想要在php中检索数据 [英] want to retrieve the data in php

问题描述

我正在建立一个餐厅的网站.在数据库中，我创建了两个表

I am making a website of a resturant. in database I have created two tables

1-food(id,food_name,cat_id(F.K),)
2-category(cat_id,cat_name)

我想在导航栏中显示类别，然后单击类别就必须显示该类别下受人尊敬的食物.

示例我的表中有以下条目

I want to display the category in navigation bar and on clicking the category it has to show the respected food that comes under the category.

example I have the following entries in my table

food_name=Zinger Burger,Chicken Burger,Sea Food
cat_name=FastFood,Fish

如何在菜单栏上将类别显示为超链接，并在单击时显示数据库中的食物，在单击快餐时它将显示生姜和鸡肉汉堡.

以下是我到目前为止已完成的工作

how can I show the category on menu bar as a hyperlink and on clicking the shows the food from database that is onclicking the fastfood it shows the zinger and chicken burger.

below is my work which I have done so far

?php  
 $query="Select * from category";
  $result=mysql_query($query,$connection);
  while($category=mysql_fetch_array($result))
  {?>
	   <ul>
        <li>
     <?php  $category["cat_name"]; <br mode="hold" /?>  
   
echo "<a href="\"show.php\"">  ".$category['cat_name'] ." </a>";
}?&gt;
</li>
</ul>
<?php <br mode="hold" /?>
//$food="Select * from food where cat_id={$category["cat_id"]} ";
        $query = "SELECT * ";
		$query .= "FROM subjects ";
		$query .= "WHERE cat_id=" . $category["cat_id"] ." ";

$result1=mysql_query($query,$connection);
  
while($foodname=mysql_fetch_array($result1))
   {
	echo"<li> {$foodname["food_name"]}</li>";
	}
echo "";

?>

推荐答案

query = " 从类别中选择* ;

query="Select * from category";

result = mysql_query(

result=mysql_query(

query，

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