从多个表中抓取信息

Question

目前我能够从食谱表中拉出食谱名称，但我希望能够从配料表中获取所需的配料。我知道这是关于JOINS的，但我是JOINS的新手。

这是成分表

这是recipeingredients表，这主要有两个键，以便我能够多种成分分配到一个配方

这是配方表

这是搜索脚本

<?php 
    $query = $_GET['query']; 
    // gets value sent over search form 

    $min_length = 3; 


    if(strlen($query) >= $min_length){ 

     $query = htmlspecialchars($query); 


     $query = mysql_real_escape_string($query); 
     // makes sure nobody uses SQL injection 

     $raw_results = mysql_query("SELECT * FROM recipes 
      WHERE (`recipename` LIKE '%".$query."%') OR (`ingredients` LIKE '%".$query."%')") or die(mysql_error()); 





     if(mysql_num_rows($raw_results) > 0){ // if one or more rows are returned do following 

      while($results = mysql_fetch_array($raw_results)){ 
      // $results = mysql_fetch_array($raw_results) puts data from database into array, while it's valid it does the loop 

       echo "<p>Recipe:".$results['recipename']."</p><p>Ingredients:".$results['ingredients']."<p>Instructions:".$results['instructions']."</p>"; 
       // posts results gotten from database(title and text) you can also show id ($results['id']) 
      } 

     } 
     else{ // if there is no matching rows do following 
      echo "No results"; 
     } 

    } 
    else{ // if query length is less than minimum 
     echo "Minimum length is ".$min_length; 
    } 
?> 

个

成分采样数据

recipeingredients样本数据

配方表的样本数据

Answer 1

SELECT 
    r.*, 
    i.* 
FROM recipe AS r 
INNER JOIN recipeingredients AS ri 
    ON ri.recipeid = r.recipeid 
INNER JOIN ingredients AS i 
    ON i.ingredientid = ri.ingredientid 
WHERE r.recipename = 'Beans On Toast' 

这会给你配方及其配方。

EDITS

下面介绍如何做到这一点。

$query =" SELECT 
       r.*, 
       i.* 
      FROM recipe AS r 
      INNER JOIN recipeingredients AS ri 
       ON ri.recipeid = r.recipeid 
      INNER JOIN ingredients AS i 
       ON i.ingredientid = ri.ingredientid 
      WHERE r.recipename = 'Beans On Toast'"; 

$raw_results = mysqli_query($query) or die(mysqli_error());