[SOLVED] Sql Error

[Stealth549]

Hey everyone, getting a sql error that I cannot pin down.

 

 

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ' newspaper.id FROM subs , newspaper WHERE subs.news_id = newspaper.id AND newspa' at line 1

 

sql query

$sql = "SELECT subs.news_id, subs.cust_id, subs.sub_id, newspaper.name, newspaper.$day, newspaper.cost, newspaper.id FROM subs , newspaper WHERE subs.news_id = newspaper.id AND newspaper.$day = 'on'";  

[jaymc]

Could it be the comma?

 

$sql = "SELECT subs.news_id, subs.cust_id, subs.sub_id, newspaper.name, newspaper.$day, newspaper.cost, newspaper.id FROM subs , newspaper WHERE subs.news_id = newspaper.id AND newspaper.$day = 'on'"; 

 

 

$sql = "SELECT subs.news_id, subs.cust_id, subs.sub_id, newspaper.name, newspaper.$day, newspaper.cost, newspaper.id FROM subs, newspaper WHERE subs.news_id = newspaper.id AND newspaper.$day = 'on'"; 

 

If not, then try this

 

 

$sql = "SELECT subs.news_id, subs.cust_id, subs.sub_id, newspaper.name, newspaper.$day, newspaper.cost, newspaper.id FROM subs INNER JOIN newspaper ON subs.news_id = newspaper.id AND newspaper.$day = 'on'"; 

[Stealth549]

there is still an error with all that. I have another query that is similar but works, all that is different is that i dont have newspaper.cost.

 

$sql = "SELECT subs.news_id, subs.cust_id, subs.sub_id, newspaper.name, newspaper.$day, newspaper.id  FROM subs , newspaper WHERE subs.news_id = newspaper.id AND newspaper.$day = 'on'";

[jaymc]

Yeh, the error mysql has gave also shows it doesnt like the field newspaper.cost for what ever reason

 

Are you sure it exists... what type of data is in there, what is the collation compared to your other ones

 

Im not really sure on this one the query does look fine

[Stealth549]

I still get the error if I take that out, The field is varchar( and it stores numbers

[fenway]

Echo the query.

[Stealth549]

Echo the query.

 

SELECT subs.news_id, subs.cust_id, subs.sub_id, newspaper.name, newspaper., newspaper.id FROM subs , newspaper WHERE subs.news_id = newspaper.id AND newspaper. = 'on'

[fenway]

"Newspaper." would be the error.  $day is undefined?

[Stealth549]

ah, thanks.