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Syntax fix

"'CurrentUserID()' OR FIND_IN_SET('office', sharegrpall) OR FIND_IN_SET('" . CurrentUserID() . "', sharegrp)) "

What am I missing here?
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Starquest321
Asked:
Starquest321
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3 Solutions
 
Starquest321Author Commented:
There is some issue with syntax
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Dan CraciunIT ConsultantCommented:
Can you post the whole query?
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Starquest321Author Commented:
It's actually part of a select statement . . . embed
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Dan CraciunIT ConsultantCommented:
OK. Can you post the php code?
Cause your original code does not really look like php.
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Starquest321Author Commented:
The code is generated by itself . . . I just create the filter. . . so I just need the part between the quotes
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Dan CraciunIT ConsultantCommented:
I still don't get it.

In php a variable has a $ in front, so sharegrpall and sharegrp have to be constants.
Apart from the fact that it's common practice to name the constants in ALL CAPS, are they defined as constants?

CurrentUserID() has to be a function. Where is it defined and what does it return?
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Kim WalkerWeb Programmer/TechnicianCommented:
The first 'CurrentUserID()' is in single quotes so it is probably being handled as a string. The second CurrentUserID() appears to be a function that generates a string that's being concatenated with the the strings before and after it because it is between two dots or concatenate operators.
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Starquest321Author Commented:
you are in the right direction. . .

userid is an int 11
where as the other two fields are varchar 300 fields.  . .

The error seems to be when I added my self the userid
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Kim WalkerWeb Programmer/TechnicianCommented:
Either way, none of this is PHP. It appears that you're using PHP to generate an SQL statement. But you say you're creating a "filter" and you've posted a screen capture of what might be an error that's being generated.

What are you trying to communicate with in PHP, a database? What kind of database? You may want the topic administrator to add the kind of database to the question topic so experts in that topic will know to look at the question.

you are in the right direction. . .
What leads you to believe I'm in the right direction? Have you made a change that lead to a different error? Please post the changed code and how it changed the results.
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Julian HansenCommented:
Based on your other post

http://www.experts-exchange.com/Programming/Languages/SQL_Syntax/Q_28044665.html

shargrpall is a db field that (I assume) contains a comma separated string that could contain the word 'office' or the CurrentUserID()

What you have not made clear is what you are trying to do - the inclusion of CurrentUserId() as a quoted string in the filter is creating confusion - as you are using it as a concatenated value in the same string.

Which is it to be - evaluated by PHP as a function call.

...' . CurrentUserId() . '...

Or as a MySQL interpreted function as
'CurrentUserID() OR ...'

I suspect the former in which case you need to add the item to your filter string as a PHP function call - not as a string

$filter .= CurrentUserId() . ' ... rest of filter';

Having said that - you need to provide more information - you have not made clear what the problem is, what the error is or any other information that would enable a quick solution to the problem. The more information you provide the easier it will be to help you.
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Starquest321Author Commented:
Perfect. You all helped. Figured it out!
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