Hitting an “array” & "sql" wall, spl

Anon

I think I might have hit a dead end, but cannot believe it since there's always an answer.

I’ve stored some values in a field in a db as a varchar. I’m working with:

{“1", “2", ...etc }

Trying to grab from the id column based on the id I have...

My question is:

How can I pull a record set from the db table based on an id that needs to be matched against an array value stored in the db as a varchar.

I don’t think like would work, but I am thinking there must be a way in sql to say WHERE “there is a id that matches this id in the string?

Any help or hints will be awarded with my appreciation. =) Thanks.

Anon

Does your field contain only one number in each field (why store it as a varchar if it's a number?)? and you want to mach it against an array of values? You'd have to use IN. It works like this:

SELECT * FROM tbl WHERE ID IN (1,2,3,4)

Diego

Anon

Appreciate your feedback, thanks.

I'm trying the in sql clause but am having sytax issues, can you give me an example of a in clause used? I couldn't find one online.

As for the number value array, I figured a varchar would be more flexible and easier to work with than an int. What's your opinion?

Thanks a million.

I want to match an id ( int ) against an array value in a column in a table, the array value holds multiple numbers.

I think your in clause will work, just having problems formating it.

Anon

Yeah, what I don't get is where is the one ID that you mention? DB or PHP? and where is the array of values? DB or PHP? If the one value is in the DB, and the array in PHP, then you can use the IN clause...if it's the other way around, I don't know yet...

Here's the IN function:

http://www.mysql.com/documentation/mysql/bychapter/manual_Reference.html#Comparison_Operators

Diego

Anon

thank you very much, and yes,

the array is in the db, and the value to match against the array in the db is php which i think from what i've read leaves me still against the wall. =(

thanks for your help, i shall continue my search.