I have two tables in mysql that contain details about users such as email, first name and last name.
我在mysql中有两个表,其中包含有关用户的详细信息,如电子邮件,名字和姓氏。
and I made a way to welcome the user that login but the problem is that some value in the two tables are similar which cause duplication, and I want to know how to differ between values from different tables.
我提出了一种方法来欢迎用户登录,但问题是两个表中的某些值是相似的,这会导致重复,我想知道如何区分来自不同表的值。
I tried this code but it didn't work:
我试过这段代码,但它不起作用:
$query= "SELECT * FROM Table1 WHERE email = '".$_SESSION['Email']."' NOT IN (SELECT email = '".$_SESSION['Email']."' FROM Table2)";
2 个解决方案
#1
1
Your query is:
您的查询是:
SELECT *
FROM Table1
WHERE email = '".$_SESSION['Email']."' NOT IN (SELECT email = '".$_SESSION['Email']."' FROM Table2)
The expression email = '".$_SESSION['Email']."'
is a boolean expression -- essentially 0
or 1
.
表达式email ='“。$ _ SESSION ['Email']。”'是一个布尔表达式 - 基本上是0或1。
SELECT *
FROM Table1
WHERE email = '".$_SESSION['Email']."' and
email NOT IN (SELECT email from Table2 where email = '".$_SESSION['Email']."')
Because the comparisons are the same, this is more efficiently written as:
因为比较是相同的,所以更有效地写为:
SELECT *
FROM Table1
WHERE email = '".$_SESSION['Email']."' and
not exists (SELECT email from Table2 where email = '".$_SESSION['Email']."')
Having an index on email
in both tables will speed performance.
在两个表中都有一个电子邮件索引可以提高性能。
#2
0
Have the two tables the same email? Why?
这两个表有相同的电子邮件吗?为什么?
Ok here is maybe a solution if you want only from table 1 the user data from you session email try this:
好的,这里可能是一个解决方案,如果你只想从表1中获得会话电子邮件中的用户数据,请尝试以下方法
$query = "SELECT t1.*, t2.* FROM table t1, table t2 WHERE t1.email <> t2.email AND t1.email = '".$_SESSION["Email"]."'";
Recommend: Try to select what you really need... email, username, name etc... not SELECT *
推荐:尝试选择你真正需要的东西......电子邮件,用户名,姓名等...不是SELECT *
#1
1
Your query is:
您的查询是:
SELECT *
FROM Table1
WHERE email = '".$_SESSION['Email']."' NOT IN (SELECT email = '".$_SESSION['Email']."' FROM Table2)
The expression email = '".$_SESSION['Email']."'
is a boolean expression -- essentially 0
or 1
.
表达式email ='“。$ _ SESSION ['Email']。”'是一个布尔表达式 - 基本上是0或1。
SELECT *
FROM Table1
WHERE email = '".$_SESSION['Email']."' and
email NOT IN (SELECT email from Table2 where email = '".$_SESSION['Email']."')
Because the comparisons are the same, this is more efficiently written as:
因为比较是相同的,所以更有效地写为:
SELECT *
FROM Table1
WHERE email = '".$_SESSION['Email']."' and
not exists (SELECT email from Table2 where email = '".$_SESSION['Email']."')
Having an index on email
in both tables will speed performance.
在两个表中都有一个电子邮件索引可以提高性能。
#2
0
Have the two tables the same email? Why?
这两个表有相同的电子邮件吗?为什么?
Ok here is maybe a solution if you want only from table 1 the user data from you session email try this:
好的,这里可能是一个解决方案,如果你只想从表1中获得会话电子邮件中的用户数据,请尝试以下方法
$query = "SELECT t1.*, t2.* FROM table t1, table t2 WHERE t1.email <> t2.email AND t1.email = '".$_SESSION["Email"]."'";
Recommend: Try to select what you really need... email, username, name etc... not SELECT *
推荐:尝试选择你真正需要的东西......电子邮件,用户名,姓名等...不是SELECT *