将PHP变量插入SQL Select查询输出

Question

I have a dynamic php variable that I would like to join to every row in a column of an sql query. I know that the following works to add as static variable:

$output = "SELECT
'example' as extra_column,
order_date as order_date
FROM $order_table"

would return:

extra_column     order_date
example          01/01/2017
example          02/01/2017
example          02/01/2017

but I cant figure out how to make the field in the extra column dynamic, for example I've tired

$output = "SELECT
'"$dynamic_variable"' as extra_column,
order_date as order_date
FROM $order_table"

but this breaks the query, anyone got any ideas?

Answer 1

尝试使用点将字符串连接起来。

$output = "SELECT '".$dynamic_variable."' as extra_column, order_date as order_date FROM $order_table"

Answer 2

You have syntax error.

There should be dots when you concat string and variable.

$dynamic_variable = 'example';

$output = "SELECT
'" . $dynamic_variable . "' as extra_column,
order_date as order_date
FROM $order_table";