storing the selected value of a drop-down list

Question

Venom Rush 21 Posting Whiz

16 Years Ago

What I'd like to do is store the "selected" value of the drop down box in a DB. At the moment the selected value for the drop down list is pulled from the DB. Now what I want to happen is if I don't want to change that specific option from the drop-down list then the variable gets put back into the DB. At the moment if I don't select anything then my the value that is put back into the DB is 0.

My code is as follows:

The drop-down list

<?php	
// open connection to MySQL server
$connection = mysql_connect('localhost', 'username', 'password')
or die ('Unable to connect!');
			
//select database
mysql_select_db('database') or die ('Unable to select database!');
						
//create and execute query
$query = "SELECT day(farmFullStartDate) FROM rates";
$result = mysql_query($query) or die ('Error in query: $query. ' . mysql_error());
									
//create selection list
echo "<select name='startDay' class='eventDay' id='startDay'>\n";
while($row = mysql_fetch_row($result))
  {
    echo "<option selected='selected'>$row[0]</option>";
  }
echo  "<option>01</option>";
echo  "<option>02</option>";
echo  "<option>03</option>";
echo  "<option>04</option>";
echo  "<option>05</option>";
echo  "<option>06</option>";
echo  "<option>07</option>";
echo  "<option>08</option>";
echo  "<option>09</option>";
echo  "<option>10</option>";
echo  "<option>11</option>";
echo  "<option>12</option>";
echo  "<option>13</option>";
echo  "<option>14</option>";
echo  "<option>15</option>";
echo  "<option>16</option>";
echo  "<option>17</option>";
echo  "<option>18</option>";
echo  "<option>19</option>";
echo  "<option>20</option>";
echo  "<option>21</option>";
echo  "<option>22</option>";
echo  "<option>23</option>";
echo  "<option>24</option>";
echo  "<option>25</option>";
echo  "<option>26</option>";
echo  "<option>27</option>";
echo  "<option>28</option>";
echo  "<option>29</option>";
echo  "<option>30</option>";
echo  "<option>31</option>";
echo "</select>";
?>

And this is the php that updates the DB

// open connection to MySQL server
$connection = mysql_connect('localhost', 'username', 'password')
or die ('Unable to connect!');
		
//select database
mysql_select_db('database') or die ('Unable to select database!');
	
//define variables
$startDay = $_POST['startDay'];
$startMonth = $_POST['startMonth'];
$startYear = $_POST['startYear'];
$startDate = $startYear."".$startMonth."".$startDay;
$endDay = $_POST['endDay'];
$endMonth = $_POST['endMonth'];
$endYear = $_POST['endYear'];
$endDate = $endYear."".$endMonth."".$endDay;
$luxury = $_POST['luxury'];
$orchid = $_POST['orchid'];
	
//create query
$query = "UPDATE rates SET farmFullStartDate = '$startDate', farmFullEndDate = '$endDate', farmFullLuxury = '$luxury', farmFullOrchid = '$orchid'";
//$query = "INSERT INTO rates (farmFullStartDate, farmFullEndDate, farmFullLuxury, farmFullOrchid) VALUES ('$startDate', '$endDate', '$luxury', '$orchid')";
	
//execute query
$result = mysql_query($query)
or die ("Error in query: $query. " . mysql_error());
	
// close connection
mysql_close($connection);

Any help would be greatly appreciated.

P.S. If I haven't explained myself well enough then here's an example situation:

I select 05 from startDay and hit submit. The page refreshes and 5 is displayed in the startDay drop-down list. Now I don't want to change startDay but I want to change other fields in the form so I do so and hit submit. What I have changed gets sent to the DB and 0 gets sent to the DB for startDay. What should be happening is it should be sending 5 to the DB.

php

4 Contributors
7 Replies
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1 Year Discussion Span
Latest Post 15 Years Ago Latest Post by Venom Rush

All 7 Replies

ezb 0 Newbie Poster

16 Years Ago

The first problem I can see if the options have no values, so there for they can not post anything?

<option value="4">4</option>

Not sure if that will sort it but worth a try

JeniF 0 Junior Poster in Training

16 Years Ago

I have built a dynamic select box for something like this.
Here I populate the select options with information from a table:
(I also created the "sticky" to hold the users input upon error checking

<?php
$result = mysql_query("SELECT * FROM facilities ORDER BY Street1 ASC") or 
die(mysql_error());
$sticky= '';
if (isset($_POST['facility']))
$sticky = ($_POST['facility']);
$pulldown1 = '<select name="facility">';
$pulldown1 .= '<option></option>';
while($row = mysql_fetch_array($result))
{
if($row['FacilitiesID'] == $sticky) {
$pulldown1 .= "<option selected value=\"{$row['FacilitiesID']}\">
								{$row['Street1']}&nbsp;-&nbsp;
{$row['City']}&nbsp;-&nbsp;
{$row['Name']}
</option>\n";
} else {
$pulldown1 .= "<option value=\"{$row['FacilitiesID']}\">
{$row['Street1']}&nbsp;-&nbsp;
{$row['City']}&nbsp;-&nbsp;
{$row['Name']}
</option>\n";
}
}
$pulldown1 .= '</select>';
echo $pulldown1;	
?>

Then all I do is place the value in the insert statement

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Venom Rush 21 Posting Whiz · Answer 1 · 2007-11-13T15:25:37+00:00

The first problem I can see if the options have no values, so there for they can not post anything?
Not sure if that will sort it but worth a try

The values get posted without the value specified. So if I select <option>04</option> then it gets posted to the database.

My problem is that <option selected='selected'>$row[0]</option> isn't being posted to the DB. And adding a value to it doesn't help. 0 still gets processed.

Venom Rush 21 Posting Whiz · Answer 2 · 2007-11-13T15:52:43+00:00

I've come up with an idea. What I could do is use an if statement in the code that posts the variales to the DB. Something along the lines of the following:

if (farmHouse.startDay.value = $row[0]) {
$startDay = $row[0];
}
else {
$startDay = $_POST['startDay'];
}

I know this won't work. I have no idea how to point to the form using php as I'm somewhat new to it.

Could anyone help me with this please.

Venom Rush 21 Posting Whiz · Answer 3 · 2007-11-13T21:01:03+00:00

Thanks for your response JeniF.

Managed to sort my problem out with the following code:

if ($_GET['startDay'] = $row[0]) {
    $day = $_GET['startDay'];
    if ($day < 10) {
        $startDay = "0".$day;
    } else {
    $startDay = $_GET['startDay'];
    }
} else {
$startDay = $_POST['startDay'];
}

mad0 0 Newbie Poster · Answer 4 · 2009-02-23T09:41:49+00:00

i know this thread if quite old, but i just wanted to say thanks for lingering web pages, i almost tore my hair out trying to figure this out and literally just as I was about to call it a day, found quite the easy solution, so thanks JeniF and Venom Rush and Daniweb. i created an account just for this today. THANKS!

Venom Rush 21 Posting Whiz · Answer 5 · 2009-02-23T13:21:01+00:00

i know this thread if quite old, but i just wanted to say thanks for lingering web pages, i almost tore my hair out trying to figure this out and literally just as I was about to call it a day, found quite the easy solution, so thanks JeniF and Venom Rush and Daniweb. i created an account just for this today. THANKS!

Welcome to the community and glad this thread helped someone other than myself ;)

storing the selected value of a drop-down list

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