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Best Way To Show Selected Option In List Box


I need a way so that we can show selected option in list box without using if condition in loop since some times we have lot of options and we have too much if conditions to check.




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I have a drop down list with the town options 'Bury' and 'Ipswich' as
shown below.

When selecting one of the options, its value is passed to variable
$townsearch.

How do I change the drop down list so the option selected last time is
then the current one shown in the drop down box?

<body>

<?php $townsearch = $_get['town']; ?>

<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="get">

<select size="1" name="town">

<option>ipswich</option>
<option>bury</option>
&nbsp;
</select>&nbsp;

<input type="submit" value="GO" />

</form>

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I've got a little text editor thing on my site, and to open a file, you input the file name and choose the folder from a drop-down list and click the Open button. The default filename is "index" and that file keeps track of all the files contained in that folder. It works okay. What's annoying is that regardless of what folder you're messing around in, the drop down list always goes back to showing the default folder name as the page is reloaded. I want it to be preselected to the folder it's in.

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I have a drop down list which I want to be set to what was selected when the form is submitted. As the form is processed by the same page, it returns to the preset default. How would I go about making this selection box set the option selected on the next page?

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Given a dropdown form how can I output the selected option when the submit button is pressed.

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[green ]
[blue ]
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I have a page that is intendend to update a table field on DB i have a drop down menu wich displays all the position fields available on db and a text area where i will insert the new data.

now the problem: I'm using the query SELECT * FROM table WHERE position = $position
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Having major issues with this simple task, and I cant work out why its
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Basically, ive got a drop down box with the added bit of php:

<select name="title" id="title">
<?php
if($_SESSION['title'] != "") {
echo '<option value="'. $_SESSION['title'] .'"
selected="selected"></option>'
}
else {
echo "<option value="" selected></option>";
}
?>

<option>Mr</option>
<option>Mrs</option>
<option>Miss</option>
<option>Ms</option>
<option>Dr</option>
</select>

Basically, if the user submits a form, the title is stored in the
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Yet, when they return to the page, there value they selected is not
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correct, and this is whats shown:

<select name="title" id="title">
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Multiple Select Box With Selected Option ...
first of all I have three tables (tournament_game, umpire_game, umpires). Updating tournament_game umpires, each game may have from two to four umpires, and those umpires can work from one to n number of games, that's why I'm using connection table umpire_game. I'd like to get selected multiple select box of umpires in game, it could be two to four options, with all umpires listed as options example:

<select name=umpire_game[] size=10 multiple>
<option value=$umpire_id>$umpire_name</option>
<option value=$umpire_id selected>$umpire_name</option>
<option value=$umpire_id selected>$umpire_name</option>
<option value=$umpire_id>$umpire_name</option>
<option value=$umpire_id selected>$umpire_name</option>
<option value=$umpire_id selected>$umpire_name</option>

PHP Code:

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How To Get Form To Remember Selected Option?
I am now trying to make a form remember which color (a select list) they selected before pressing any of the submit buttons. I entered this code into the form:

<select name="color">
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The submit is a GET type. For example, if the user selects blue and then clicks on either one of the "submit" buttons, I want the form to remember/show the blue option. Right now the select list always goes back to red, the first choice. I tried doing a <?php echo $red; ?> and <?php echo $blue; ?> for the option value, but this does not work since the select list needs a value.

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I have a normal drop down menu:

<select name="status" class="db_list_text">
                    <option value="All" selected="selected">All</option>
                    <option value="True">Active</option>
                    <option value="False">Inactive</option>
                  </select>
The idea for this is for the user to select either true or false and then to click submit to find data that matches either one. The coding is in the same page so the form basically just reloads the page and grabbes whatever info is requested. How can I with PHP, make it that the dropdown stays on the selected option when the page is reloaded?

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Get The Option Selected Once I Have Submitted The Form.
I am using the following code to loop 12 months backwards from the current month in a drop down list - using the get method.

I'm having difficulties trying to get the option selected once I have submitted the form.  For example, I am using this to filter months, so once a month has been selected and submitted, the first option in my drop down list is displayed and not the one that was selected and submitted. So basically, I need to be able to select a month > hit submit and when the form returns, have the drop down list show me the selected item. Code:

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I have a dir listing appearing in a list field of a form as showwn below PHP Code:

<?php
    $current_dir = 'images/upload/'
    $dir = opendir($current_dir);
        
        $list = "<select name="listbox">";
        while ($file = readdir($dir))
        {
           
            $list .= "<option value="$file">$file</option>";
        }
        $list .= "</select>";
        {
        echo $list;
        }
        closedir($dir);

?>

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I then need to print to screen the currently selected option, but on the same page- no going to a new page etc. PHP Code:

$host = "localhost";
$user = "root";
$password = "!";
$database = "rio_test";

$connection = mysql_connect($host,$user,$password)
or die ("Couldn't connect to server.");
$db = mysql_select_db($database, $connection)
or die ("Couldn't select database.");

echo "<select name="ticketName" id="ticketName">";
echo "<option>IPTIS Ticket Name</option>";
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I am having problem getting the logic to work to get a multiple option box to get populated from the database and select the ones which are already in the database. Here is my code: PHP Code:

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Automatic Selected Option + Select Menu
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PHP/MySQL: Query Based On Selected Form Option
Using PHP and MySQL. Trying to put a list of categories into a drop down
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<form name="form" action="<? print $_SERVER['PHP_SELF']?>" method="get">
<select name="subject">
<option value=""></option>
<option value="field1">Field 1</option>
<option value="field2">Field 2</option>
</select>

<input type="submit" name="Submit" />
</form>

Then I want to process it so the MySQL query gets done depending on what
was selected. I came up with this:

//connect to database
mysql_connect("$dbhost","$dbuser","$dbpass");
mysql_select_db("$dbase") or die("Unable to select database");

// Build SQL Query
if (isset($_GET['subject']))
{
switch($_GET['subject'])
{
case 'field1':
$query = "select * from tips where category = 'field1'";
break;
case 'field2':
$query = "select * from tips where text like 'field2' ";
break;
default:
echo 'No subject found'
}
}

$results=mysql_query($query);
$numrows=mysql_num_rows($results);

etc etc etc

But that switch doesn't seem to work. Anyone have a suggestion as to how
I can code this to do the MySQL query based on the subject?

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I have a query that shows all appointments of a chosen day using the following format:

$eventid=2007-05-21

WHERE starttime LIKE '".$eventid."%'

I have a calendar and am trying to make it show all records for the selected month, or display all records from a specific day if a day is selected from the calendar.

So far I have it where I can show records from specific day, but want selected months records as default.

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In my pages i have a select list. For example the visitor has
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Let's call this script script1.php
In this script i want to check if there are no errors, i mean i want to see if
the visitor choose a valid location (USA, FR, IT or SW ) and not the default
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Now if the user select didn't select a location (equivalent to selected the
default value) the same script after is executed will will echo an error
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Now there is one more thing to ask. If the visitor goes back from script2.php
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I have a database with 1 to many and the many is a list with multiple
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$dresscat = specdresscat($dresstypeid); //this is the query to get the
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$drcatrow = pg_fetch_array($dresscat);// the array for the detail
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<select name="stages[]" multiple="multiple">
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while ($allstagerows=pg_fetch_array($allstages)) {
$eachstage= $allstagerows["stageid"];
$specstage= $drcatrow["stageid"]; // this is the point I need help
if ($specstage == $eachstage){ ?>
<option value=<?php echo $eachstage;?> selected="selected"><?php echo
allstagerows["stage"]?></option>
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?>
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I frequently work with forms where menus are pulled from a database. When a user enters data in the form and then saves it, if it is retrieved later, the form shows the saved data, including the drop-down menus.

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Both of the below scripts work as intended, but how do I make the mySql query results work with my existing script?  Yes, I know I must replace the "$options = array" part in the top piece of code, but I need a little more guidance than that. Code:

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i have problem with the following function by where i was not able to get the $id display from database. and i have no problem with $data. is there anything got to do with primary key? shouldn't be right? then it would be sytax error? but i checked for so many times but still can't figure out what's wrong with it.

id tinyint(2) - primary key
data char(100 )

function writeOptionList( $table,$field )
{
global $connect;

$result = mysql_query( "SELECT $field FROM $table", $connect );
if (! $result )
{
print "failed to open $table<p>";
return false;
}

while ( $a_row = mysql_fetch_array( $result ) ){
print "<option value="$a_row[id]"";

print ">$a_row[data]
"; } }

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The next page reads the form value as "" instead of "None" but only for the first menu. The other three menus work fine. I know this isn't really php related, but this is usually such a helpful forum that I thought I'd post here before I start to search for some helpful html forum. Answers would be greatly appreciated, but so would corrections for my terminology (which I think I am confused about) or a good link to a place where I should be asking this question.

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<?php
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<li><a href="filelocation2">filename 1</li>
<li><a href="filelocation3">filename 1</li>
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etc.?

So basically it creates a list of links with the contents in that directory,
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mbxno: 1114 msgno: 0412141623 msglist: 1
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mbxno: 1114 msgno: 0412141623 msglist: 1
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mbxno: 2189 msgno: 0412141408 msglist: 1
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How should I be doing this? If nothing else, what kind of join will work with a long list compared to a value from a short list?

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I need to produce a list of tables currently within my database and list them in a drop down box.

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