MySQL查询执行失败

Question

为什么此代码在数据库中不起作用？

$sql = 
UPDATE users_information INNER JOIN users ON users.id = user_id SET following_count = :following_count WHERE user_id = :user_id;

$ data：

["following_count" => "following_count + 1" , "user_id" => "1273"]

执行return false

Answer 1

following_count + 1是表达式，您不能绑定表达式，只能绑定值。因此，将查询重写为：

$sql = "
UPDATE users_information INNER JOIN users ON users.id = user_id 
SET following_count = following_count + 1 
WHERE user_id = :user_id";

在$data中，您仅传递user_id：

$data = ["user_id" => "1273"];

此外，我不清楚使用联接的意义是什么，为什么不进行简单查询：

$sql = "
UPDATE users_information 
SET following_count = following_count + 1 
WHERE user_id = :user_id";

Answer 2

[“ following_count” =>“ following_count + 1”，“ user_id” =>“ 1273”]。

  

“ following_count +1”是一个字符串。

将其设为整数。或者，如果它是一个字符串，则在数据库中更改类型 作为varchar。

数据应类似于以下内容：

$following_count = $following_count + 1;
["following_count" => "some_number(from following_count value)" , "user_id" => "1273"]