-----------------------------------------
Table Name: product_cat
-----------------------------------------
id - int(11) - auto_increment
- -
cat_name - varchar(30) -
- -
branch_code- varchar(30) -
- -
date - datetime -
-----------------------------------------
Table Name: product_brand
-----------------------------------------
id - int(11) - auto_increment
- -
parent_id - int(11) -
- -
branch_name - varchar(40) -
- -
category - varchar(40) -
我通过此查询在product_brand
表中应用了外键(alter table Product_brand add约束ta1外键(parent_id)引用product_cat(id)
)
我正在通过以下方式在 product_brand
表中插入数据
<table>
<form action="index.php?tanzil=brand_added" method="post">
<tr>
<td>Brand Name:</td>
<td>
<input type="text" name="brand_name" />
</td>
</tr>
<tr>
<td>Category:</td>
<td>
<select name="cat_name">
<?php
db_connect();
$q="select * from product_cat order by cat_name";
$rs=mysql_query($q);
for($i=0;$i<mysql_num_rows($rs);$i++){
$rd=mysql_fetch_object($rs);
echo"<option value='$rd->id'>$rd->cat_name</option>";
}
?>
</select>
</td>
</tr>
<tr>
<td>
<input type="submit" value="submit" />
</td>
</tr>
</form>
</table>
并通过这种方式向数据库插入数据
db_connect();
$name_of_cat = $_POST['cat_name'];
$name_of_brand = $_POST['brand_name'];
$g ="insert into product_brand (parent_id,brand_name,category) values ('$name_of_cat','$name_of_brand','$name_of_cat')";
if(mysql_query($g)){
echo "Done";
}
else{
echo"Some error occured";
}
和 通过这种方式显示数据
$query= "select * from product_brand";
$result= mysql_query($query);
while($row=mysql_fetch_object($result))
{
echo $row['brand_name'];
}
但我没有从parent_id的product_cat表引用中获取cat_name名称
我怎样才能得到它?
最佳答案
JOIN
两个表:
SELECT
p.id,
p.branch_name,
c.cat_name
FROM product_Brand AS p
INNER JOIN Product_cat AS c ON c.id = c.parent_id;
关于php - 如何根据id显示两个表中的数据,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/15547502/