport :P

[penders]

lol (Still a very informative regex post I might add! :)