Re: [PHP-DB] Select statement only returns 1 record

2002-01-30 Thread Miles Thompson 

Check the manual for the mysql_fetch_array() function, it shows you how to 
extract data from $result. You will use the while loop (as shown in the 
manual's example) to fill your combo box / selct list /drop down menu, 
whatever we're calling that creature today.

Miles

At 12:45 PM 1/30/2002 -0600, Todd Williamsen wrote:
I am trying to get data from two columns, FirstName and Last name and
displaying all the records LastName, FirstName in a drop down menu.

The weird thing is that it only displays one record.  I thought the table
was hosed, but its not.  I tried it through another database and still
doesn't work.

Here is the code:

?
$db = mysql_connect($dbserver, $dbuser, $dbpass);
mysql_select_db($dbname,$db);
$sortby = name ASC;
$sql=SELECT * FROM webl_players ORDER BY $sortby;
$result=mysql_query($sql,$db);
$row = mysql_fetch_array($result);
$FirstName = $row[FirstName];
$LastName = $row[LastName];
?

select name=canidate
   option selected
   ?php echo $LastName, $FirstName; ?
   /option

I would list the whole page but its long!  There is another SQL statement at
the top of the page:

?php

$db = mysql_connect($dbserver, $dbuser, $$dbpass);
mysql_select_db($dbname, $db);
$sql = SELECT * FROM Canidate SORT BY 'LastName';
$result = mysql_query($sql);
?



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RE: [PHP-DB] Select statement only returns 1 record

2002-01-30 Thread Rick Emery 

?
$db = mysql_connect($dbserver, $dbuser, $dbpass);
mysql_select_db($dbname,$db);
$sortby = name ASC;
$sql=SELECT * FROM webl_players ORDER BY $sortby;
$result=mysql_query($sql,$db);

print select name=canidate;
while( $row = mysql_fetch_array($result) )
{
$FirstName = $row['FirstName'];
$LastName = $row['LastName'];
$id=$row['id'];
print option value=$id $LastName, $FirstName/option;
}
?
print /SELECT;

-Original Message-
From: Todd Williamsen [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, January 30, 2002 12:45 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Select statement only returns 1 record


I am trying to get data from two columns, FirstName and Last name and
displaying all the records LastName, FirstName in a drop down menu.

The weird thing is that it only displays one record.  I thought the table
was hosed, but its not.  I tried it through another database and still
doesn't work.

Here is the code:

?
$db = mysql_connect($dbserver, $dbuser, $dbpass);
mysql_select_db($dbname,$db);
$sortby = name ASC;
$sql=SELECT * FROM webl_players ORDER BY $sortby;
$result=mysql_query($sql,$db);
$row = mysql_fetch_array($result);
$FirstName = $row[FirstName];
$LastName = $row[LastName];
?

select name=canidate
  option selected
  ?php echo $LastName, $FirstName; ?
  /option

I would list the whole page but its long!  There is another SQL statement at
the top of the page:

?php

$db = mysql_connect($dbserver, $dbuser, $$dbpass);
mysql_select_db($dbname, $db);
$sql = SELECT * FROM Canidate SORT BY 'LastName';
$result = mysql_query($sql);
?



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Re: [PHP-DB] Select statement only returns 1 record

2002-01-30 Thread biorn 

You will need to put the option tag in a loop to get all of the records in 
the table.

select name=canidate (this can go anywhere but in while loop)

?
$db = mysql_connect($dbserver, $dbuser, $dbpass);
mysql_select_db($dbname,$db);
$sortby = name ASC;
$sql=SELECT * FROM webl_players ORDER BY $sortby;
$result=mysql_query($sql,$db);
$num=mysql_numrows($result);
$i=0;
while ($i  $num) {
$row = mysql_fetch_array($result);
$FirstName = $row[FirstName];
$LastName = $row[LastName];

echo option value='$LastName, $FirstName'?php echo '$LastName, 
$FirstName'?/option;
$i++;
}
?

Also, without the value in the option tag, you will not be passing anything 
to the form.

HTH

Maureen Biorn


Todd Williamsen [EMAIL PROTECTED] said:

 I am trying to get data from two columns, FirstName and Last name and
 displaying all the records LastName, FirstName in a drop down menu.
 
 The weird thing is that it only displays one record.  I thought the table
 was hosed, but its not.  I tried it through another database and still
 doesn't work.
 
 Here is the code:
 
 ?
 $db = mysql_connect($dbserver, $dbuser, $dbpass);
 mysql_select_db($dbname,$db);
 $sortby = name ASC;
 $sql=SELECT * FROM webl_players ORDER BY $sortby;
 $result=mysql_query($sql,$db);
 $row = mysql_fetch_array($result);
 $FirstName = $row[FirstName];
 $LastName = $row[LastName];
 ?
 
 select name=canidate
   option selected
   ?php echo $LastName, $FirstName; ?
   /option
 
 I would list the whole page but its long!  There is another SQL statement at
 the top of the page:
 
 ?php
 
 $db = mysql_connect($dbserver, $dbuser, $$dbpass);
 mysql_select_db($dbname, $db);
 $sql = SELECT * FROM Canidate SORT BY 'LastName';
 $result = mysql_query($sql);
 ?
 
 
 
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RE: [PHP-DB] Select statement only returns 1 record

2002-01-30 Thread Todd Williamsen 

Nope, that doesn't work

-Original Message-
From: Rick Emery [mailto:[EMAIL PROTECTED]] 
Sent: Wednesday, January 30, 2002 1:01 PM
To: 'Todd Williamsen'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Select statement only returns 1 record


?
$db = mysql_connect($dbserver, $dbuser, $dbpass);
mysql_select_db($dbname,$db);
$sortby = name ASC;
$sql=SELECT * FROM webl_players ORDER BY $sortby;
$result=mysql_query($sql,$db);

print select name=canidate;
while( $row = mysql_fetch_array($result) )
{
$FirstName = $row['FirstName'];
$LastName = $row['LastName'];
$id=$row['id'];
print option value=$id $LastName, $FirstName/option;
}
?
print /SELECT;

-Original Message-
From: Todd Williamsen [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, January 30, 2002 12:45 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Select statement only returns 1 record


I am trying to get data from two columns, FirstName and Last name and
displaying all the records LastName, FirstName in a drop down menu.

The weird thing is that it only displays one record.  I thought the
table
was hosed, but its not.  I tried it through another database and still
doesn't work.

Here is the code:

?
$db = mysql_connect($dbserver, $dbuser, $dbpass);
mysql_select_db($dbname,$db);
$sortby = name ASC;
$sql=SELECT * FROM webl_players ORDER BY $sortby;
$result=mysql_query($sql,$db);
$row = mysql_fetch_array($result);
$FirstName = $row[FirstName];
$LastName = $row[LastName];
?

select name=canidate
  option selected
  ?php echo $LastName, $FirstName; ?
  /option

I would list the whole page but its long!  There is another SQL
statement at
the top of the page:

?php

$db = mysql_connect($dbserver, $dbuser, $$dbpass);
mysql_select_db($dbname, $db);
$sql = SELECT * FROM Canidate SORT BY 'LastName';
$result = mysql_query($sql);
?



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