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skyhighweb

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Everything posted by skyhighweb

  1. Still On submenu issue

    Am able to use this code to bind a.id = :auc_id in order to display certain result needed for menu code, but when i introduce it in the submenu i dont get any result below are the codes for menu and submenu' MENU works fine, display whats need to be showned thanks to a.id = :auc_id $query = "SELECT s.team_id, s.teams AS teams1, ss.team_id, ss.teams AS teams2, a.id, a.team1, a.team2 FROM " . $DBPrefix . "auctions a LEFT JOIN " . $DBPrefix . "sports s ON (s.team_id = a.team1) LEFT JOIN " . $DBPrefix . "sports ss ON (ss.team_id = a.team2) WHERE a.id = :auc_id and a.id = a.id"; $params = array(); $params[] = array(':auc_id', $id, 'int'); $db->query($query, $params); ?> <script type="text/javascript" src="js/dropdownjquery.js"></script> <script type="text/javascript"> $(document).ready(function() { $("#menu").change(function() { $(this).after('<div id="loader"><img src="images/loading.gif" alt="loading subcategory" /></div>'); $.get('loadsubcat.php?menu=' + $(this).val(), function(data) { $("#sub_cat").html(data); $('#loader').slideUp(200, function() { $(this).remove(); }); }); }); }); </script> <form name="bid"> <label for="category">Select Winner</label> <select name="willwin" id="menu"> <?php if ($db->numrows() > 0){ while ($row = $db->fetch()) { ?> <option value=''></option> <option value="<?php echo $row["team1"]; ?>"><?php echo $row["teams1"]; ?></option> <option value="<?php echo $row["team2"]; ?>"><?php echo $row["teams2"]; ?></option> <?php } } ?> SUBMENU displays all the details in bid table column auction bidder willwin willlose 38 4 3 2 39 4 2 4 39 5 4 2 the result should have a display for column ((auction) 38) willwin (3) but instead i get 3,2,4 <?php include('config.php'); $menu = $_GET['menu']; $query = mysql_query("SELECT a.id, s.team_id, s.teams, u.nick, b.id, b.willwin, b.willlose, b.bidder FROM vs_bids b LEFT JOIN " . $DBPrefix . "vs_users u ON (u.id = b.bidder) LEFT JOIN " . $DBPrefix . "vs_auctions a ON (a.id = b.auction) LEFT JOIN " . $DBPrefix . "vs_sports s ON (s.team_id = b.willwin) WHERE willlose='$menu' and a.id = b.auction and b.bidder NOT IN ('b.tagged') and b.tagged IN ('b.bidder')"); while($row = mysql_fetch_array($query)) { echo "<option value=''></option>"; echo "<option value='$row[team_id]'>$row[nick]...$row[teams]</option>"; } ?> i cant seem to introduce b.auction = :auc_id which will help in displaying the correct result, but once have converted it like the menu above but doesnt seem to work, so i converted it back. please need help with the issue here, thanks.
  2. Still On submenu issue

    decided for now to do things a bit manual i wrote this code to check if a user is selecting the same team as the bidder they are tagging to return error not want i want fully but still get the job done
  3. Still On submenu issue

    yeah take care
  4. Still On submenu issue

    and it didnt work well in the previous example thats y am asking how can it b done, saying it to me wont help
  5. Still On submenu issue

    <option value='1','autualext'>{TEAM1}</option> u mean something like that?
  6. Still On submenu issue

    u mean like this <option value='1','{TEAM1}'>{TEAM1}</option> ?if yes didnt work
  7. Still On submenu issue

    thats just the thing when ever i try to convert it like i did the first part it displaying result i even was trying a back up plan for the now i hit a road block the code below <div class="form-group"> <label class="col-sm-5 control-label">Select Winner</label> <div class="col-sm-7"> <select name="willwin" class="form-control" id="first"> <!-- BEGIN select_team --> <option value='1'>select_team</option> <option value='2'>{select_team.TAGTEAM1}</option> <option value='3'>{select_team.TAGTEAM2}</option> </select><!-- END select_team --> <br> <select name="willlose" class="form-control" id="first_selected"> </div> </div> <script type="text/javascript"> $(document).ready(function() { var dropdown = { 1 : ['Select a winner to trigger'], 2 : ['{select_team.TAGTEAM2}'], 3 : ['{select_team.TAGTEAM1}'] } $('#first_selected').html( '<option>'+dropdown[1].join('</option><option>')+'</option>' ); $('#first').on('change',function() { $('#first_selected').html( '<option>'+dropdown[$(this).val()].join('</option><option>')+'</option>' ); }); }) </script> when ever i submit the form the value get inserted but the value that insert are the numbers <option value='2'> and <option value='3'> instead of what beside them, it a swap code and those values is what been used by the javascript to swap if only the codes could be change to make the javascript use something else instead of val to swap, hope u understand
  8. Still On submenu issue

    yeah have tried and tried but it just wont display result anytime i tried, i could only convert the first part but the second part wont, i guess am missing something or not including something, but no way for me to know what am doing wrong
  9. php drop down sub menu

    hi programmers how u doing? i need help creating a sub menu but with a twist MENU Submenu- team1 on click will display dropdown list of team2 lists team2 on click will display dropdown list of team1 lists if a user select (team1) it should drop down list of [team2] and if the user select (team2) it should drop list of [team1]. the code below is only a menu without submenu need ur assistance, thanks alot // team select $query = "SELECT a.id, a.team1, a.team2, b.auction, b.bidder, b.tagged, b.willwin FROM " . $DBPrefix . "auctions a LEFT JOIN " . $DBPrefix . "bids b ON (b.auction = a.id) WHERE a.id = :auc_id and a.id = a.id group by a.id"; $params = array(); $params[] = array(':auc_id', $id, 'int'); $db->query($query, $params); $TPL_team_list = '<select name="willwin" class="form-control">' . "\n"; while ($row = $db->fetch()) { $TPL_team_list .= "\t" . ' <option value="' . $row[''] . '" ' . $selected . '>' . $row[''] . '</option> <option value="' . $row['team1'] . '" ' . $selected . '>' . $row['team1'] . '</option> <option value="' . $row['team2'] . '" ' . $selected . '>' . $row['team2'] . '</option> ' . "\n"; } $TPL_team_list .= '</select>' . "\n"; { $template->assign_block_vars('tag_bidder', array( 'TEAM' => $TPL_team_list, )); $i++; } Tables : auctions columns: id(int), team1, team2 3 chelsea ajax Tables : bids columns: auction(int) bidder willwin 3 9 chelsea 3 5 ajax 3 7 chelsea 3 2 ajax 3 6 chelsea 3 4 chelsea
  10. php drop down sub menu

    ok, like i said i drop the code, i was interfereing with the loading speed and wasnt work well i tried another one still working on that one.
  11. php drop down sub menu

    tried all that, aam dropping the code.
  12. php drop down sub menu

    <?php $query = "SELECT a.id, a.team1, a.title, a.team2, s.team_id, s.teams, u.nick, b.id, b.willwin, b.willlose, b.bidder, b.auction FROM bids b LEFT JOIN " . $DBPrefix . "users u ON (u.id = b.bidder) LEFT JOIN " . $DBPrefix . "auctions a ON (a.id = b.auction) LEFT JOIN " . $DBPrefix . "sports s ON (s.team_id = b.willwin) WHERE b.willlose = :menu and a.id = :auc_id and b.bidder NOT IN ('b.tagged') and b.tagged IN ('b.bidder')"; $params = array(); $params[] = array(':auc_id', $id, 'int'); $params[] = array(':menu', $menu, 'int'); $db->query($query, $params); while ($row = $db->fetch()) { echo "<option value=''></option>"; echo "<option value='$row[team_id]'>$row[nick]...$row[teams]</option>"; } ?>
  13. php drop down sub menu

    tried that didnt work i got this other code been working on the menu part works fine but the submenu after converting to mysqli refuse to work $query = "SELECT s.team_id, s.teams AS teams1, ss.team_id, ss.teams AS teams2, a.id, a.team1, a.team2, b.auction, b.bidder, b.tagged, b.willwin, b.willlose FROM " . $DBPrefix . "auctions a LEFT JOIN " . $DBPrefix . "bids b ON (b.auction = a.id) LEFT JOIN " . $DBPrefix . "sports s ON (s.team_id = a.team1) LEFT JOIN " . $DBPrefix . "sports ss ON (ss.team_id = a.team2) WHERE a.id = :auc_id and a.id = a.id group by a.id"; $params = array(); $params[] = array(':auc_id', $id, 'int'); $db->query($query, $params); ?> <script type="text/javascript" src="js/dropdownjquery.js"></script> <script type="text/javascript"> $(document).ready(function() { $("#menu").change(function() { $(this).after('<div id="loader"><img src="images/loading.gif" alt="loading subcategory" /></div>'); $.get('bid.php?menu=' + $(this).val(), function(data) { $("#sub_cat").html(data); $('#loader').slideUp(200, function() { $(this).remove(); }); }); }); }); </script> <form name="bid"> <label for="category">Select Winner</label> <select name="willwin" id="menu"> <?php if ($db->numrows() > 0){ while ($row = $db->fetch()) { ?> <option value=''></option> <option value="<?php echo $row["team1"]; ?>"><?php echo $row["teams1"]; ?></option> <option value="<?php echo $row["team2"]; ?>"><?php echo $row["teams2"]; ?></option> <?php } } ?> </select> <br/><br/><br/> <label>Tag Bettor</label> <select name="sub_cat" id="sub_cat"></select>
  14. hiding a value

    hi how do i hide the (team_id) value in the below form <option value="' . $row['team_id'] . '' . $row['teams'] . '" ' . $selected . '>' . $row['teams'] . '</option> thanks
  15. hiding a value

    so this is like an example i believe right
  16. hiding a value

    hi i kinda want it to hide hide as in the code shows result like this 4 Chelsea Chelsea i want the 4 to hide but still be able to be submitted.
  17. Left join from two columns

    i need to connect team1 and team2 to a table called sports so i can get there names which is in column called (teams) how do i do that i already connected sports to another table as u can see in the code, how do i go about this, thanks. $query = "SELECT s.team_id, s.teams, a.current_bid, a.current_bid_id, a.id, a.team1, a.title, a.team2, a.ends, a.starts, u.nick, b.bid, b.tagged, b.bidder, b.willwin, b.quantity, p.bid As proxybid, b.id As bid_id FROM " . $DBPrefix . "bids b LEFT JOIN " . $DBPrefix . "proxybid p ON (p.itemid = b.auction AND p.userid = b.bidder) LEFT JOIN " . $DBPrefix . "sports s ON (s.team_id = b.willwin) LEFT JOIN " . $DBPrefix . "auctions a ON (a.id = b.auction) LEFT JOIN " . $DBPrefix . "users u ON (u.id = b.tagged)
  18. Left join from two columns

    ok done did it this way just incase someone wants to know how $query = "SELECT sss.team_id, sss.teams AS teams2, ss.team_id, ss.teams AS teams1, s.team_id, s.teams, a.current_bid, a.current_bid_id, a.id, a.team1, a.title, a.team2, a.ends, a.starts, u.nick, b.bid, b.tagged, b.bidder, b.willwin, b.quantity, p.bid As proxybid, b.id As bid_id FROM " . $DBPrefix . "bids b LEFT JOIN " . $DBPrefix . "proxybid p ON (p.itemid = b.auction AND p.userid = b.bidder) LEFT JOIN " . $DBPrefix . "sports s ON (s.team_id = b.willwin) LEFT JOIN " . $DBPrefix . "auctions a ON (a.id = b.auction) LEFT JOIN " . $DBPrefix . "sports ss ON (ss.team_id = a.team1) LEFT JOIN " . $DBPrefix . "sports sss ON (sss.team_id = a.team2) LEFT JOIN " . $DBPrefix . "users u ON (u.id = b.tagged)
  19. am having tpl isuue

    how do i convert this <select name="willwin" id="willwin"> <?php if ($db->numrows() > 0){ while ($row = $db->fetch()) { ?> <option value=''></option> <option value="<?php echo $row["willwin"]; ?>"><?php echo $row["team1"]; ?></option> <option value="<?php echo $row["willlose"]; ?>"><?php echo $row["team2"]; ?></option> <?php } } ?> </select> <br/><br/> <label>Tag Bettor</label> <select name="sub_cat" id="sub_cat"></select> to look like this $TPL_team_list = '<select name="willwin" class="form-control">' . "\n"; while ($row = $db->fetch()) { $TPL_team_list .= "\t" . ' <option value="' . $row[''] . '" ' . $selected . '>' . $row[''] . '</option> <option value="' . $row['team1'] . '" ' . $selected . '>' . $row['team1'] . '</option> <option value="' . $row['team2'] . '" ' . $selected . '>' . $row['team2'] . '</option> ' . "\n"; } $TPL_team_list .= '</select>' . "\n"; { $template->assign_block_vars('tag_bidder', array( 'TEAM' => $TPL_team_list, )); $i++; } i cant intergrate the above form (doesnt respond when place directly) in tpl unless converted
  20. am having tpl isuue

    yeah i understand what u saying i keep trying it but doesnt burge
  21. am having tpl isuue

    thansk for replying i use a tpl as frontend display, but the code above the first one cant work in tpl unless convertered this is the full code below $query = "SELECT a.id, a.team1, a.team2, b.auction, b.bidder, b.tagged, b.willwin, b.willlose FROM " . $DBPrefix . "auctions a LEFT JOIN " . $DBPrefix . "bids b ON (b.auction = a.id) WHERE a.id = :auc_id and a.id = a.id group by a.id"; $params = array(); $params[] = array(':auc_id', $id, 'int'); $db->query($query, $params); ?> <script type="text/javascript" src="js/dropdownjquery.js"></script> <script type="text/javascript"> $(document).ready(function() { $("#willwin").change(function() { $(this).after('<div id="loader"><img src="images/loading.gif" alt="loading subcategory" /></div>'); $.get('loadsubcat.php?willwin=' + $(this).val(), function(data) { $("#sub_cat").html(data); $('#loader').slideUp(200, function() { $(this).remove(); }); }); }); }); </script> <form name="bid"> <label for="category">Select Winner</label> <select name="willwin" id="willwin"> <?php if ($db->numrows() > 0){ while ($row = $db->fetch()) { ?> <option value=''></option> <option value="<?php echo $row["willwin"]; ?>"><?php echo $row["team1"]; ?></option> <option value="<?php echo $row["willlose"]; ?>"><?php echo $row["team2"]; ?></option> <?php } } ?> </select> <br/><br/> <label>Tag Bettor</label> <select name="sub_cat" id="sub_cat"></select> </form> it only works inside php files and not tpl below is an example of a php with tpl connector // team $query = "SELECT a.id, a.team1, a.team2, b.auction, b.bidder, b.tagged, b.willwin, b.willlose FROM " . $DBPrefix . "auctions a LEFT JOIN " . $DBPrefix . "bids b ON (b.auction = a.id) WHERE a.id = :auc_id and a.id = a.id group by a.id"; $params = array(); $params[] = array(':auc_id', $id, 'int'); $db->query($query, $params); $TPL_team_list = '<select name="willwin" class="form-control">' . "\n"; while ($row = $db->fetch()) { $TPL_team_list .= "\t" . ' <option value="' . $row[''] . '" ' . $selected . '>' . $row[''] . '</option> <option value="' . $row['team1'] . '" ' . $selected . '>' . $row['team1'] . '</option> <option value="' . $row['team2'] . '" ' . $selected . '>' . $row['team2'] . '</option> ' . "\n"; } $TPL_team_list .= '</select>' . "\n"; { $template->assign_block_vars('tag_bidder', array( 'TEAM' => $TPL_team_list, )); $i++; } hope u understand what am trying to accomplish
  22. php drop down sub menu

    yeah true, but fixed anyway.
  23. php drop down sub menu

    ok so after checking em am finally getting close i have them connected to one database and converted them from conn to db, the menu works fine but the sub menu disappears when i select a menu, the code below plz chk for me what am doing wrong thanks, i might not have converted all or missing a code or something $query = "SELECT a.id, a.team1, a.team2, b.auction, b.bidder, b.tagged, b.willwin FROM " . $DBPrefix . "auctions a LEFT JOIN " . $DBPrefix . "bids b ON (b.auction = a.id) WHERE a.id = :auc_id and a.id = a.id group by a.id"; $params = array(); $params[] = array(':auc_id', $id, 'int'); $db->query($query, $params); ?> <select name="country" id="sports-list" onchange="changeSelect( this.value )"> <option value="1">Select Team</option> <?php if ($db->numrows() > 0) { // output data of each row while ($row = $db->fetch()) { ?> <option value="<?php echo $row["willwin"]; ?>"><?php echo $row["team1"]; ?></option> <option value="<?php echo $row["willlose"]; ?>"><?php echo $row["team2"]; ?></option> <?php } } ?> </select> </br></br></br><div id="subcats"> <select name="bidder" id="bids-list"> <option value='1'>Select Number</option> </select></div> <?php $willlose = mysqli_real_escape_string($_POST['willlose']); if($willlose!='') { $query = "SELECT b.*, u.nick, u.rate_sum FROM " . $DBPrefix . "bids b LEFT JOIN " . $DBPrefix . "users u ON (u.id = b.bidder) WHERE willlose=".$willlose." and b.bidder NOT IN ('b.tagged') and b.tagged IN ('b.bidder') and b.auction = :auc_id order by b.willwin"; $params = array(); $params[] = array(':auc_id', $id, 'int'); $db->query($query, $params); $options = "<option value=''>Select Name</option>"; while ($row = $db->fetch()) { $options .= "<option value='".$row['auction']."'>".$row['nick']."</option>"; } echo $options; } ?> <script src="js/dropdown/jquery-1.3.0.min.js"></script> <script> $('#sports-list').on('change', function(){ var willlose = this.value; $.ajax({ type: "POST", data:'willlose='+willlose, success: function(result){ $("#bids-list").html(result); } }); }); </script>
  24. php drop down sub menu

    no i meant, using the previous form field, i can insert values into the table using $db without the drop down values which i iinsert maunaly from the database backend into the row, but with the $conn i can only view those values along side the drop down values. u know what is there a way i can message u the script so u see for ur self along side what am trying to do?
  25. php drop down sub menu

    it produces the correct result, the problem now is the script am using as u can see is different from what have been using, have tried changing things around so the script can connect to a db and not conn i have two seperate connection, db is the one have been using while conn is the one that came along with the drop menu code
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