Re: [PHP] creating dropdown lists from fetched arrays.

2002-02-25 Thread William Lovaton 

El dom, 24-02-2002 a las 17:39, Matthew Darcy escribió:
 Hi,
 
 I want to create a dropdown list with options from a table.
 
 ie
 HTML
 BODY
 
 ?php
 
 $sql_select = select * from dropdown_options;
 $results = mysql_query($sql_select);
 
 while ($row = mysql_fetch_array($results);
 {
   echo option VALUE=$row[col1] NAME=option1
 }
 
 /BODY
 /HMTL
 
 I know this is basic but it is to give you an idea of what I want.
 
 I have tried to find an example of this in the book I am using to learn to
 no result.
 
 I am guessing this is how it works from the info I have read from the book.
 
 Can someone put me on the right path.
 
 Thanks,
 
 Matt.

With that semi-colon at the end of the while you are doing an empty
loop... so when the script gets to the option... stuff... the array is
empty.

I would do something like this:

HTML
BODY
select name=myOptions
?php
$sql_select = select * from dropdown_options;
$results = mysql_query($sql_select);

while ($row = mysql_fetch_array($results){
?
option value=? print $row[id] ?? print $row[name] ?/option
?
}
?
/select
/BODY
/HMTL


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RE: [PHP] creating dropdown lists from fetched arrays.

2002-02-25 Thread Jon Haworth 

 while ($row = mysql_fetch_array($results){

ITYM

while ($row = mysql_fetch_array ($results)) {

Cheers
Jon

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[PHP] creating dropdown lists from fetched arrays.

2002-02-24 Thread Matthew Darcy 

Hi,

I want to create a dropdown list with options from a table.

ie
HTML
BODY

?php

$sql_select = select * from dropdown_options;
$results = mysql_query($sql_select);

while ($row = mysql_fetch_array($results);
{
echo option VALUE=$row[col1] NAME=option1
}

/BODY
/HMTL

I know this is basic but it is to give you an idea of what I want.

I have tried to find an example of this in the book I am using to learn to
no result.

I am guessing this is how it works from the info I have read from the book.

Can someone put me on the right path.

Thanks,

Matt.


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