连接mysql表以显示一个表字段,但根据类似字段的值选择另一个表字段 [英] join mysql tables to show one tables field but select another tables field based on the value of similar fields
问题描述
我有以下表格(与我的问题相关)
类别和产品
字段加入
类别 - 短,图片
产品 - 简短,图片
短=类别名称简称
image =图片名称
基本上我想根据产品中已经设置的图片选择一个类别的图像(所以当我选择我的下拉列表应该只有为该类别的产品设置的图像)
我的代码下面有两个问题 - 一个没有显示已经选中的类别中的字段和两个列出所有图像,而不仅仅是短字段值匹配的位置
代码:
<?php
$ qry = mysql_query(SELECT * FROM product,$ con);
if(!$ qry)
{
die(Query Failed:。mysql_error());
}
?>
选择类别图像
< select name =imageid =image>
< option value =>< / option>
<?php
//首先,定义所需的默认值
$ default_value = $ row ['image'];
while($ row = mysql_fetch_array($ qry))
{
//然后你可以在你的while循环中选择那个选择
$ selected =($ row ['image'] == $ default_value)? '选中':'';
echo< option value ='。 $ row ['image']。 '。 $ selected。 > 中。 $ row ['image']。 < /选项> 中;
}
?>
< / select>
qry = mysql_query(SELECT * FROM product,
CON);
if(!
qry)
{
die(Query Failed:。mysql_error());
}
?>
选择类别图像
< select name =imageid =image>
< option value =>< / option>
<?php
//首先,定义所需的默认值
i have the following tables (relevant to my question)
category and product
fields to join
category - short, image
product - short, image
short = short form of category name
image = image name
basically i want to select an image for category based on the images already set in product (so when i select my drop down should only have the images that were set for products for that category)
my code below has two problems - one it doesn't show the already selected field in category and two it lists all the images and not just where the short field value match
code:
<?php $qry=mysql_query("SELECT * FROM product", $con); if(!$qry) { die("Query Failed: ". mysql_error()); } ?> Select Category Image <select name="image" id="image"> <option value=""></option> <?php // First, define the desired default value $default_value = $row['image']; while($row=mysql_fetch_array($qry)) { // Then you can mark that one as selected in your "while" loop $selected = ($row['image'] == $default_value) ? ' selected' : ''; echo "<option value='" . $row['image'] . "'" . $selected . ">" . $row['image'] . "</option>"; } ?> </select>解决方案qry=mysql_query("SELECT * FROM product",
con); if(!
qry) { die("Query Failed: ". mysql_error()); } ?> Select Category Image <select name="image" id="image"> <option value=""></option> <?php // First, define the desired default value
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