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Code (PHP)
<?php
$query = mysql_query("SELECT brandname_id, FROM tbl_brandname");
while( $row = mysql_fetch_array($query) ){
$query2 = mysql_query("SELECT COUNT(product_id) FROM tbl_product WHERE brandname_id ={$row[0]}");
$p_number = mysql_result($query2, 0);
echo $row[1].' '.$p_number, '<br>';
}
?>
ถ้าคุณเก็บสินค้าทุกชิ้นไว้ใน tbl_product นะ
http://gunner.freetzi.com
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| Date :
2009-06-05 22:09:23 |
By :
pjgunner |
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ขอบคุณมากครับ ท่านเทพ ขอหยิบไปลองเล่นดูนะครับ ขอบคุณครับ
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| Date :
2009-06-05 23:01:54 |
By :
4902520053 |
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ขอบคุณครับ
ขอศึกษาด้วยคน
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| Date :
2009-06-07 19:26:52 |
By :
zeaous |
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ลองแบบนี้ดู
Code (PHP)
<?php
$query = mysql_query("SELECT DISTINCT Count(tbl_brandname.brandname_id) AS `Sum`, tbl_brandname.name FROM tbl_brandname Inner Join tbl_product ON tbl_brandname.brandname_id = tbl_product.brandname_id GROUP BY tbl_product.brandname_id, tbl_brandname.name");
while( $row = mysql_fetch_array($query) ){
echo $row[$tbl_brandname.name]."--''.$row[$Sum]."<br>";
?>
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| Date :
2009-06-08 15:53:20 |
By :
nongjen |
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โทษทีผิดนิดนึง
Code (PHP)
<?php
$query = mysql_query("SELECT DISTINCT Count(tbl_brandname.brandname_id) AS `Sum`, tbl_brandname.name FROM tbl_brandname Inner Join tbl_product ON tbl_brandname.brandname_id = tbl_product.brandname_id GROUP BY tbl_product.brandname_id, tbl_brandname.name");
while( $row = mysql_fetch_array($query) ){
echo $row[tbl_brandname.name]."--''.$row[Sum]."<br>";
?>
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| Date :
2009-06-08 16:05:05 |
By :
nongjen |
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------SQL Server นะครับ ลองดู
select count(n.brandname),n.name
from tbl_brandname as n
inner join
tbl_product as p
on n.brandname = p.brandname_id
group by sum(n.brandname),n.name
having count(*) > 1
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| Date :
2009-07-26 22:48:18 |
By :
Guriko10 |
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ลองดูแบบนี้
Select a.brandname_id,a.name,count(*) as CNT
from tbl_brandname a
inner join tbl_product b
on a.brandname_id = b.brandname_id
group by a.brandname_id,a.name
ได้หรือป่าวน่า
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| Date :
2009-07-27 09:39:15 |
By :
taobsd |
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ลองดูแบบนี้
Select a.brandname_id,a.name,count(*) as CNT
from tbl_brandname a
inner join tbl_product b
on a.brandname_id = b.brandname_id
group by a.brandname_id,a.name
ได้หรือป่าวน่า
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| Date :
2009-07-27 09:39:44 |
By :
taobsd |
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