我有两张桌子
表personal_info具有以下信息
+--------+--------+--------+--------+
| id | fname | lname | email |
+--------+--------+--------+--------+
| | | | |
+--------+--------+--------+--------+
表
other_info具有以下信息+--------+-----------------+--------------+----------------+
| id | university_name | course_name | business_name |
+--------+-----------------+--------------+----------------+
| | | | |
+--------+-----------------+--------------+----------------+
现在，我想使用PHP来连接表并从两个表中获取信息，我决定这样做
<?php
include ('config.php');
$con = mysqli_connect($host, $user, $pass, $db) or die ('Cannot Connect : '.mysqli_error());
$sql = "select * from personal_info, other_info "; // Here i want to join the two tables to echo results
$result = mysqli_query($con,$sql) or die("Error: ".mysqli_error($con));
while( $row = mysqli_fetch_array( $result, MYSQLI_ASSOC ) ){
echo "$row['fname'] ."".$row['lname'] ."".$row['university_name'] ."".$row['course_name'] ."".$row['businessname'] ";
}
?>
如何加入表格并获取数据？
最佳答案
您需要首先在other_info表中设置一个外键，以便可以通过该键进行联接。
然后这样的选择应该工作
select fname, lname, university_name, course_name, business_name
from personal_info t1
inner join other_info t2 on (t1.id = t2.personal_id)
t2.personal_id是表other_info中的外键列
关于php - PHP MYSQL连接两个表并回显数据，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/45276589/