Re: [PHP] php'ed mysql query

2001-01-16 Thread Toby Butzon
quot;Sam Masiello" <[EMAIL PROTECTED]>; "php" <[EMAIL PROTECTED]> Sent: Tuesday, January 16, 2001 3:00 PM Subject: Re: [PHP] php'ed mysql query > > If you are going to do a comparison from one of the fields in your table, > > doesn't the field name need to

Re: [PHP] php'ed mysql query

2001-01-16 Thread Christopher Allen
> If you are going to do a comparison from one of the fields in your table, > doesn't the field name need to be on the left hand side of the comparison? > So instead of "where '$zip1' >= zip", shouldn't it be "zip < '$zip1'" (et al > for the other comparisons in your query) . > Didn't matter. Is t

Re: [PHP] php'ed mysql query

2001-01-16 Thread Ignacio Vazquez-Abrams
On Tue, 16 Jan 2001, Christopher Allen wrote: > > Hello, > Can anyone tell me why this query errors out? It works from the mysql > client command line just fine > > > $query1 = "select * from zip_base where '$zip1' >= zip && '$zip1' <= CONCAT > (SUBSTRING(zip , '1' , LENGTH (zip) - LENGTH (

RE: [PHP] php'ed mysql query

2001-01-16 Thread Sam Masiello
comparisons in your query) . HTH Sam Masiello Systems Analyst Chek.Com (716) 853-1362 x289 [EMAIL PROTECTED] -Original Message- From: Christopher Allen [mailto:[EMAIL PROTECTED]] Sent: Tuesday, January 16, 2001 2:25 PM To: php Subject:[PHP] php'ed mysql query Hello

RE: [PHP] php'ed mysql query

2001-01-16 Thread Brian Paulson
: Tuesday, January 16, 2001 12:25 PM To: php Subject: [PHP] php'ed mysql query Hello, Can anyone tell me why this query errors out? It works from the mysql client command line just fine $query1 = "select * from zip_base where '$zip1' >= zip && '$zip1' <

[PHP] php'ed mysql query

2001-01-16 Thread Christopher Allen
Hello, Can anyone tell me why this query errors out? It works from the mysql client command line just fine $query1 = "select * from zip_base where '$zip1' >= zip && '$zip1' <= CONCAT (SUBSTRING(zip , '1' , LENGTH (zip) - LENGTH (range) ), range)"; Christopher C. M. Allen -- PHP Gen