[RESOLVED] selected value in combo box

[kopul784]

I am trying to retrieve data from a database and print it into a form so it can be edited and resubmitted. I know how to print the result in a standard text box but not how to do this for a combo box.
I created an array with the same values as stored in the enum field in the database and this displays fine but how can i get the form to display the value stored in the database?
Any help would be great!

	$q = "SELECT FirstName, Surname, PostID, ReportsTo, Team, HoursWorkedPerWeek, Reference, Team, CentreName, FundingStream, Salary, StartDate
	FROM EmploymentDetails WHERE EmployeeID=$ID";

$r = @mysqli_query($dbc, $q);

if (mysqli_num_rows($r) == 1)  {
	$row = mysqli_fetch_array($r, MYSQLI_ASSOC);

$team = array(1 => 'Management', ' General Advice', 'Legal Services Commission', ' Primary Care Advice Link', 'Financial Inclusion Fund');

echo'

<form action="EmploymentDetails.php" method="post">

<form>
<p>Reports To: <input type="text" name="ReportsTo" size="15" value="' .$row['ReportsTo'].'"></p>
<p>Hours Per Week: <input type="text" name="HoursWorkedPerWeek" size="5" value="' .$row['HoursWorkedPerWeek']. '"></p>

<p>Team:<select name="Team">';
	foreach ($team as $k => $teams) {
	echo "<option value=\"$teams\">$teams</option>\n";
	}
	echo '</select>';
	echo '</p>

<p>Reference: <input type="radio" name="reference" value="pending">Pending
	      <input type="radio" name="reference" value="applied">Applied
	      <input type="radio" name="reference" value="recieved">Recieved</p>
<p>Salary: <input type="text" size="8" name="Salary" value="' . $row['Salary'] . '"></p>
<p>Start Date: <input type="text" size="10" name="StartDate" value="' .$row['StartDate']. '"></p>
<input type="hidden" name="submitted" value="TRUE"/>
<input type="hidden" name="ID" value="' .$ID. '"/>
<input type="submit" name="submit" value="Submit"/>
</form>';

[nemonoman]

I assume you can retrieve (or have retrieved) the previously selected teams from the database. This solution will work for both a single value (1-1 relation) or multiple values( 1- many). Call this data $savedteam. It can be a single value or an array.

$teamoptions='';
    foreach ($team as $k => $teams) { 
    $teamoptions.="<option value=\"$teams\">$teams</option>\n"; 
    } 
//here's where you highlight the selections:
$teamoptions=str_replace("<option value=\"$savedteam", "<option selected value=\"$savedteam",$teamoptions);
echo $teamoptions;
//then continue as before

[kopul784]

Thanks for your help, i put the code in that you suggested but its still not showing the right value when the page loads up. Im guessing i have not set your code up right?

	$q = "SELECT FirstName, Surname, PostID, ReportsTo, Team, HoursWorkedPerWeek, Reference, Team, CentreName, FundingStream, Salary, StartDate
	FROM EmploymentDetails WHERE EmployeeID=$ID";

$r = @mysqli_query($dbc, $q);

if (mysqli_num_rows($r) == 1)  {
	$row = mysqli_fetch_array($r, MYSQLI_ASSOC);

$team = array(1 => 'Management', ' General Advice', 'Legal Services Commission', ' Primary Care Advice Link', 'Financial Inclusion Fund');


$savedteam = $row['CentreName'];

echo'

<form action="employmentDetails.php" method="post">

<form>

<p>Team:<select name="Team">';
	$teamoptions='';
	foreach ($team as $k => $teams) {
	 $teamoptions.="<option value=\"$teams\">$teams</option>\n"; 
	}
	$teamoptions=str_replace("<option value=\"$savedteam", "<option selected value=\"$savedteam",$teamoptions);
	echo $teamoptions;
	echo '</select>';

[nemonoman]

If (and only if) the precise value of $savedteam appears in $teamoptions the code will work. If $savedteam varies from $teamoptions (like an extra space, etc.) then it will not work. Success means you'll see <option selected value= in the return

[kopul784]

Resolved now, instead of storing an identical array in the table i changed it to a string and it works perfectly.