PHP MYSQL列表中的表数据问题

Question

below is my PHP code to display all the data in my database in a table..

if I use this code below to login and select DB rather than using PDO:

 mysql_connect("localhost","root","12345678") or die(mysql_error());
   mysql_select_db("clubresults") or die(mysql_error());
   $query=mysql_query("SELECT * from events ORDER By EventID ASC") or die(mysql_error());

The output gives me a table with all the data listed correctly.

However If I use PDO.. it gives me this Error:

Warning: mysql_num_fields() expects parameter 1 to be resource, object given in C:\xampp\htdocs\clubresults\listevents.php on line 52

Warning: mysql_fetch_row() expects parameter 1 to be resource, object given in C:\xampp\htdocs\clubresults\listevents.php on line 58

Below is the full php code to grab data to and place in a table (Doesn't work) anyone able to point out where i have a mistake? because I can't seem to figure it out myself.. All suggestions are really appreciated! Thanks!

$pdo = new PDO('mysql:host=localhost;dbname=clubresults', 'root', '12345678');
        #Set Error Mode to ERRMODE_EXCEPTION.
        $pdo->setAttribute( PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);  


   $query=$pdo->query('SELECT * from events ORDER By EventID ASC');


   <<<<<<<<<<<<<<<<THIS BIT WORKS IF NOT USING PDO>>>>>>>>>>
$numfields = mysql_num_fields($query);
       print("<table border=\"1\">\n<tr>\n");
       for ($i=0; $i<$numfields; $i++) {
          printf("<th>%s</th>\n", mysql_field_name($query,$i));
       }
       print("</tr>\n");
       while ($row = mysql_fetch_row($query)) {
          print("<tr>\n");
          for ($i=0; $i<sizeof($row); $i++) {
             printf("<td><a href=\"index.php?ID=%s\">%s</a></td>\n", $row[0],$row[$i]);
          }
          print("</tr>\n");
       }
       print("</table>\n");

Answer 1

You won't be able to use mysql_num_fields() or mysql_fetch_row() with PDO -- they're part of a different API, the old MySQL extension.

Instead, since you have called PDO::query() , call $query->fetch() . Use the first row's array keys to build your table heading.

$query = $pdo->query('SELECT * from events ORDER By EventID ASC');
$rowset = array();

if ($query) {
  while ($row = $query->fetch(PDO::FETCH_ASSOC)) {
    // Build array of rows
    $rowset[] = $row;
  }    

  // Output header first
  $headrow = $rowset[0];
  print("<table border=\"1\">\n<tr>\n");
  // Use $rowset[0] to write the table heading
  foreach ($headrow as $col => $val) {
    printf("<th>%s</th>\n", $col);
  }
  print("</tr>");

  // Then output table rows.
  // Outer loop iterates over row
  foreach ($rowset as $row) {
     print("<tr>");
     // Inner loop iterates over columns using $col => $val
     foreach ($row as $col => $val) {
        // We don't know your column names, but substitute the first column (the ID) for FIRSTCOL here
        printf("<td><a href=\"index.php?ID=%s\">%s</a></td>\n", $row['FRISTCOL'],$val);
     }
     print("</tr>");
  }
}
print("</table>");

Addendum

Generally, I would recommend against using SELECT * and dynamically writing out your column headings. Instead, be explicit about the columns you request in your SELECT statement, and be explicit about them in the table you write out. That way, you can be deterministic about the order the columns appear in, and you won't ever get caught by having added columns to the table which should not have appeared in output on screen.