mysqli使用mysql变量报错

Question

我终于在这里创建了我的帐户。你帮助我很久了，但这次我在这里找不到答案。

我是 MySqli 的新手，我正在尝试运行在 phpMyAdmin 中工作的脚本。

当我尝试使用“mysql 变量”时，就会出现问题。代码和错误如下。

$consulta = "SET @var= (SELECT ticket_id FROM ost_ticket WHERE number = 410037);

SELECT DISTINCT a.number, a.created, b.address, 
SELECT DISTINCT title FROM ost_ticket_thread a WHERE a.ticket_id = @var AND a.source LIKE  'API')title, 
SELECT DISTINCT body FROM ost_ticket_thread a WHERE a.ticket_id = @var AND a.source LIKE  'API')body,
d.state, 
c.body resposta
FROM ost_ticket a
LEFT JOIN ost_user_email b ON b.user_id = a.user_id
LEFT JOIN ost_ticket_thread c ON c.ticket_id = a.ticket_id
LEFT JOIN ost_ticket_status d ON d.id = a.status_id
WHERE a.ticket_id = @var AND c.id = (SELECT MAX( a.id ) FROM ost_ticket_thread a WHERE a.ticket_id = @var ) ";


$resultado = $MySQLi->query($consulta) OR trigger_error($MySQLi->error, E_USER_ERROR);

$cont = 0;

while ($informacao = $resultado->fetch_object()) {


$data[$cont] = array(
'numprot'      =>      $informacao->number,
'email'     =>      $informacao->address,
'assunto'     =>      $informacao->title,
'status'     =>      $informacao->state,
'body'     =>      $informacao->body,
'resposta'     =>      $informacao->resposta,
'dtcriacao'     =>      $informacao->created,
);

$cont++;

}

echo "{'lista':".json_encode($data)."}";

正如我所说，它在 phpMyAdmin 中运行良好，但在 mysqli 查询函数中出现以下错误。

Fatal error: You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'SELECT distinct a.number,a.created,b.address, (select distinct title from ost_t' at line 3 in /home/u605553750/public_html/JSONResposta.php on line 22

第 22 行匹配在

 $resultado = $MySQLi->query($consulta) OR trigger_error($MySQLi->error, E_USER_ERROR);

在 MySqli 上运行它的正确方法是什么？

谢谢 布鲁诺

Answer 1

好像你忘记了那行的发件人。

SELECT DISTINCT a.number,a.created,b.address [FROM YOURTABLE]<-add this, (... 

经过澄清，我想我更了解您的意图，在这种情况下，您应该确保您的子查询返回一个 select_expression。注意到“AS”了吗？

SELECT DISTINCT a.number,a.created,b.address, 
(SELECT DISTINCT title FROM ost_ticket_thread a WHERE a.ticket_id = @var AND a.source LIKE  'API') AS title

http://dev.mysql.com/doc/refman/5.7/en/select.html