Converting mysql code to PDO gives no output

Question

I am trying to convert the code given here to PDO using OOP approach. This is what I've got so far:

comments.PHP:

public function loadComments() {
              $sql = "SELECT * FROM `comments`
                      WHERE
                      `comments`.`ImageID`  = :imageid ;";

            try
            {
                $imageid = $_REQUEST['imageid'];



                $query = $this->_db->prepare($sql);
                $params = array(':imageid' => $imageid);
                $query->execute($params); 

                for ($x = 0, $row = $query->fetch(PDO::FETCH_ASSOC); $x < $row; $x++) {
                 $comments[$x] = array("name" => $row["name"], "comment" => $row["comment"], "date" => $row["date"]);       
                    }

                $response = $_GET["jsoncallback"] . "(" . json_encode($comments) . ")";
                echo $response;
                return TRUE;

            }
            catch(Exception $ex)
            {
                return FALSE;
            }
    }

Firebug throws the undefined variable: comments error.

This is the original code:

  $query = mysql_query("SELECT
                          * FROM `comments`
                          WHERE
                          `comments`.`ImageID`  = '$imageid' ;");

        //loop through and return results
      for ($x = 0, $numrows = mysql_num_rows($query); $x < $numrows; $x++) {
            $row = mysql_fetch_assoc($query);

            $comments[$x] = array("name" => $row["name"], "comment" => $row["comment"], "date" => $row["date"]);            
        }

        //echo JSON to page
        $response = $_GET["jsoncallback"] . "(" . json_encode($comments) . ")";
        echo $response;

Where have I gone wrong?

Michael Berkowski · Accepted Answer · 2011-05-20 19:58:00Z

2

You are using $x < $row when I think you intend to use $x < $numrows

for ($x = 0, $row = $query->fetch(PDO::FETCH_ASSOC); $x < $row; $x++)
                                                          ^^^^^
$numrows = $query->rowCount();
for ($x = 0, $row = $query->fetch(PDO::FETCH_ASSOC); $x < $numrows; $x++)

This whole loop could be better written this way:

while ($row = $query->fetch(PDO::FETCH_ASSOC)) {
  $comments[] = array("name" => $row["name"], "comment" => $row["comment"], "date" => $row["date"]);       
}

There's no need for the $x counter if you use the $comments[] syntax, since that will append each new row with a numeric key onto the array.

answered May 20, 2011 at 19:58

Michael Berkowski

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Comments

Michael · Accepted Answer · 2011-05-20 19:59:21Z

0

Your for loop is wrong. You need to get the number of rows, then call $query->fetch() on every loop iteration:

$numrows = //...
for ($x = 0; $x < $numrows; $x++) {
  $row = $query->fetch(PDO::FETCH_ASSOC);
  $comments[$x] = array("name" => $row["name"], "comment" => $row["comment"], "date" => $row["date"]);       
}

answered May 20, 2011 at 19:59

Michael

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Comments

George Cummins · Accepted Answer · 2011-05-20 20:00:28Z

0

The following line has a syntax error (comma instead of semicolon):

for ($x = 0, $row = $query->fetch(PDO::FETCH_ASSOC); $x < $row; $x++) {

It should be:

for ($x = 0; $row = $query->fetch(PDO::FETCH_ASSOC); $x < $row; $x++) {

answered May 20, 2011 at 20:00

George Cummins

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1 Comment

Michael Berkowski Over a year ago

That comma syntax is actually valid. php.net/manual/en/control-structures.for.php

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