How do I show all the results from an SQL statement? - php

Question

I wanted to display all of my shirts but I always get the same error
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /homepages/36/d362778419/htdocs/home/specials.php on line 34
Here is the code I have currently in place
<?php
require "connect2.php";
$sql = mysql_query("SELECT * FROM shirts WHERE all='1'");
$runrows = mysql_fetch_array($sql);
$title = $runrows['title'];
$picture = $runrows['picture'];
$newp = $runrows['newp'];
$date = strftime("%b %d, %Y %l:%M %p" ,strtotime($runrows['date']));
$oldp=$runrows['oldp'];
$viewl=$runrows['viewl'];
$shirtl = $runrows['shirtl'];
echo "";
?>
<div class="specialsListBoxContents centeredContent back" style="width:25%;"><div class="product-col" > <div class="img">
<a href="detail.php?id=<?php echo $id; ?>"><img src="<?php echo $picture; ?>" alt="<?php echo $title; ?>" title=" <?php echo $title; ?> " width="190"
height="160" /></a> </div> <div class="prod-info"> <div class="wrapper">
<div class="price"> <strong><span class="normalprice"><?php echo $oldp; ?
></span><br /><span class="productSpecialPrice"><?php echo $newp; ?></span></strong> </div>
<div class="button"><a href="detail.php?id=<?php echo $id; ?>"><img src="images/button_add_to_cart.gif" alt="Add to Cart" title=" Add to Cart " width="54"
height="49" /></a></div> </div> </div> </div></div>
<?php
?>

Answer 1

mysql_*() functions return boolean FALSE if they fail, which means your query call did not succeed. Add this code to find out the reason why:
$sql = mysql_query("SELECT * FROM shirts WHERE all='1'");
if ($sql === FALSE) {
die(mysql_error());
}

Answer 2

Here's some sample code:
<?php
$sql = mysql_query("SELECT * FROM shirts WHERE `all`='1'");
if (!$sql ) {
echo "Could not successfully run query from DB: " . mysql_error();
exit;
}
if (mysql_num_rows($sql) == 0) {
echo "No rows found, nothing to print so am exiting";
exit;
}
// While a row of data exists, put that row in $row as an associative array
// Note: If you're expecting just one row, no need to use a loop
while ($row = mysql_fetch_assoc($sql))
{
?>
<a href="detail.php?id=<?php echo $row['id']; ?>"><img src="<?php echo $row['picture']; ?>" alt="<?php echo echo $row['title']; ?>" title=" <?php echo $row['title']; ?> " width="190" height="160" />
<?php
}
?>
And here is more info. Notice the exiting php mode inside the while.