Question

<?php 
include 'connect.php';
$query = "SELECT * FROM (
SELECT CONCAT(customer_name, ' ', address, ' ', street, ' ', phone_1, ' ', phone_2, ' ', phone_3, ' ', phone_4) as `mysearch`
FROM customers) base
WHERE `mysearch` LIKE '%102%'";

$result = mysql_query($query);
while ($row = mysql_fetch_array($result)) {
echo "customer name: " . $row['customer_name'];    
}
?>

查询正在运行，但我错过了能够回显结果的内容。它在phpmyadmin中运行时返回结果。数据库中有2个条目应该返回我的实际输出如下：

客户名称：客户名称：

表示它找到了2个条目，但我无法让它们回显。 customer_name与数据库中的字段名称匹配。

Answer 1

如果您希望在结果中包含customer_name，则必须在子查询中将其选为分隔列：

SELECT 
    * 
    FROM (
        SELECT
            customer_name,
            CONCAT(customer_name, ' ', address, ' ', street, ' ', phone_1, ' ', phone_2, ' ', phone_3, ' ', phone_4) as `mysearch`
        FROM 
            customers
    ) base
WHERE 
    `mysearch` LIKE '%102%'

没有看到查询结果回显

1 个答案: