为DateTime准备PHP代码

Question

So, I have an SQL query result loaded into an array. Now, I want to display it in a html table. This the code I used to use.

$table = "<table><thead><tr><th>".implode('</th><th>', array_keys(current($res)))."</th></tr></thead><tbody>";
for($i = 0; $i < count($res); ++$i) {
  $table .= "<tr>";
    for($j = 0; $j < count($res[$i]); ++$j) {
      $table .= "<td>".$res[$i][$j]."</td>"; //here is the issue if it's date
    }
  $table .= "</tr>";
}
$table .= "</tbody></table>";
echo $table;

If a certain value is Date or DateTime then I get the error Catchable fatal error: Object of class DateTime could not be converted to string.

Here is my next step which is causing the exact same error. I don't really understand why.

$table = "<table><thead><tr><th>".implode('</th><th>', array_keys(current($res)))."</th></tr></thead><tbody>";
    for($i = 0; $i < count($res); ++$i) {
      $table .= "<tr>";
        for($j = 0; $j < count($res[$i]); ++$j) {
          try {
            $table .= "<td>".$res[$i][$j]."</td>"; //the issue is here again, so try doesn't seem to work as expected
          }
          catch (Exception $e){
            $table .= "<td>".$res[$i][$j]->format('Y-m-d H:i:s')."</td>";
          }
        }
      $table .= "</tr>";
    }
    $table .= "</tbody></table>";
    echo $table;

So, let's say I want to write a general function for any query result in which I can't tell in advance what record has what type. How can I prepare the code to handle dates?

Answer 1

You can check whether its a date time object using is_a function and handle it

if (is_a($res[$i][$j], 'DateTime')) {
// handle code
 }