Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result

Question

coubury 0 Newbie Poster

17 Years Ago

Hello.

Im getting this error on a product page on my website

Quote

'Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/supremew/public_html/catalog/osc/products_new.php on line 89'

?>
    <tr>
    <td width="<?php echo SMALL_IMAGE_WIDTH + 10; ?>" valign="top" class="main"><?php echo '<a href="' . tep_href_link(FILENAME_PRODUCT_INFO, 'products_id=' . $products_new['products_id']) . '">' . tep_image(DIR_WS_IMAGES . $products_new['products_image'], $products_new['products_name'], SMALL_IMAGE_WIDTH, SMALL_IMAGE_HEIGHT) . '</a>'; ?></td>
    <td valign="top" class="main"><?php echo '<a href="' . tep_href_link(FILENAME_PRODUCT_INFO, 'products_id=' . $products_new['products_id']) . '"><b><u>' . $products_new['products_name'] . '</u></b></a><br>' . TEXT_DATE_ADDED . ' ' . tep_date_long($products_new['products_date_added']) . '<br>' . TEXT_MANUFACTURER . ' ' . $products_new['manufacturers_name'] . '<br><br>' . TEXT_PRICE . ' ' . $products_price; ?></td>
    <!--Start JLG Disable Buy Now and Shopping Cart-->
    <?php
    ?>
    <tr>
    <td width="<?php echo SMALL_IMAGE_WIDTH + 10; ?>" valign="top" class="main"><?php echo '<a href="' . tep_href_link(FILENAME_PRODUCT_INFO, 'products_id=' . $products_new['products_id']) . '">' . tep_image(DIR_WS_IMAGES . $products_new['products_image'], $products_new['products_name'], SMALL_IMAGE_WIDTH, SMALL_IMAGE_HEIGHT) . '</a>'; ?></td>
    <td valign="top" class="main"><?php echo '<a href="' . tep_href_link(FILENAME_PRODUCT_INFO, 'products_id=' . $products_new['products_id']) . '"><b><u>' . $products_new['products_name'] . '</u></b></a><br>' . TEXT_DATE_ADDED . ' ' . tep_date_long($products_new['products_date_added']) . '<br>' . TEXT_MANUFACTURER . ' ' . $products_new['manufacturers_name'] . '<br><br>' . TEXT_PRICE . ' ' . $products_price; ?></td>
    <!--Start JLG Disable Buy Now and Shopping Cart-->
    <?php
    $nononline_items = array();
    $sql = "SELECT products_id FROM products WHERE disable_addtobasket='Y'";
    $result = mysql_query($sql);
    while($id = mysql_fetch_array($result,MYSQL_ASSOC)){
    array_push($nononline_items,$id['products_id']);
    }
    if (in_array($products_new['products_id'],$nononline_items)){
    echo ("<td align=\"right\">". NON_ONLINE_ITEM ."</td>");
    }
    else{
    echo "<td align=\"right\" valign=\"middle\" class=\"main\">";
    echo '<a href="' . tep_href_link(FILENAME_PRODUCTS_NEW, tep_get_all_get_params(array('action')) . 'action=buy_now&products_id=' . $products_new['products_id']) . '">' . tep_image_button('button_in_cart.gif', IMAGE_BUTTON_IN_CART) . '</a>';
    echo "</td>";
    }
    ?>
    <!--End JLG Disable Buy Now and Shopping Cart-->
    $sql = "SELECT products_id FROM products WHERE disable_addtobasket='Y'";
    $result = mysql_query($sql);
    while($id = mysql_fetch_array($result,MYSQL_ASSOC)){
    array_push($nononline_items,$id['products_id']);
    }
    if (in_array($products_new['products_id'],$nononline_items)){
    echo ("<td align=\"right\">". NON_ONLINE_ITEM ."</td>");
    }
    else{
    echo "<td align=\"right\" valign=\"middle\" class=\"main\">";
    echo '<a href="' . tep_href_link(FILENAME_PRODUCTS_NEW, tep_get_all_get_params(array('action')) . 'action=buy_now&products_id=' . $products_new['products_id']) . '">' . tep_image_button('button_in_cart.gif', IMAGE_BUTTON_IN_CART) . '</a>';
    echo "</td>";
    }
    ?>
    <!--End JLG Disable Buy Now and Shopping Cart-->

In red is line 89

Can anyone help? I really have no idea what to edit

Thanks.

mysql php

2 Contributors
6 Replies
238 Views
22 Hours Discussion Span
Latest Post 17 Years Ago Latest Post by Menster

Dani AI

Generated 8 Months Ago

That warning means the value you pass to mysql_fetch_array() is not a valid result resource — in practice mysql_query() returned false (a failure) and you handed that boolean to mysql_fetch_array(). was right that the result set is the starting point. Below are focused, practical steps to find and fix the problem, plus a safer pattern to avoid the warning.

Start checklist (fast):

Confirm your DB connection actually succeeded before running queries.
Find the query that builds the list of products you use to populate $nononline_items and check it immediately after mysql_query() — don’t call mysql_fetch_* unless the query succeeded.
Remove the duplicated query block in the posted code (you run the same query twice); build the array once and reuse it.
Avoid displaying raw DB errors to visitors; log them instead.

Safer pattern (example — switch to mysqli when possible):

$conn = mysqli_connect(DB_HOST, DB_USER, DB_PASS, DB_NAME);
if (!$conn) { error_log('DB connect error: ' . mysqli_connect_error()); }

$res = mysqli_query($conn, $sql);
if (!$res) {
  error_log('Query failed: ' . mysqli_error($conn) . ' SQL: ' . $sql);
} else {
  $nononline_items = array();
  while ($row = mysqli_fetch_assoc($res)) {
    $nononline_items[] = $row['products_id'];
  }
  mysqli_free_result($res);
}

Extra notes tied to this thread: ’s file is osCommerce-based (products_new.php). If you prefer the platform API, use the built-in DB wrapper (the tepdb helpers) so you stay consistent with osCommerce’s connection and error handling. Finally, if the site runs PHP 7+, the old `mysql_extension is removed — plan to migrate tomysqliorPDO` to avoid future breakage.

All 6 Replies

Menster 38 Junior Poster

17 Years Ago

Hi there,
The problem that you are most likely facing is that the result set your database is returning is null. Please checkout http://www.daniweb.com/forums/thread201892.html, it should set you straight.

Menster 38 Junior Poster

17 Years Ago

Please look for pre-solved problems before reposting them.

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coubury 0 Newbie Poster · Answer 1 · 2009-07-06T20:24:01+00:00

Do i just paste this code in my php files?

1.
      $query = 'SELECT * FROM `security` LIMIT 0 , 30';
   2.
      $result = mysql_query($query, $conn);
   3.
      if (mysql_numrows($result) == 0)
   4.
      {
   5.
      echo "Result set is EMPTY";
   6.
      } else if (!$result) {
   7.
      echo "Result set is NULL";
   8.
      } else {
   9.
      echo "Result set is VALID";
  10.
      $row = mysql_fetch_assoc($result);
  11.
      $a = array_search($user, $row);
  12.
      echo $a;
  13.
      }

Menster 38 Junior Poster · Answer 2 · 2009-07-06T20:31:24+00:00

Pretty much, all you need to do is replace the sql with your own and put all of your data handling where I've specified:

$query = '[B]YOUR SQL HERE[/B]';
      $result = mysql_query($query, $conn);
      if (mysql_numrows($result) == 0)
      {
      echo "Result set is EMPTY";
      } else if (!$result) {
      echo "Result set is NULL";
      } else {
      echo "Result set is VALID";
      $row = mysql_fetch_row($result);
      [B]YOUR DATA HANDLING HERE[/B]      
      }

Good luck.

coubury 0 Newbie Poster · Answer 3 · 2009-07-06T20:37:19+00:00

Sorry, Im still a novice, what do you mean by data handling?

I take it my 'SQL' is the database?

Menster 38 Junior Poster · Answer 4 · 2009-07-07T12:32:07+00:00

Data handling is what you want to do to your data one you've extracted it from the database, display it in a table, perform calculations etc.

Your SQL is the statement you execute to get the data out of the database (eg "SELECT * FROM table_name WHERE condition bla bla bla)

On a side note, osCommerce is one of the most complex third party programs I have had the chance to work with, and if you are a php novice it might be a better idea to cut your teeth on something a little simpler, or at least avoid customising it as much as possible.

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result

Dani AI

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