RE: [PHP-DB] Re: Display Mysql Result in drop down list
Hi:
Well, when i bring out the page with the drop down list it was able to display
all tutors' names from tutor_name column. Anyway here's a review of my code
(snip) again before i continue:
---
snip
$sql = INSERT INTO class (class_code, tutor_name, edu_level, timetable_day,
timetable_time)
VALUES
('$class_code','$tutor_name','$edu_level','$timetable_day','$timetable_time');
?//retrieve data from DB display in dynamic drop down ?
SELECT class=textarea name=tutor_name /
?
$sql = mysql_query(SELECT DISTINCT tutor_name FROM tutor );
while ($row = mysql_fetch_array($sql))
{
print OPTION VALUE=\$tutor_name\ SELECTED .$row
[tutor_name]. /option;
}
$result = $db-query($sql);
?
/select
?
while($selected_tutor_name == $tutor_name)
echo $_POST[tutor_name];
?
/snip
---
so when i submit the form, i am suppose to echo the values i have entered into
the field and then INSERT the values into DB (Queries stated above). However i
was able to echo all other values eg. class_code, edu_level, etc...but
not tutor_namesame thing happen when i do an INSERT, all other values
are inserted into DB but not $tutor_namewhy is this so???Really need some
help here...Anyway i have already specify a name to be reference :
SELECT class=textarea name=tutor_name
and then I also did an echo of tutor_name being selected:
while($selected_tutor_name == $tutor_name)
echo $_POST[tutor_name];
All help are greatly appreciated =)
Irin.
---
On Tue, 23 Dec 2003 16:00:04 -0500, Aleks @ USA.net [EMAIL PROTECTED] wrote:
---
Hi,
I am looking at your code and have a question, in your code you have
print OPTION VALUE=\$tutor_name\ SELECTED .$row [tutor_name].
/option;
Doesn't the SELECTED piece mark everything listed as selected?? When you
bring up the page with the drop down list and open the source, what does it
show??
I handle this a little differently, I create my drop down list as follows:
=== Code used
// First I retrieve all customer information
$BB = mysql_query(SELECT * FROM customer ORDER BY CID );
// Then I create the drop down list
SELECT style=WIDTH: 410px size=1 name=Customer
?
// Creates the Customer dropdown with the $id number
while ($Site = mysql_fetch_array($BB))
{
$Sid = $Site[CID];
$Sname = htmlspecialchars($Site[Customer]);
$SCity = htmlspecialchars($Site[City]);
$SState = htmlspecialchars($Site[State]);
$SCountry = htmlspecialchars($Site[Country]);
if($Sid == $Customer)
{ $add = ' selected'; } else { $add = ''; }
echo(option value='$Sid'$add$Sname nbsp;nbsp;nbsp; $SCity
nbsp;nbsp;nbsp; $SState nbsp;nbsp;nbsp; $SCountry /option\n);
}
?
/select
=== Code used
This will create a drop down list where the value of the option is the sites
ID. In another part of this form, if there is a value set for $Customer,
then it will set the $add to selected. This will then make the drop down
list auto select the customer.
The code above is passed to another form for processing and is were the
insert data occurs. Using this method I can echo the form data easily
Hope this helps...
Aleks
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Sent: Tuesday, December 23, 2003 4:33 AM
To: David Robley
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Re: Display Mysql Result in drop down list
Hi:
then now i'm trying to select a value from the drop down list, echo the
value i have selected and lastly INSERT the value into DB with the following
:
Below's a snippet of code:
snip
//retrieve all tutor_name from DB and display in drop down list
SELECT class=textarea name=tutor_name / ?
$sql = mysql_query(SELECT tutor_name FROM tutor ); while ($row =
mysql_fetch_array($sql)) { print OPTION VALUE=\$tutor_name\ SELECTED
.$row [tutor_name]. /option; } $result = $db-query($sql);
?
/select
/snip
snip
***INSERT selected values into
DB** $tutor_name = $_POST[tutor_name];
$sql = INSERT INTO class (class_code, tutor_name, edu_level, timetable_day,
timetable_time)
VALUES
('$class_code','$tutor_name','$edu_level','$timetable_day','$timetable_time'
);
//execute query statement
$result = $db-query($sql);
/snip
snip
*echo the value i have
selected***
?
if($selected_tutor_name == $tutor_name)
echo $_POST[tutor_name];
?
/snip
Problem: I was unable to echo the value i selected from drop down as well as
INSERT into DB
[PHP-DB] Re: Display Mysql Result in drop down list
In article [EMAIL PROTECTED],
[EMAIL PROTECTED] says...
Hi all,
Right now I'm trying to retrieve one of the column tutor_name and display al
the tutor's name in a drop down list. The problem now is, the drop down list
only manage to display 1 record from that row instead of all tutor's name
under tutor_name column...wonder where the problem lies??
Hope to get some help soon.
Below is a snip of the code:
Drop me a msg if anyone needs the entire code.
SELECT NAME=tutor_name class=textarea
?
$sql = mysql_query(SELECT DISTINCT tutor_name FROM tutor );
if ($row = mysql_fetch_array($sql))
replace the above with
while ($row = mysql_fetch_array($sql))
If will be true, but will only produce one result. You want to iterate
through all results, which is why you use while. It will return as many
results as are available.
{
print OPTION VALUE=\$tutor_name\ SELECTED .$row
[tutor_name]. /option;
}
$result = $db-query($sql);
?
/select
Thanks in advance =)
Irin.
--
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A: Because it messes up the order in which people normally read text.
Q: Why is top-posting such a bad thing?
A: Top-posting.
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Re: [PHP-DB] Re: Display Mysql Result in drop down list
Hi:
then now i'm trying to select a value from the drop down list, echo the value
i have selected and lastly INSERT the value into DB with the following :
Below's a snippet of code:
snip
//retrieve all tutor_name from DB and display in drop down list
SELECT class=textarea name=tutor_name /
?
$sql = mysql_query(SELECT tutor_name FROM tutor );
while ($row = mysql_fetch_array($sql))
{
print OPTION VALUE=\$tutor_name\ SELECTED .$row
[tutor_name]. /option;
}
$result = $db-query($sql);
?
/select
/snip
snip
***INSERT selected values into DB**
$tutor_name = $_POST[tutor_name];
$sql = INSERT INTO class (class_code, tutor_name, edu_level, timetable_day,
timetable_time)
VALUES
('$class_code','$tutor_name','$edu_level','$timetable_day','$timetable_time');
//execute query statement
$result = $db-query($sql);
/snip
snip
*echo the value i have selected***
?
if($selected_tutor_name == $tutor_name)
echo $_POST[tutor_name];
?
/snip
Problem: I was unable to echo the value i selected from drop down as well as
INSERT into DB...wonder where have i gone wrong???
Reali need some help here...all help are greatly appreciated =)
Thanks in advance.
Irin.
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Re: Display Mysql Result in drop down list
Hi,
I am looking at your code and have a question, in your code you have
print OPTION VALUE=\$tutor_name\ SELECTED .$row [tutor_name].
/option;
Doesn't the SELECTED piece mark everything listed as selected?? When you
bring up the page with the drop down list and open the source, what does it
show??
I handle this a little differently, I create my drop down list as follows:
=== Code used
// First I retrieve all customer information
$BB = mysql_query(SELECT * FROM customer ORDER BY CID );
// Then I create the drop down list
SELECT style=WIDTH: 410px size=1 name=Customer
?
// Creates the Customer dropdown with the $id number
while ($Site = mysql_fetch_array($BB))
{
$Sid = $Site[CID];
$Sname = htmlspecialchars($Site[Customer]);
$SCity = htmlspecialchars($Site[City]);
$SState = htmlspecialchars($Site[State]);
$SCountry = htmlspecialchars($Site[Country]);
if($Sid == $Customer)
{ $add = ' selected'; } else { $add = ''; }
echo(option value='$Sid'$add$Sname nbsp;nbsp;nbsp; $SCity
nbsp;nbsp;nbsp; $SState nbsp;nbsp;nbsp; $SCountry /option\n);
}
?
/select
=== Code used
This will create a drop down list where the value of the option is the sites
ID. In another part of this form, if there is a value set for $Customer,
then it will set the $add to selected. This will then make the drop down
list auto select the customer.
The code above is passed to another form for processing and is were the
insert data occurs. Using this method I can echo the form data easily
Hope this helps...
Aleks
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Sent: Tuesday, December 23, 2003 4:33 AM
To: David Robley
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Re: Display Mysql Result in drop down list
Hi:
then now i'm trying to select a value from the drop down list, echo the
value i have selected and lastly INSERT the value into DB with the following
:
Below's a snippet of code:
snip
//retrieve all tutor_name from DB and display in drop down list
SELECT class=textarea name=tutor_name / ?
$sql = mysql_query(SELECT tutor_name FROM tutor ); while ($row =
mysql_fetch_array($sql)) { print OPTION VALUE=\$tutor_name\ SELECTED
.$row [tutor_name]. /option; } $result = $db-query($sql);
?
/select
/snip
snip
***INSERT selected values into
DB** $tutor_name = $_POST[tutor_name];
$sql = INSERT INTO class (class_code, tutor_name, edu_level, timetable_day,
timetable_time)
VALUES
('$class_code','$tutor_name','$edu_level','$timetable_day','$timetable_time'
);
//execute query statement
$result = $db-query($sql);
/snip
snip
*echo the value i have
selected***
?
if($selected_tutor_name == $tutor_name)
echo $_POST[tutor_name];
?
/snip
Problem: I was unable to echo the value i selected from drop down as well as
INSERT into DB...wonder where have i gone wrong???
Reali need some help here...all help are greatly appreciated =) Thanks in
advance.
Irin.
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