我如何在一条语句中使用两个mysql表

Question

So, I've put in a favourite images table, and I can't figure out how to get it working properly

Basically, this is the setup:

ImageFavs
FavID, UserID, ImgID

ImageList
ImgID, SiteID

I'd like it to select 20 or so images from the favourites table, but only ones that match the siteid in the image list table

This is the code I have at the moment, but it dawned on me it'd select 20 images from favourites, then only display them if the matching site was actually checked.

#select matching sites
for($i=0;$i<count($sites)-1;$i++){
    $siteinfo = explode("-",$sites[$i]);
    $siteid = $siteinfo[0];
    $sitegroup = $siteinfo[1];
    $selection[$siteid]="exists";
    if($i!=0){
        $sqlextra .= " OR ";
    }
    else{
        $sqlextra = "AND (";
    }
    $sqlextra .= "SiteID='".$siteid."'";
}
if(!empty($sqlextra)){
    $sqlextra .= ")";
}
else{
    $sqlextra = "AND SiteID='-1'";
}

#show favourites
if($_GET['f']==1){
    $sql="SELECT * FROM ImageFavs WHERE UserID='".$_SESSION['User_ID']."' AND Active = '1' ORDER BY RAND() LIMIT 20";
    #(rand is just me being lazy, eventually I'll figure out how to separate it onto pages)
    $result = mysql_query($sql);
    $num = mysql_numrows($result);
    if($num>0){
        while ($row=mysql_fetch_array($result, MYSQL_BOTH)){
            if(empty($sqlextra2)){
                $sqlextra2 = " AND (";
            }
            else{
                $sqlextra2 .= " OR ";
            }
            $sqlextra2 .= "ImgID='".$row['ImgID']."'";
        }
        $sqlextra2 .= ")";
    }
}
#don't show favourites
if(empty($sqlextra2)){
    $sqlextra2 = " ORDER BY RAND() LIMIT 20";
}
$sql="SELECT * FROM ImageList WHERE Valid='1' ".$sqlextra.$sqlextra2;

This output $sql from this seems like it could be so much neater though, an example of it is like this

SELECT * FROM ImageList WHERE Valid='1' AND (SiteID='6') AND (ImgID='5634' OR ImgID='8229' OR ImgID='9093' OR ImgID='5727' OR ImgID='7361' OR ImgID='5607' OR ImgID='7131' OR ImgID='5785' OR ImgID='7339' OR ImgID='5849' OR ImgID='7312' OR ImgID='5641' OR ImgID='8921' OR ImgID='7516' OR ImgID='7284' OR ImgID='5873' OR ImgID='8905' OR ImgID='7349' OR ImgID='7392' OR ImgID='8725')

Also, while I'm here, would there be a non resource heavy way to count the number of favourites for a user per website?

It's not for anything big, just messing around on a personal website to see what I can do.

Answer 1

您想使用“ JOIN”

SELECT * FROM ImageFavs LEFT JOIN ImageList ON ImageFavs.ImgID = ImageList.ImgID WHERE ImageList.SiteID = <your_site_id>

Answer 2

You can INNER JOIN your two tables to get the results you want. INNER is used when you want results from both tables. It's best to use aliases to keep your tables straight.

SELECT l.*
FROM ImageFavs f
    INNER JOIN ImageList l ON f.ImgID = l.ImgID
WHERE l.SiteID = [your site ID]
    AND f.UserID='" . $_SESSION['User_ID'] . "' 
    AND f.Active = '1' 
    ORDER BY RAND() LIMIT 20

To get a count by site you can use GROUP BY . I think this should get you that count

SELECT COUNT(f.ImgID)
FROM ImageFavs f
    INNER JOIN ImageList l ON f.ImgID = l.ImgID
WHERE f.UserID='" . $_SESSION['User_ID'] . "' 
    AND f.Active = '1' 
GROUP BY l.SiteID

Answer 3

This works-

//Assuming $site_id contains the site ID/
$query = "select *.IF from ImageFavs as IF, ImageList as IL where IL.ImgId = IF.ImgId and IL.SiteId = $site_id LIMIT 20"