php string parsing

[keith]

I am trying to build a where clause for mysql based on different form
inputs but php seems to have a problem with the following:

$whrClause = "WHERE keywords LIKE '%$keywords%'"

I thought with the double quotes that the single-quote and percent
signs would remain unchanged and the variable $keywords would be
expanded to its value, but I guess I am wrong because i get a parsing
error... Any help for a php newbie?

[Jochen Daum]

Hi,

On 17 Mar 2004 19:38:56 -0800, ke**********@cox.net (keith) wrote:

I am trying to build a where clause for mysql based on different form
inputs but php seems to have a problem with the following:

$whrClause = "WHERE keywords LIKE '%$keywords%'"

semicolon?

HTH, Jochen
I thought with the double quotes that the single-quote and percent
signs would remain unchanged and the variable $keywords would be
expanded to its value, but I guess I am wrong because i get a parsing
error... Any help for a php newbie?

--
Jochen Daum - Cabletalk Group Ltd.
PHP DB Edit Toolkit -- PHP scripts for building
database editing interfaces.
http://sourceforge.net/projects/phpdbedittk/

[IndyTim]

Jochen Daum wrote:

Hi,

On 17 Mar 2004 19:38:56 -0800, ke**********@cox.net (keith) wrote:

I am trying to build a where clause for mysql based on different form
inputs but php seems to have a problem with the following:

$whrClause = "WHERE keywords LIKE '%$keywords%'"

semicolon?

HTH, Jochen

I thought with the double quotes that the single-quote and percent
signs would remain unchanged and the variable $keywords would be
expanded to its value, but I guess I am wrong because i get a parsing
error... Any help for a php newbie?

Try:

$whrClause = "WHERE keywords LIKE %'$keywords'%"

You want to pass the literal of $keywords, not the literal of %$keywords%.

IndyTim

[keith]

Jochen Daum <jo*********@cabletalk.co.nz> wrote in message news:<sb********************************@4ax.com>. ..

Hi,

On 17 Mar 2004 19:38:56 -0800, ke**********@cox.net (keith) wrote:

I am trying to build a where clause for mysql based on different form
inputs but php seems to have a problem with the following:

$whrClause = "WHERE keywords LIKE '%$keywords%'"

semicolon?

HTH, Jochen

I thought with the double quotes that the single-quote and percent
signs would remain unchanged and the variable $keywords would be
expanded to its value, but I guess I am wrong because i get a parsing
error... Any help for a php newbie?

YEP, lol - the switch from vbscript to php is killing me. Thanks for the help.