如何将mysql表中的两个值放入一个变量中？答案

【问题标题】：How can I put two values from mysql table into one variable?如何将mysql表中的两个值放入一个变量中？
【发布时间】：2018-07-22 15:13:24
【问题描述】：

这是我的查询的样子：

$pdo = $db->prepare('SELECT *,
s.name as name,
s.firstname as firstname,
FROM projects pj
LEFT JOIN school s ON pj.teacher=s.id 
WHERE project_id = :project_id ');  
$pdo->execute(array('project_id' => $project_id));
while ($row = $pdo->fetch(PDO::FETCH_ASSOC)) {
    echo $row["firstname"].' '.$row["name"];
}

我想知道是否有办法将两个值放入一个变量中。我想像这样：

$pdo = $db->prepare('SELECT *,
s.firstname.''.name as teacher,
FROM projects pj
LEFT JOIN school s ON pj.teacher=s.id 
WHERE project_id = :project_id ');  
$pdo->execute(array('project_id' => $project_id));
while ($row = $pdo->fetch(PDO::FETCH_ASSOC)) {
    echo $row["teacher"];
}

【问题讨论】：

你可以使用 concat..

标签： mysql select pdo

【解决方案1】：

如果你使用的是mysql，你可以concat来拼接：

SELECT *,
CONCAT(s.firstname, s.name) as teacher,
FROM projects pj
LEFT JOIN school s ON pj.teacher=s.id 
WHERE project_id = :project_id

http://www.mysqltutorial.org/sql-concat-in-mysql.aspx

【讨论】：

这将输出 firstnamename 作为你忘记包含空格的教师列。
如果你看原来的问题，没有空格，所以我在做同样的事情s.firstname.''.name as teacher

【解决方案2】：

你可以使用 MySQL 的 CONCAT：

$pdo = $db->prepare('SELECT *,
CONCAT(s.firstname, ' ', s.name) AS teacher
FROM projects pj
LEFT JOIN school s ON pj.teacher=s.id 
WHERE project_id = :project_id ');  

$pdo->execute(array('project_id' => $project_id));
while ($row = $pdo->fetch(PDO::FETCH_ASSOC)) {
    echo $row["teacher"];
}

【讨论】：