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Apr 15th, 2007, 10:12 PM
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#1 |
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Programmer
Join Date: Mar 2007
Posts: 33
Rep Power: 0
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Combo Box selected item - from DB
Hey guys,
I have a database which contains a list of game names, i am reading it and using it to create a combo box in a webpage. When a user submits an item, and then views the same page again, i want that item to be the one automatically selected in the combo box. i know that this creates a selectable item: <option value="3">Age Of Empires 2</option> and that this: <option value="3" selected>Age Of Empires 2</option> will force that item to be the automatically selected item. but i cannot get this to work dynamically based on the database reads, this is the code (at its current state, i have mashed with it that much just seeing what effect different things have) but havent really gotten any closer: <?php session_start(); ?>
<html>
<head>
<title>Update Your Games</title>
<body>
<Form method="post" action="AddGames.php">
<fieldset>
<legend>Update Your Games - TEST</legend>
<?php
include ("Functions.php");
Global $Option;
$Username = safe_output($_SESSION['username']);
$GamesList = mysql_query("SELECT * FROM games ORDER BY GameName");
$ViewMyGames = mysql_query("SELECT Game1,Game2,Game3,Game4,Game5 FROM members WHERE Username='$Username'");
If (!$GamesList) {
Print "Unable to retrieve games list";
} Else {
$a = mysql_fetch_array($GamesList);
$ChosenGames = mysql_fetch_array($ViewMyGames);
print sizeof($a);
While ($row = mysql_fetch_array($GamesList)) {
If ($ChosenGames['Game1'] == $row['GameID']) {
$Option = $Option.'<option value="'.$row['GameID'].'" selected>'.$row['GameName'].'</option><br/>';
} ElseIf ($ChosenGames['Game2'] == $row['GameID']) {
$Option = $Option.'<option value="'.$row['GameID'].'" selected>'.$row['GameName'].'</option><br/>';
} ElseIf ($ChosenGames['Game3'] == $row['GameID']) {
$Option = $Option.'<option value="'.$row['GameID'].'" selected>'.$row['GameName'].'</option><br/>';
} ElseIf ($ChosenGames['Game4'] == $row['GameID']) {
$Option = $Option.'<option value="'.$row['GameID'].'" selected>'.$row['GameName'].'</option><br/>';
} ElseIf ($ChosenGames['Game5'] == $row['GameID']) {
$Option = $Option.'<option value="'.$row['GameID'].'" selected>'.$row['GameName'].'</option><br/>';
} Else {
$Option = $Option.'<option value="'.$row['GameID'].'">'.$row['GameName'].'</option><br/>';
}
$i = 1;
While ($i <= 5) {
Print "<p>Game $i:<br/>";
Print '<select name="game'.$i.'"><br>';
Print $Option;
Print "</select></p>";
$i = $i + 1;
}
}
}
mysql_close($con);
?>
</p>
<input type="submit" name="action" value="send" />
</p>
</fieldset>
</form>
<?php
print_menu();
?>
</body>
</html>If someone has a nice simple example script for me to go by, that would be much appreciated. thanks /tAK |
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Apr 17th, 2007, 11:36 PM
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#2 |
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Programmer
Join Date: Mar 2007
Posts: 39
Rep Power: 0
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[php]<?php
$var = $_POST['var']; echo '<select name="var">'; for ($i = 0; $i < 10; $i++) { $selected = ($var == $i) ? 'selected="selected"' : ''; echo "<option value=\"$i\" $selected>$i</option>"; } echo '</select>'; ?>[/php] The $selected part is I believe what you want. This example uses a $_POST variable though. |
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Apr 18th, 2007, 10:32 PM
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#3 |
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Programmer
Join Date: Mar 2007
Posts: 33
Rep Power: 0
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seems to do what i wanted..
it also appears to be close to what i was trying, with a few exceptions.. hopefully i can sort it all out now that i have seen a script capable of this on its own. i'll be sure to drop back in and let you know how i went Cheers /tAK |
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Apr 23rd, 2007, 12:53 AM
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#4 |
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Programmer
Join Date: Mar 2007
Posts: 33
Rep Power: 0
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Righto, i said i would let you know how i went, so here goes..
with great thanks to Styx, i got it working for 1 combo box, i then stuck the while loop for reading the array of data into a for loop, and was getting 5 drop downs as needed, BUT !! when you do a mysql query: $query = mysql_query("SELECT * FROM table"); and then fetch it in a while loop: while($var = mysql_fetch_array($query)) { print_r($var); } it somehow closes off the $query, so putting that WHILE loop into a FOR loop means that the other X number of times it runs, it doesnt output anything. how to solve this?, well.. easy enough.. it goes something like this: for ($i = 1; $i < 6; $i++) { $query = mysql_query("SELECT * FROM table"); while($var = mysql_fetch_array($query)) { print_r($var); } } You then get 5 outputs of the array, and i just did the basic IF statements etc.. eitherway, from what i can tell this is giving the database a right royal flogging, by hitting it with 5 requests for the same chunk of data when i should be able to grab it and hold it.. shouldn't i? |
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Apr 23rd, 2007, 10:08 AM
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#5 |
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Programmer
Join Date: Mar 2007
Posts: 39
Rep Power: 0
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[php]$query = mysql_query("SELECT * FROM table");
for ($i = 0; $i < ($var = mysql_fetch_array($query)); $i++) { print_r($var); }[/php] |
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Apr 23rd, 2007, 10:29 AM
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#6 |
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Resident Grouch
 
Join Date: Jun 2005
Posts: 6,453
Rep Power: 10
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Try not to forget your code tags, Tak.
__________________
Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers |
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