JavaScript正则表达式 - 在填充的表格单元格后删除空表格单元格

Question

I've been banging my head against this one for a week, just can't get it to work, please can you help? I process user input, transforming it into a HTML table. For each empty cell, I create a cell that contains 1 vertical tab character which I then use to (try to) do the following: I need to remove 1 table cell (that contains 1 vertical tab character) only if it appears after a table cell that contains 0 or more non-vertical-tab characters.

For example...

<table>
<tr>
<td>1</td>
<td></td> // this cell should be removed
<td></td>
<td></td>
<td></td>
<td>some text</td>
<td></td> // this cell should be removed
<td></td>
<td></td>
<td></td>
<td>4973as</td>
<td></td> // this cell should be removed
<td></td>
<td></td>
<td></td>
<td>90-16</td>
<td>sdf;.s'df'f</td>
<td>£%$Dcgcfcgf</td>
<td></td> // this cell should be removed
<td></td>
<td></td>
</tr>
</table>

I'm trying to do this in JavaScript and a lookbehind seemed to be the answer, but JavaScript's regex engine doesn't support lookbehinds. So I tried this in PHP which worked perfectly (in regex101.com) with the data in $testInput...

/(?<=<td>[^\x0B]*<\/td>)<td>\x0B<\/td>/g

...until I realised PHP doesn't allow the global modifier. I tried working around this by using preg_replace_all() - which doesn't exist, but preg_match_all() does, which you can use in a loop to execute preg_replace(). Problem with this? Warning: preg_match(): Compilation failed: lookbehind assertion is not fixed length at offset 19

$testInput = "<table id=\"testInput\"><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><td>4</td><td></td><td>4</td><td></td><td>4</td><td></td><td>4</td><td></td><td>4</td><td></td><td>4</td><td></td><td>4</td><td></td><td>4</td><td></td></tr><tr><td></td><td>4</td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td></td><td>4</td><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td>4</td><td></td><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td>4</td><td></td><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td>4</td><td></td><td></td><td></td><td>4</td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr></table>";

$pattern = "/(?<=<td>[^\x0B]*<\/td>)<td>\x0B<\/td>/";

    pregReplaceAll($pattern, "", $testInput);

function pregReplaceAll($find, $replacement, $s) {
    while (preg_match($find, $s)) {
        $s = preg_replace($find, $replacement, $s);
    }
    echo $s;
}

The above script results in an echo of the input with no replace having occurred.

I've been plugging away at this daily and my brain's gone numb... it seems so simple and yet so complicated.

My last resort is find a solution in C#/ASP but that's another new thing to learn... not that that's a bad thing, but JavaScript/PHP can't be this crippled in 2017 can it??

Answer 1

preg_replace does replace all occurrences, so you should stick with that. Indeed, a look behind cannot have a * in it, so that wont work. But you can use a capture group for the part you want to keep, and inject it again with $1 :

$pattern = "/(<td>[^\x0B]*<\/td>)<td>\x0B<\/td>/";
$result = preg_replace($pattern, "$1", $testInput);
echo $result;