Question

我做错了什么。它没有回音。期望输出是苹果和梨，val好和0。首先，我将获得table_1上价格为4的所有水果，然后我加入table_2，排除任何水果标记= 4，排除香蕉，然后我加入table_3，得到val如果因子匹配56，所以苹果匹配，和梨因子不匹配，所以返回0但是我的代码没有回音，没有看到错误，但只是回音没什么。

＆＃13;

/*
table_1
fruit  price
apple   4
apple   5
banana  4
banana  5
pear    4

table_2
fruit   mark
apple    5
apple    6
banana   4
banana   7
pear     6

table_3
fruit   factor  val
apple    56     good
apple    60      OK
banana   89     good
banana   90     good
pear     56      bad
*/
$pri=4;
$sql = $wpdb->get_results( $wpdb->prepare("
				 SELECT 
				 b.fruit,c.val
				 FROM table_1 a,
				 LEFT JOIN table_2 b
				  ON b.fruit=a.fruit and b.mark != '4'
				 LEFT JOIN table_3 c
				  ON c.fruit=a.fruit AND c.factor = '56'
                WHERE a.price=%d
				 ",$pri));
print_r(sql);

＆＃13;

Answer 1

在FROM之后删除（，）（SELECT DISTINCT fruit FROM table_1 WHERE price =％d） a，

 /*table_1
 fruit  price
 apple   4
 apple   5
 banana  4
 banana  5
 pear    4

table_2
fruit   mark
apple    5
apple    6
banana   4
banana   7
pear     6

table_3
fruit   factor  val
apple    56     good
apple    60      OK
banana   89     good
banana   90     good
pear     56      bad
*/
$pri=4;
$sql = $wpdb->get_results( $wpdb->prepare("
                 SELECT 
                 b.fruit,c.val
                 FROM (SELECT DISTINCT fruit FROM table_1 WHERE price=%d) a
                 LEFT JOIN table_2 b
                  ON b.fruit=a.fruit and b.mark != '4'
                 LEFT JOIN table_3 c
                  ON c.fruit=a.fruit AND c.factor = '56'
                 ",$pri));
print_r(sql);

输出是：

fruit   val     
apple   good
apple   good
pear    bad
banana  NULL

加入三个表mysql php，但没有回音

1 个答案: