PHP：简单的Dom Parser查找第N个元素类存在

Question

I am using the PHP Simple DOM Parser for parsing the HTML Page, Now i am lacking in particular point of how to find the nth element class should be a particular class

For Example:

<table>
<tr>
<th class="h1">ONE</td>
<th class="h2">TWO</td>
<th class="h3">THREE</td>
</tr>
<tr>
<td class="one">Apple</td>
<td class="two">Orange</td>
<td class="null">N/A</td>
</tr>
<tr>
<td class="one">Apple</td>
<td class="null">N/A</td>
<td class="three">Banana</td>
</tr>
</table>

The table looks something like this , so i am traversing the table via tr

foreach ($demo->find("tr") as $val) 
{
   if(is_object($val->find('td.null', 0))
    {
      echo "FOUND";
    }
}

But the above foreach loop returns "FOUND" if td.null exist. But I need to find if 2nd element td class is null i need to return as TWO, If third td element class is null i need to return as Three

I hope so that you understand that what i am asking for , so please help me of how to find the nth element class is null

Answer 1

First, what I would do is also iterate each td's thru foreach. So that you'll be able to get which index number key it falls into. (Note that of course its indexing is zero based so it actually starts at 0).

Then inside the inner loop, just check if the class is null, then map it in the corresponding word 1 = one, 2 = two, etc...

Rough example:

$map = array(1 => 'one', 2 => 'two', 3 => 'three');
foreach ($demo->find('tr') as $tr) { // loop each table row
    // then loop each td
    foreach($tr->find('td') as $i => $td) { // indexing starts at zero
        if($td->class == 'null') { // if its class is null
            echo $map[$i+1]; // map it to its corresponding word equivalent
        }
    }
}

So in this case, this would output three and then two. Inside the second table row, the null lands on the third, inside the third row it lands into the second.