php：对于一个mysql表中的每个记录，显示第二个表中的所有记录

Question

I have 2 tables, one with email addresses, and one with a number of rows per email address:

Table 1:
1@gmail.com
2@gmail.com
etc

Table 2:
1@gmail.com,value111,value112,value113
1@gmail.com,value121,value122,value123
2@gmail.com,value211,value212,value213
etc.

I want to send each of the email addresses their values, so I get the email:

$query = "SELECT email FROM email_table";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result)){
$email = $row['email'];

and then, I need to get the values for each email. I tried using foreach but apparently not in a correct way Can anybody help?

Many thanks!

Answer 1

$query = "SELECT * FROM email_table AS table1 LEFT JOIN table2 ON table2.email = table1.email";

--edit-- after clarification, you want to get the info from table 2 attached to the email address in table 1 and send an email to table 1...

$query = "SELECT table1.email, table2.* FROM email_table AS table1 LEFT JOIN table2 ON table2.email = table1.email";

Your result here is the email field from table 1 with all information from table 2 joined on the same email address. You don't need both emails returned here so in your field list, put something like this instead of table2.* :

SELECT table1.email, table2.field1, table2.field2, table2.field3

Then you can use PHP's mail to send to the email you get from table1.

Answer 2

You are wanting to do a JOIN . So your query will look like this:

SELECT * FROM table2 JOIN table1 ON (table2.email = table1.email);

Coding Horror has a great explanation of all the join types between the tables.