RE: [PHP-DB] Drop Down Menus
Some body asked for it!
-Original Message-
From: chintan [mailto:[EMAIL PROTECTED]
Sent: Jueves, 05 de Mayo de 2005 07:36 a.m.
To: php-db@lists.php.net
Subject: [PHP-DB] Drop Down Menus
hey guys i wrote this code from another code of zend.php
Can anyone tell me how do i update the value of second menu and third one?
can i do that without javascript?
?php
session_start();
header(Cache-control: private);
$link = mysql_connect(localhost, chintan,hellomysql)
or die(Could not connect);
mysql_select_db(chintan) or die(Could not select database);
$XX = No data availabe!;
echo form name=HingeType action=$PHP_SELF method=POST;
$_SESSION['ItemType'] = $_REQUEST['ItemType'];
echo select name=ItemType tabindex=1;
$aku = mysql_query(SELECT ItemType FROM Products WHERE
ItemName='Hinges' group by ItemType);
while ($row = mysql_fetch_array($aku))
{
$colom_name=$row[ItemType];
echo option value=$colom_name$colom_name/option;
}
if ($colom_name=)
{
print ($XX);
}
echo /select;?
input type=Submit value=Update onclick=Call Test_Click()br
?php
echo $_SESSION['ItemType'];
$_SESSION['Thickness'] = $_REQUEST['Thickness'];
echo select name=Thickness tabindex=2\n;
$ak = mysql_query(SELECT Thickness FROM Products WHERE
ItemName='Hinges' and ItemType='$_SESSION[ItemType]'
group by Thickness);
while ($row = mysql_fetch_array($ak))
{
$colom_name2=$row[Thickness];
echo option value=$colom_name2$colom_name2/option\n;
}
if ($colom_name2=)
{
print ($XX);
}
echo /select;
echo input type=submit value='Update'br;
echo $_SESSION['Thickness'];
$_SESSION['SizeinMM'] = $_REQUEST['SizeinMM'];
echo select name=SizeinMM tabindex=3\n;
$ak = mysql_query(SELECT SizeinMM FROM Products WHERE ItemName='Hinges'
and ItemType='$_SESSION[ItemType]'
and Thickness='$_SESSION[Thickness]');
while ($row = mysql_fetch_array($ak))
{
$colom_name3=$row[SizeinMM];
echo option value=$colom_name3$colom_name3/option\n;
}
if ($colom_name3=)
{
print ($XX);
}
echo /select\n;
echo input type=submit value='Update'br\n;
echo $_SESSION['SizeinMM'];
echo /form;
$ak = mysql_query(SELECT Rates,Ratesforsspin FROM Products WHERE
ItemName='Hinges'
and ItemType='$_SESSION[ItemType]' and Thickness='$_SESSION[Thickness]'
and SizeinMM='$_SESSION[SizeinMM]');
while ($row = mysql_fetch_array($ak))
{
$colom_name4=$row[Rates];
$colom_name5=$row[Ratesforsspin];
}
echo brbRates for M.S.Pin:/b\t$colom_name4br;
echo bRates for S.S.Pin:/b\t$colom_name5br;
?
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RE: [PHP-DB] Drop Down Menus
to do it without js, you'll need to roundtrip it to the server, check for
the values that you have and run a query to generate the next set of items
for the drop down. And do this for each time that you need a menu generated.
Bastien
From: chintan [EMAIL PROTECTED]
To: php-db@lists.php.net
Subject: [PHP-DB] Drop Down Menus
Date: Thu, 05 May 2005 18:05:32 +0530
hey guys i wrote this code from another code of zend.php
Can anyone tell me how do i update the value of second menu and third one?
can i do that without javascript?
?php
session_start();
header(Cache-control: private);
$link = mysql_connect(localhost, chintan,hellomysql)
or die(Could not connect);
mysql_select_db(chintan) or die(Could not select database);
$XX = No data availabe!;
echo form name=HingeType action=$PHP_SELF method=POST;
$_SESSION['ItemType'] = $_REQUEST['ItemType'];
echo select name=ItemType tabindex=1;
$aku = mysql_query(SELECT ItemType FROM Products WHERE ItemName='Hinges'
group by ItemType);
while ($row = mysql_fetch_array($aku))
{
$colom_name=$row[ItemType];
echo option value=$colom_name$colom_name/option;
}
if ($colom_name=)
{
print ($XX);
}
echo /select;?
input type=Submit value=Update onclick=Call Test_Click()br
?php
echo $_SESSION['ItemType'];
$_SESSION['Thickness'] = $_REQUEST['Thickness'];
echo select name=Thickness tabindex=2\n;
$ak = mysql_query(SELECT Thickness FROM Products WHERE ItemName='Hinges'
and ItemType='$_SESSION[ItemType]'
group by Thickness);
while ($row = mysql_fetch_array($ak))
{
$colom_name2=$row[Thickness];
echo option value=$colom_name2$colom_name2/option\n;
}
if ($colom_name2=)
{
print ($XX);
}
echo /select;
echo input type=submit value='Update'br;
echo $_SESSION['Thickness'];
$_SESSION['SizeinMM'] = $_REQUEST['SizeinMM'];
echo select name=SizeinMM tabindex=3\n;
$ak = mysql_query(SELECT SizeinMM FROM Products WHERE ItemName='Hinges'
and ItemType='$_SESSION[ItemType]'
and Thickness='$_SESSION[Thickness]');
while ($row = mysql_fetch_array($ak))
{
$colom_name3=$row[SizeinMM];
echo option value=$colom_name3$colom_name3/option\n;
}
if ($colom_name3=)
{
print ($XX);
}
echo /select\n;
echo input type=submit value='Update'br\n;
echo $_SESSION['SizeinMM'];
echo /form;
$ak = mysql_query(SELECT Rates,Ratesforsspin FROM Products WHERE
ItemName='Hinges'
and ItemType='$_SESSION[ItemType]' and Thickness='$_SESSION[Thickness]' and
SizeinMM='$_SESSION[SizeinMM]');
while ($row = mysql_fetch_array($ak))
{
$colom_name4=$row[Rates];
$colom_name5=$row[Ratesforsspin];
}
echo brbRates for M.S.Pin:/b\t$colom_name4br;
echo bRates for S.S.Pin:/b\t$colom_name5br;
?
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RE: [PHP-DB] Drop Down Menus
FROM THE MANUAL:
mysql_fetch_row() fetches one row of data from the result associated with
the specified result identifier. The row is returned as an array. Each
result column is stored in an array offset, starting at offset 0.
Therefore:
print(option value=\$row[0]\$row[0]/option\n);
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, February 12, 2002 10:10 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Drop Down Menus
Hi wonder if anyone knows what I am doing wrong here.
I have a drop down selection menu that is generated from a mysql database.
In the database we have over 15 fields one of them contains text for the
catergory that the entry belongs to. I have used the following code to
generate my drop down menu but when i view it in the browser the drop down
menu has not listed the categories instead we have blank entries for each
selection.
? mysql_connect(localhost,user,password);
mysql_select_db(database);
$sql = select distinct category from books ORDER BY category ASC;
$makes_result = mysql_query($sql);
print (select name=\category\\n);
print(option selected value=\\Please select a
Category/option\n);
while($row = mysql_fetch_row($makes_result))
{
print(option value=\$row[1]\$row[1]/option\n);
}
print(/select); ?
Thanks for the help.
Barry
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RE: [PHP-DB] Drop Down Menus
Hi
the row index starts at 0 not 1 so you need $row[0], as you have only one
field $row[1] will be empty.
HTH
Peter
---
Excellence in internet and open source software
---
Sunmaia
www.sunmaia.net
[EMAIL PROTECTED]
tel. 0121-242-1473
---
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
Sent: 12 February 2002 16:10
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Drop Down Menus
Hi wonder if anyone knows what I am doing wrong here.
I have a drop down selection menu that is generated from a mysql
database. In the database we have over 15 fields one of them
contains text for the catergory that the entry belongs to. I have
used the following code to generate my drop down menu but when i
view it in the browser the drop down menu has not listed the
categories instead we have blank entries for each selection.
? mysql_connect(localhost,user,password);
mysql_select_db(database);
$sql = select distinct category from books ORDER BY category ASC;
$makes_result = mysql_query($sql);
print (select name=\category\\n);
print(option selected value=\\Please select a
Category/option\n);
while($row = mysql_fetch_row($makes_result))
{
print(option value=\$row[1]\$row[1]/option\n);
}
print(/select); ?
Thanks for the help.
Barry
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Re: [PHP-DB] Drop Down Menus
here is an example from my page:
?
$connection = mysql_connect($databaseserver, $databaseuser, $databasepass)
or die(Can't connect to DB);
$db = mysql_select_db($databasename, $connection);
$sql= SELECT id, State FROM $statetable ORDER BY State ASC;
$result= mysql_query($sql, $connection);
print select name=\State\;
while ($row = mysql_fetch_array($result))
{
$id = $row[id];
$State = $row[State];
print option$State/option;
}
print /SELECT
?
Peter Lovatt [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Hi
the row index starts at 0 not 1 so you need $row[0], as you have only one
field $row[1] will be empty.
HTH
Peter
---
Excellence in internet and open source software
---
Sunmaia
www.sunmaia.net
[EMAIL PROTECTED]
tel. 0121-242-1473
---
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
Sent: 12 February 2002 16:10
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Drop Down Menus
Hi wonder if anyone knows what I am doing wrong here.
I have a drop down selection menu that is generated from a mysql
database. In the database we have over 15 fields one of them
contains text for the catergory that the entry belongs to. I have
used the following code to generate my drop down menu but when i
view it in the browser the drop down menu has not listed the
categories instead we have blank entries for each selection.
? mysql_connect(localhost,user,password);
mysql_select_db(database);
$sql = select distinct category from books ORDER BY category ASC;
$makes_result = mysql_query($sql);
print (select name=\category\\n);
print(option selected value=\\Please select a
Category/option\n);
while($row = mysql_fetch_row($makes_result))
{
print(option value=\$row[1]\$row[1]/option\n);
}
print(/select); ?
Thanks for the help.
Barry
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