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OPTION problem

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Join Date: Dec 2007

Posts: 28

Reputation: abhi287 is an unknown quantity at this point

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abhi287

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OPTION problem

Feb 27th, 2008

Hi

i m trying to get the values from database and put it into the select options..
i m using code-

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
<?php
$con = mysql_connect("localhost","cdccpl","d123");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
					  mysql_select_db("cdccpl_aus", $con);
$result = mysql_query("SELECT * FROM bak_bill_cust");
 while( $row = mysql_fetch_array( $result ) ):
 $option = $row["cust_code"];
 
echo "<option value='$row[cust_code]'>$row[cust_code]</option>";
endwhile;
 mysql_close($con);?>
<?php
$con = mysql_connect("localhost","cdccpl","d123");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
					  mysql_select_db("cdccpl_aus", $con);
$result = mysql_query("SELECT * FROM bak_bill_cust");
 while( $row = mysql_fetch_array( $result ) ):
 $option = $row["cust_code"];

echo "<option value='$row[cust_code]'>$row[cust_code]</option>";
endwhile;
 mysql_close($con);?>

but i m getting displayes with the values as--

Help with Code Tags

PHP Syntax (Toggle Plain Text)

Customer auswidestonemanpnicholsonllwmplnewuserguestanftasturnbullanglic

means values getting displayed one after another and not in option..

why so..
plz help

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Join Date: Nov 2007

Posts: 84

Reputation: johnsquibb is an unknown quantity at this point

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johnsquibb johnsquibb is offline

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Junior Poster in Training

Re: OPTION problem

Feb 27th, 2008

echo your option tags inside of a select tag like this:

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
<select name="test" id="test">
	<option value="value 1">option 1</option>
	<option value="value 2">option 2</option>
	<option value="value 3">option 3</option>
</select>
<select name="test" id="test">
	<option value="value 1">option 1</option>
	<option value="value 2">option 2</option>
	<option value="value 3">option 3</option>
</select>

The End

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Join Date: Apr 2005

Posts: 1,429

Reputation: ShawnCplus is a glorious beacon of light

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ShawnCplus ShawnCplus is offline

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Code Monkey

Re: OPTION problem

Feb 27th, 2008

Well firstly

 Help with Code Tags 
 php Syntax (Toggle Plain Text) 
$result = mysql_query("SELECT * FROM bak_bill_cust");
 while( $row = mysql_fetch_array( $result ) ):
 $option = $row["cust_code"];
 
echo "<option value='$row[cust_code]'>$row[cust_code]</option>";
endwhile;
$result = mysql_query("SELECT * FROM bak_bill_cust");
 while( $row = mysql_fetch_array( $result ) ):
 $option = $row["cust_code"];

echo "<option value='$row[cust_code]'>$row[cust_code]</option>";
endwhile;

>>>

 Help with Code Tags 
 php Syntax (Toggle Plain Text) 
$result = mysql_query("SELECT * FROM bak_bill_cust");
while( $row = mysql_fetch_assoc( $result ) )
{
    $option = $row["cust_code"];
    echo "<option value='$option'>$option</option>";
}
$result = mysql_query("SELECT * FROM bak_bill_cust");
while( $row = mysql_fetch_assoc( $result ) )
{
    $option = $row["cust_code"];
    echo "<option value='$option'>$option</option>";
}

I actually wasn't even aware PHP could use that format for while loops but I wouldn't advise its use given that it completely breaks the Perl-like syntax and just makes for ugly code.

Last edited by ShawnCplus; Feb 27th, 2008 at 3:32 pm.

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Join Date: Nov 2007

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Reputation: nav33n is a jewel in the rough

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nav33n

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Re: OPTION problem

Feb 27th, 2008

•

I actually wasn't even aware PHP could use that format for while loops

Yeah.. I wasn't aware too..

Ignorance is definitely not bliss!

*PM asking for help will be ignored*

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Join Date: Aug 2007

Posts: 52

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JeniF

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Re: OPTION problem

Feb 27th, 2008

Not sure if your porblem is solved yet, but here is something I use all the time for drop downs.
First get your connection string going. preferably using an include.
then use the following code, adapting statement to your needs:

 Help with Code Tags 
 php Syntax (Toggle Plain Text) 
<?php
$result = mysql_query("SELECT * FROM users ORDER BY username ASC") or 
die(mysql_error());
	$sticky= '';
	if (isset($_POST['id']))
	$sticky = ($_POST['id']);
	$pulldown1 = '<select name="id">';
	$pulldown1 .= '<option></option>';
	while($row = mysql_fetch_array($result))
		{
		if($row['id'] == $sticky) {
		$pulldown1 .= "<option selected value=\"{$row['id']}\">
			{$row['username']}&nbsp;-&nbsp;
                                                {$row['id']}				
                                                </option>\n";
                                } else {
                                $pulldown1 .= "<option value=\"{$row['id']}\">					{$row['username']}&nbsp;-&nbsp;
                                                {$row['id']}
                                                </option>\n";
                                           }
                                }
                               $pulldown1 .= '</select>';
                                echo $pulldown1;	
            ?>
<?php
$result = mysql_query("SELECT * FROM users ORDER BY username ASC") or 
die(mysql_error());
	$sticky= '';
	if (isset($_POST['id']))
	$sticky = ($_POST['id']);
	$pulldown1 = '<select name="id">';
	$pulldown1 .= '<option></option>';
	while($row = mysql_fetch_array($result))
		{
		if($row['id'] == $sticky) {
		$pulldown1 .= "<option selected value=\"{$row['id']}\">
			{$row['username']}&nbsp;-&nbsp;
                                                {$row['id']}				
                                                </option>\n";
                                } else {
                                $pulldown1 .= "<option value=\"{$row['id']}\">					{$row['username']}&nbsp;-&nbsp;
                                                {$row['id']}
                                                </option>\n";
                                           }
                                }
                               $pulldown1 .= '</select>';
                                echo $pulldown1;	
            ?>

This works well for holding the users selected values when error checking as well...
hope this helps

I keep hitting "escape", but I'm still here!!!!
:}

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Join Date: Dec 2007

Posts: 28

Reputation: abhi287 is an unknown quantity at this point

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abhi287

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Re: OPTION problem

Feb 28th, 2008

Hi all..
thanks for reply..

Problem is solved..

Last edited by abhi287; Feb 28th, 2008 at 12:29 am.

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Join Date: Jun 2007

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kkeith29

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Nearly a Posting Virtuoso

Re: OPTION problem

Feb 28th, 2008

try

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
$sql = "SELECT * FROM bak_bill_cust";
$query = mysql_query($sql) or die('Error: ' . mysql_error());
$select = '<select name="test">';
$select .= '<option value=""></option>';
while ($row = mysql_fetch_assoc($query)) {
$select .= '<option value="' . $row['cust_code'] . '">' . $row['cust_code'] . '</option>';
}
$select .= '</select>';
echo $select;
$sql = "SELECT * FROM bak_bill_cust";
$query = mysql_query($sql) or die('Error: ' . mysql_error());
$select = '<select name="test">';
$select .= '<option value=""></option>';
while ($row = mysql_fetch_assoc($query)) {
$select .= '<option value="' . $row['cust_code'] . '">' . $row['cust_code'] . '</option>';
}
$select .= '</select>';
echo $select;