PHP/mysql加入

Question

我需要得到3个表的值，从我需要得到aff_id其中v_id = 5，从第二我需要得到用户id在哪里aff_id = first的aff_id，从第三我需要获取用户名，电子邮件，身份证其中ID =秒的aff_id。我无法编写正确的mysql查询来获取数据，请帮助我获取它。这是我的错代码

SELECT * FROM wp_vendor_affiliates WHERE vendor_id = 21 LEFT JOIN wp_affiliate_wp_affiliates 

SELECT wp_vendor_affiliates.affiliate_id , 
wp_affiliate_wp_affiliates.user_id 
FROM wp_vendor_affiliates 

INNER JOIN wp_affiliate_wp_affiliates INNER JOIN 

请hetp我，并更正我的查询。由于FOT帮助和支持

Answer 1

使用此代码

$id1 = 5; 
$query1 = "SELECT * FROM table1 WHERE v_id='{$id1}'"; 
$result1 = mysqli_query($con,$query1); 
$row1 = mysqli_fetch_array($result1); 
$id2 = $row1['aff_id']; 

$query2 = "SELECT * FROM table2 WHERE aff_id='{$id2}'"; 
$result2 = mysqli_query($con,$query2); 
$row2 = mysqli_fetch_array($result2); 
$id3 = $row2['aff_id']; 

$query3 = "SELECT * FROM table3 WHERE id='{$id3}'"; 
$result3 = mysqli_query($con,$query3); 
$row3 = mysqli_fetch_array($result3); 
$id2 = $row3['aff_id']; 

这里$ con是你的数据库

$con= mysqli_connect("localhost","root","","data_base"); 

我这里使用的连接到本地主机的连接。你必须改变最肯定的。它有点长，但基本是