php - MYSQL 查询忽略 php 中的结果

Question

编辑这个问题，因为它已经缩小了范围。如果我按照下面手动运行 SQL 查询，即使我知道 nfw_lead_messages_sent.reply = '2' 至少有 5 个应该显示的结果，它也不会显示任何结果。

SELECT
CONCAT(title,' ',first_name,' ',last_name) AS lead_name,
nfw_leads.mobile,
nfw_leads.lead_date,
nfw_leads.service_type,
nfw_leads.state,
nfw_lead_messages_sent.reply
FROM
nfw_users
LEFT JOIN nfw_leads ON nfw_leads.allocate_to = nfw_users.id_num
LEFT JOIN nfw_lead_messages_sent ON nfw_lead_messages_sent.sent_to_id = nfw_leads.allocate_to
WHERE
nfw_users.id_num = '3513'
GROUP BY
    nfw_leads.first_name,
    nfw_leads.last_name,
    nfw_leads.mobile,
    nfw_leads.service_type

HAVING nfw_lead_messages_sent.reply = '2'
Order by
    lead_date ASC

如果我手动运行 SQL 查询 nfw_lead_messages_sent.reply = '1' 则会显示结果。请参阅下面的内容，了解适用于 = '1' 但当我将其更改为 = '2' 时不起作用

SELECT
CONCAT(title,' ',first_name,' ',last_name) AS lead_name,
nfw_leads.mobile,
nfw_leads.lead_date,
nfw_leads.service_type,
nfw_leads.state,
nfw_lead_messages_sent.reply
FROM
nfw_users
LEFT JOIN nfw_leads ON nfw_leads.allocate_to = nfw_users.id_num
LEFT JOIN nfw_lead_messages_sent ON nfw_lead_messages_sent.sent_to_id = nfw_leads.allocate_to
WHERE
nfw_users.id_num = '3513'
GROUP BY
    nfw_leads.first_name,
    nfw_leads.last_name,
    nfw_leads.mobile,
    nfw_leads.service_type

HAVING nfw_lead_messages_sent.reply = '1'
Order by
    lead_date ASC

Answer 1

首先将 $data 转换为整数

   $data = intval($_GET["data"])); 

然后按如下方式使用

   if($data === 1) {
   } elseif($data === 2) {
   } elseif(empty($data)) {
   }