My query is throwing an error

[Jacobb123]

I am running the following query but keep getting this error message:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/jacobb2/public_html/includes/people.class.php on line 56

The query runs fine and displays the info that is requested but always has the error above it. I hope someone can help me out with this

here is what is on line 56:
function get_num_images($member_id)
{
$sql="select count(*) as a from photos where member_id = $member_id";
$res=mysql_query($sql);
print mysql_error();
$data_set=mysql_fetch_array($res);
return $data_set["a"];
}


Here is the code that is calling that function:

<?php
function selectMemberDetails($member_id){
$resultMember = mysql_query("SELECT * FROM members WHERE member_id='".$member_id."'");
$member = mysql_fetch_array($resultMember);
return $member;
}

$featureResult = mysql_query("SELECT * FROM featured_moviemaker ORDER BY id DESC LIMIT 1")or die(mysql_error());
$Feature = mysql_fetch_array($featureResult);
$Member = selectMemberDetails($Feature['member_id']);

$num_images=$people->get_num_images($Feature["member_id"]);
if($num_images==0)
{
$gender=$people->check_gender($Feature["member_id"]);
if($gender=="Male")
{
$image="<img alt='' src='images/male.gif' width='90' />";
}
else
{
$image="<img alt='' src='images/female.gif' width='90' />";
}
}
else
{
$image_url=$people->get_image($Feature["member_id"]);
$pic_name=str_replace('user_images/', '', $image_url);
$image = "<img src='image_gd/image.php?$pic_name' border='0'>";
}

if($Member["display_name"] == ""){
$Name = $Member['member_name'];
}else{
$Name = $Member["display_name"];
}

if(mysql_num_rows($featureResult) > 0){
?>
<div style="margin:8px; float:left; height:100px; width:100px;">
<a href="view_profile.php?member_id=<?php echo $Feature['member_id']; ?>">
<?php echo $image; ?>
</a> </div>
<div class="style6" style="margin:8px; float:left; width: 285px;">
<a href="view_profile.php?member_id=<?php echo $Feature['member_id']; ?>">
<?php echo $Name;?>
</a> </div>
<span class="style5" style="margin:8px; float:left; width: 285px; text-align:justify;">
<?php echo $Feature['bio']; ?></span>

<?php
}else{
echo "No Featured Movie Maker Selected";
}
?>

[_Aerospace_Eng_]

$sql="select count(*) as a from photos where member_id = '$member_id'";
If $member_id is a string it needs to be in single quotes. I would actually setup your function like this
function get_num_images($member_id)
{
$sql="select count(*) as a from photos where member_id = '$member_id'";
$res=mysql_query($sql) or die('The mysql error was: ' . mysql_error(). '<br>The query was: ' . $sql);
$data_set=mysql_fetch_array($res);
return $data_set["a"];
}
So it gives you more information about the query.

[Jacobb123]

That was it thank you so much......programming drives me nuts sometimes you forget a semi-colon or single quote and everything gets messed up...Thanks again