Re: [PHP-DB] If( Query)

2008-09-07 Thread Evert Lammerts
Is your table set up in the same way on your webserver? Maybe you
forgot to set the default value to na.gif in your item_pix column. If
not, can you send your table definition?

On Sat, Sep 6, 2008 at 9:02 PM, Chris Hale [EMAIL PROTECTED] wrote:
 I have the following function:

 function
 add_item($item_name,$item_desc,$item_price,$item_man_id,$item_cat_id,$item_pix)
   {
   connect();
   if($item_pix == )
   {
   $sql = INSERT INTO items
 (item_name,item_desc,item_price,item_man_id,item_cat_id) VALUES
 ('$item_name','$item_desc','$item_price','$item_man_id','$item_cat_id');
   }
   else {
   $sql = INSERT INTO items
 (item_name,item_desc,item_price,item_pix,item_man_id,item_cat_id) VALUES
 ('$item_name','$item_desc','$item_price','$item_pix','$item_man_id','$item_cat_id');
   }
   mysql_query($sql);
   return;
 }

 I am using the if statement because i want it so that if no picture is
 uploaded the entry is blank and the mysql database has a default entry of
 na.gif which is a picture coming soon picture.

 It works fine when i run in localy on MAMP, but if i run it on my web server
 it doesnt add the row.

 Is this a compatability error? or is there a better way to write this?

 Thanks

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[PHP-DB] mysql auto_increment

2008-09-07 Thread Chris Hale
I am writing a catologe application and i have a problem when it comes 
to the edit product part. I have a table with the catogories and a table 
with manufacturers.
Each table has a id column and a name column. The id column is set up in 
the MySQL to auto_increment, which works fine normally, but i am writing 
this script:


function edit_cat_radio($item_cat_id)
{
   connect();
   $sql = SELECT * FROM cat DISTINGT ORDER BY cat_id;
   $result = mysql_query($sql);
   $k = 1;
   while ($row = mysql_fetch_assoc($result))
   {
   extract($row);
   echo 'label for=',$cat_name,'',$cat_name,'input 
type=radio name=fcat value=',$cat_id,' id=',$cat_id,'';

   if($k == $item_cat_id)
   {
   echo 'checked';
   }
   echo ' /';
   $k++;
   }
return;   
}


This should (in theory) automatically check the radio button of the 
existing catogory. It would work fine; but what seems to mess it up is 
the auto_increment.


If i delete a catogory/manufacturer from the id's remain the same. and 
end up like this:

cat_id cat_name
1 Bridlework
2 Clippers
3 Clothing
4 Dressage Tests
5 DVD/Video/Books
9 Footwear

but if the cat_id is 9 the /while /statement doesnt repeat 9 times so 
the counter never reaches 9.


I don't know if you got all that, its hard to explain.

I would appreciate any help on how to sort this out.

Thanks

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Re: [PHP-DB] mysql auto_increment

2008-09-07 Thread Evert Lammerts
I guess this code was not copy pasted from your actual source? It
would generate mysql errors.

Anyway, I think I've an idea of what you want, so here's my two cents :

function edit_cat_radio($item_cat_id) {
connect();
$query = mysql_query(SELECT * FROM cat);
while ($row = mysql_fetch_assoc($query)) {
echo label for=\{$row['cat_name']}\{$row['cat_name']}input
type=\radio\ name=\fcat\ value=\{$row['cat_id']}\
id=\{$row['cat_id']}\  . ($item_cat_id == $row[cat_id] ?
checked=\checked\ : ) . /\n;
}
}



On Sun, Sep 7, 2008 at 11:16 PM, Chris Hale [EMAIL PROTECTED] wrote:
 I am writing a catologe application and i have a problem when it comes to
 the edit product part. I have a table with the catogories and a table with
 manufacturers.
 Each table has a id column and a name column. The id column is set up in the
 MySQL to auto_increment, which works fine normally, but i am writing this
 script:

 function edit_cat_radio($item_cat_id)
 {
   connect();
   $sql = SELECT * FROM cat DISTINGT ORDER BY cat_id;
   $result = mysql_query($sql);
   $k = 1;
   while ($row = mysql_fetch_assoc($result))
   {
   extract($row);
   echo 'label for=',$cat_name,'',$cat_name,'input type=radio
 name=fcat value=',$cat_id,' id=',$cat_id,'';
   if($k == $item_cat_id)
   {
   echo 'checked';
   }
   echo ' /';
   $k++;
   }
 return;   }

 This should (in theory) automatically check the radio button of the existing
 catogory. It would work fine; but what seems to mess it up is the
 auto_increment.

 If i delete a catogory/manufacturer from the id's remain the same. and end
 up like this:
 cat_id cat_name
 1 Bridlework
 2 Clippers
 3 Clothing
 4 Dressage Tests
 5 DVD/Video/Books
 9 Footwear

 but if the cat_id is 9 the /while /statement doesnt repeat 9 times so the
 counter never reaches 9.

 I don't know if you got all that, its hard to explain.

 I would appreciate any help on how to sort this out.

 Thanks

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 To unsubscribe, visit: http://www.php.net/unsub.php



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Re: [PHP-DB] mysql auto_increment

2008-09-07 Thread Evert Lammerts
Pastebin is so much nicer when posting code. Find the code i've sent here:

http://pastebin.com/mc5d611a

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Re: [PHP-DB] mysql auto_increment

2008-09-07 Thread Evert Lammerts
Little change - the label was not used in the right way:

http://pastebin.com/m2d98e677

On Sun, Sep 7, 2008 at 11:46 PM, Evert Lammerts
[EMAIL PROTECTED] wrote:
 I guess this code was not copy pasted from your actual source? It
 would generate mysql errors.

 Anyway, I think I've an idea of what you want, so here's my two cents :

 function edit_cat_radio($item_cat_id) {
connect();
$query = mysql_query(SELECT * FROM cat);
while ($row = mysql_fetch_assoc($query)) {
echo label 
 for=\{$row['cat_name']}\{$row['cat_name']}input
 type=\radio\ name=\fcat\ value=\{$row['cat_id']}\
 id=\{$row['cat_id']}\  . ($item_cat_id == $row[cat_id] ?
 checked=\checked\ : ) . /\n;
}
 }



 On Sun, Sep 7, 2008 at 11:16 PM, Chris Hale [EMAIL PROTECTED] wrote:
 I am writing a catologe application and i have a problem when it comes to
 the edit product part. I have a table with the catogories and a table with
 manufacturers.
 Each table has a id column and a name column. The id column is set up in the
 MySQL to auto_increment, which works fine normally, but i am writing this
 script:

 function edit_cat_radio($item_cat_id)
 {
   connect();
   $sql = SELECT * FROM cat DISTINGT ORDER BY cat_id;
   $result = mysql_query($sql);
   $k = 1;
   while ($row = mysql_fetch_assoc($result))
   {
   extract($row);
   echo 'label for=',$cat_name,'',$cat_name,'input type=radio
 name=fcat value=',$cat_id,' id=',$cat_id,'';
   if($k == $item_cat_id)
   {
   echo 'checked';
   }
   echo ' /';
   $k++;
   }
 return;   }

 This should (in theory) automatically check the radio button of the existing
 catogory. It would work fine; but what seems to mess it up is the
 auto_increment.

 If i delete a catogory/manufacturer from the id's remain the same. and end
 up like this:
 cat_id cat_name
 1 Bridlework
 2 Clippers
 3 Clothing
 4 Dressage Tests
 5 DVD/Video/Books
 9 Footwear

 but if the cat_id is 9 the /while /statement doesnt repeat 9 times so the
 counter never reaches 9.

 I don't know if you got all that, its hard to explain.

 I would appreciate any help on how to sort this out.

 Thanks

 --
 PHP Database Mailing List (http://www.php.net/)
 To unsubscribe, visit: http://www.php.net/unsub.php




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Re: [PHP-DB] If( Query)

2008-09-07 Thread Niel Archer
Hi

 I have the following function:
 
 function 
 add_item($item_name,$item_desc,$item_price,$item_man_id,$item_cat_id,$item_pix)
 {
 connect();
 if($item_pix == )
 {
 $sql = INSERT INTO items 
 (item_name,item_desc,item_price,item_man_id,item_cat_id) VALUES 
 ('$item_name','$item_desc','$item_price','$item_man_id','$item_cat_id');
 }
 else {
 $sql = INSERT INTO items 
 (item_name,item_desc,item_price,item_pix,item_man_id,item_cat_id) VALUES 
 ('$item_name','$item_desc','$item_price','$item_pix','$item_man_id','$item_cat_id');
 }
 mysql_query($sql);
 return;
 }
 
 I am using the if statement because i want it so that if no picture is 
 uploaded the entry is blank and the mysql database has a default entry 
 of na.gif which is a picture coming soon picture.
 
 It works fine when i run in localy on MAMP, but if i run it on my web 
 server it doesnt add the row.

You should be checking the mysql_query call for success and output the
error if it fails.  Something like:

   mysql_query($sql) or die('Insert failed: ' . mysql_error());

You'll now why it's failing then.  Make sure you have error reporting
enabled.

 Is this a compatability error? or is there a better way to write this?
 


--
Niel Archer



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Re: [PHP-DB] If( Query)

2008-09-07 Thread Chris Hale

Niel Archer wrote:

Hi

  

I have the following function:

function 
add_item($item_name,$item_desc,$item_price,$item_man_id,$item_cat_id,$item_pix)

{
connect();
if($item_pix == )
{
$sql = INSERT INTO items 
(item_name,item_desc,item_price,item_man_id,item_cat_id) VALUES 
('$item_name','$item_desc','$item_price','$item_man_id','$item_cat_id');

}
else {
$sql = INSERT INTO items 
(item_name,item_desc,item_price,item_pix,item_man_id,item_cat_id) VALUES 
('$item_name','$item_desc','$item_price','$item_pix','$item_man_id','$item_cat_id');

}
mysql_query($sql);
return;
}

I am using the if statement because i want it so that if no picture is 
uploaded the entry is blank and the mysql database has a default entry 
of na.gif which is a picture coming soon picture.


It works fine when i run in localy on MAMP, but if i run it on my web 
server it doesnt add the row.



You should be checking the mysql_query call for success and output the
error if it fails.  Something like:

   mysql_query($sql) or die('Insert failed: ' . mysql_error());

You'll now why it's failing then.  Make sure you have error reporting
enabled.

  

Is this a compatability error? or is there a better way to write this?





--
Niel Archer



  

I have fixed it now:

function 
add_item($item_name,$item_desc,$item_price,$item_man_id,$item_cat_id,$item_pix)

   {
   connect();
   if($item_pix == )
   {
   $sql = INSERT INTO items 
(item_name,item_desc,item_price,item_man_id,item_cat_id) VALUES 
('$item_name','$item_desc','$item_price','$item_man_id','$item_cat_id');

   }
   else {
   $sql = INSERT INTO items 
(item_name,item_desc,item_price,item_pix,item_man_id,item_cat_id) VALUES 
('$item_name','$item_desc','$item_price','$item_pix','$item_man_id','$item_cat_id');

   }
   mysql_query($sql);
   return;
}

Thanks anyway.

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