Re: [PHP] bug in code - can't find it!

2003-08-14 Thread AciD
just a little point; u can replace

by 
which is better imho.

AciD


On Wed, 6 Aug 2003 14:57:23 -0700, Jennifer Goodie 
<[EMAIL PROTECTED]> wrote:


Ok, here is my query and the bit of code I can't figure out:

$query = "SELECT * from apt_user_t a, apt_company_t b ";
$query .= "WHERE a.user_cd = b.user_cd ";
$query .= "ORDER BY a.username";
$search_results = mysql_query($query) or die("Select
Failed!");
while ($search_result2 = mysql_fetch_array($search_results))
{

TYPE="checkbox"
NAME=""
SIZE="20"
MAXLENGTH="50"
VALUE="
1){?>CHECKED"> 
}
Nothing shows up with the echo or the value.  I only included this
checkbox, but all of the other values show up fine.  Can someone give me 
a
hint?
Mysql does not prefix returned columns with table_name., so there's 
probably
no "a.retired_flag" index in your array.  A simple way to check this 
would
be to print_r($search_result2).  If you have a column named retired_flag 
in
both table a and table b and you specifically want the one from table a 
in
your result set you are going to have to alias it to a different name in
your query, i.e. "SELECT *, a.retired_flag as r_flag"





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RE: [PHP] bug in code - can't find it!

2003-08-14 Thread Jennifer Goodie

> Ok, here is my query and the bit of code I can't figure out:
>
> $query = "SELECT * from apt_user_t a, apt_company_t b ";
> $query .= "WHERE a.user_cd = b.user_cd ";
> $query .= "ORDER BY a.username";
>
>  $search_results = mysql_query($query) or die("Select
> Failed!");
> while ($search_result2 = mysql_fetch_array($search_results))
> {
>  TYPE="checkbox"
> NAME=""
> SIZE="20"
> MAXLENGTH="50"
> VALUE=" 1){?>CHECKED"> 
> }
> Nothing shows up with the echo or the value.  I only included this
> checkbox, but all of the other values show up fine.  Can someone give me a
> hint?

Mysql does not prefix returned columns with table_name., so there's probably
no "a.retired_flag" index in your array.  A simple way to check this would
be to print_r($search_result2).  If you have a column named retired_flag in
both table a and table b and you specifically want the one from table a in
your result set you are going to have to alias it to a different name in
your query, i.e. "SELECT *, a.retired_flag as r_flag"



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Re: [PHP] bug in code - can't find it!

2003-08-14 Thread John W. Holmes
AciD wrote:

just a little point; u can replace

by 
which is better imho.
and you're entitled to it, but it's not better. Shorter isn't always better.

Your code will break if enable_short_open_tags is disabled in php.ini or 
an .htaccess file. Sure, it's on by default, but some people turn it 
off, especially when dealing with XML a lot.

--
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Re: [PHP] bug in code - can't find it!

2003-08-14 Thread Chris Sherwood
 
}

well for starters checkboxes dont have a size and maxlength
 you may want to try that and see if it works...



- Original Message -
From: "Amanda McComb" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, August 06, 2003 2:47 PM
Subject: [PHP] bug in code - can't find it!


> Ok, after all of the discussion on posting, I'm afraid to post.
> Unfortunately, no one I know is a programmer, and I am a beginner.  I can
> normally figure out what I've done wrong, but I'm missing something here.
>
> Ok, here is my query and the bit of code I can't figure out:
>
> $query = "SELECT * from apt_user_t a, apt_company_t b ";
> $query .= "WHERE a.user_cd = b.user_cd ";
> $query .= "ORDER BY a.username";
>
>  $search_results = mysql_query($query) or die("Select
> Failed!");
> while ($search_result2 = mysql_fetch_array($search_results))
> {
>  TYPE="checkbox"
> NAME=""
> SIZE="20"
> MAXLENGTH="50"
> VALUE="CHECKED"> 
> }
> Nothing shows up with the echo or the value.  I only included this
> checkbox, but all of the other values show up fine.  Can someone give me a
> hint?
>
>
>
>
> --
> PHP General Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>
>


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Re: [PHP] bug in code - can't find it!

2003-08-08 Thread Chris Sherwood
silly me

upon further looking

>

>
- Original Message -
From: "Amanda McComb" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, August 06, 2003 2:47 PM
Subject: [PHP] bug in code - can't find it!


> Ok, after all of the discussion on posting, I'm afraid to post.
> Unfortunately, no one I know is a programmer, and I am a beginner.  I can
> normally figure out what I've done wrong, but I'm missing something here.
>
> Ok, here is my query and the bit of code I can't figure out:
>
> $query = "SELECT * from apt_user_t a, apt_company_t b ";
> $query .= "WHERE a.user_cd = b.user_cd ";
> $query .= "ORDER BY a.username";
>
>  $search_results = mysql_query($query) or die("Select
> Failed!");
> while ($search_result2 = mysql_fetch_array($search_results))
> {
>  TYPE="checkbox"
> NAME=""
> SIZE="20"
> MAXLENGTH="50"
> VALUE="CHECKED"> 
> }
> Nothing shows up with the echo or the value.  I only included this
> checkbox, but all of the other values show up fine.  Can someone give me a
> hint?
>
>
>
>
> --
> PHP General Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>
>


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Re: [PHP] bug in code - can't find it!

2003-08-08 Thread Marek Kilimajer
Is it realy so? I remember someone noting on this list it also work if 
short tags are off.
I have never tested it and I'm not going to restart our server now, but 
I will try it at home.

John W. Holmes wrote:

AciD wrote:

just a little point; u can replace

by 
which is better imho.


and you're entitled to it, but it's not better. Shorter isn't always 
better.

Your code will break if enable_short_open_tags is disabled in php.ini or 
an .htaccess file. Sure, it's on by default, but some people turn it 
off, especially when dealing with XML a lot.



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Re: [PHP] bug in code - can't find it!

2003-08-07 Thread Amanda McComb
Oh, yeah.  Well, I just copied and pasted that from another form.  Taken
out, it still doesn't work.  Even the echo, which is on no way related to
the checkbox.

Thanks!


On Wed, 6 Aug 2003, Chris Sherwood wrote:

>  VALUE="CHECKED">  $search_result2[a.retired_flag]?>
> }
> 
> well for starters checkboxes dont have a size and maxlength
>  you may want to try that and see if it works...
> 
> 
> 
> - Original Message -
> From: "Amanda McComb" <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> Sent: Wednesday, August 06, 2003 2:47 PM
> Subject: [PHP] bug in code - can't find it!
> 
> 
> > Ok, after all of the discussion on posting, I'm afraid to post.
> > Unfortunately, no one I know is a programmer, and I am a beginner.  I can
> > normally figure out what I've done wrong, but I'm missing something here.
> >
> > Ok, here is my query and the bit of code I can't figure out:
> >
> > $query = "SELECT * from apt_user_t a, apt_company_t b ";
> > $query .= "WHERE a.user_cd = b.user_cd ";
> > $query .= "ORDER BY a.username";
> >
> >  $search_results = mysql_query($query) or die("Select
> > Failed!");
> > while ($search_result2 = mysql_fetch_array($search_results))
> > {
> >  > TYPE="checkbox"
> > NAME=""
> > SIZE="20"
> > MAXLENGTH="50"
> > VALUE="CHECKED">  $search_result2[a.retired_flag]?>
> > }
> > Nothing shows up with the echo or the value.  I only included this
> > checkbox, but all of the other values show up fine.  Can someone give me a
> > hint?
> >
> >
> >
> >
> > --
> > PHP General Mailing List (http://www.php.net/)
> > To unsubscribe, visit: http://www.php.net/unsub.php
> >
> >
> 
> 
> 


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