使用preg_replace删除部分HTML空格

Question

I'm trying to use preg_replace to find and remove part of page in CMS.

This is example of code that I would like to have removed:

                                <dt>
                                    <label for="my_label">My label:</label>
                            </dt>
                            <dd>
                                    <select name="my_label" id="my_label">
                                            <option value="0" selected="selected">option 0</option>
                                            <option value="1">option 1</option>
                                            <option value="2">option 2</option>
                                            <option value="3">option 3</option>
                                    </select>
                            </dd>

Options can be dynamic, they depend on user settings in profile.

With help from many tips on other questions asked here I was able to make this:

preg_replace('#(?<=<label for="my_label">)(.*?)(?=<\/select>)#ms', '', $html);

It works quite nice but there's one small problem that I cannot seem to resolve - output looks like this:

                                <dt>
                                    <label for="my_label"></select>
                            </dd>

Would it be possible to remove it wit some modification to preg_replace that I wrote above? I intentionally posted code with spaces as I think that it's important to reg exp.

Answer 1

to get rid of the whole block use the following:

preg_replace('!(<dt>\s+<label for="my_label">.*?</dd>)!ms', '', $t);

\s is any whitespace .*? will match as few characters as possible before < /dd> to only remove this block (you can find more here: http://php.net/manual/de/regexp.reference.escape.php)