skyhighweb Posted May 20, 2017 Author Share Posted May 20, 2017 yeah it it returns a value Link to comment Share on other sites More sharing options...
dsonesuk Posted May 20, 2017 Share Posted May 20, 2017 Should it not be where b.willlose='.$willlose.' ' And does the values in 'willose' column have a space after the value as shown by using .' ' on the search value? Link to comment Share on other sites More sharing options...
skyhighweb Posted May 20, 2017 Author Share Posted May 20, 2017 (edited) i don't understand the question (updated) oh yeah seen it, have changed it, but still effect i still get full datas from the menu side Edited May 20, 2017 by skyhighweb Link to comment Share on other sites More sharing options...
skyhighweb Posted May 20, 2017 Author Share Posted May 20, 2017 35 minutes ago, dsonesuk said: Should it not be where b.willlose='.$willlose.' ' And does the values in 'willose' column have a space after the value as shown by using .' ' on the search value? yeah checked i remove all the b. i wasnt even joining anything, if only i could try rewrite this require_once('conn.php'); ini_set('display_errors',0); $query = "SELECT b.auction, b.bidder, b.tagged, b.willwin, b.willlose, a.id, a.team1, a.team2 FROM " . $DBPrefix . "vs_auctions a LEFT JOIN " . $DBPrefix . "vs_bids b ON (b.auction = a.id) WHERE a.id = b.auction and a.id = a.id group by b.auction"; $country_result = $conn->query($query); so it will look like this $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); so i can stay connected to my database with this config file <?php $DbHost = "localhost"; $DbDatabase = "vsb"; $DbUser = "root"; $DbPassword = "pass"; $DBPrefix = "vs_"; $main_path = "D:\\xampp\\htdocs\\vsbet\\"; $MD5_PREFIX = "ca801ec4f5c664bc6a492f942b3a07eb"; ?> that way i can simply intergrate the code smoothly , because right now it like am having two seperate connection acting differently Link to comment Share on other sites More sharing options...
dsonesuk Posted May 20, 2017 Share Posted May 20, 2017 IF $db represent connection to database, what is $conn? You just require single variable connection to a single database. For each select statement attached to $query variable you call $db->query($query); directly following it and process the results following that. IF you manually insert a value instead of a variable, does it produce correct result, if it does not then the select query is faulty. Link to comment Share on other sites More sharing options...
skyhighweb Posted May 20, 2017 Author Share Posted May 20, 2017 4 minutes ago, dsonesuk said: IF $db represent connection to database, what is $conn? You just require single variable connection to a single database. For each select statement attached to $query variable you call $db->query($query); directly following it and process the results following that. IF you manually insert a value instead of a variable, does it produce correct result, if it does not then the select query is faulty. 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 Link to comment Share on other sites More sharing options...
dsonesuk Posted May 20, 2017 Share Posted May 20, 2017 Then I dont understand?, if you insert manually the correct value in SQL statement, it works but! if the value returned by $_POST is the same as the manually entered value it does not, that just does not make sense. Link to comment Share on other sites More sharing options...
skyhighweb Posted May 20, 2017 Author Share Posted May 20, 2017 11 minutes ago, dsonesuk said: Then I dont understand?, if you insert manually the correct value in SQL statement, it works but! if the value returned by $_POST is the same as the manually entered value it does not, that just does not make sense. 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? Link to comment Share on other sites More sharing options...
dsonesuk Posted May 20, 2017 Share Posted May 20, 2017 I do not know WHAT you want now? you said you want the sub dropdown to show data from database depending what is passed back from parent dropdown, it may or may not work? then you talk about inserting? then you find a script that uses $conn? Maybe you should learn to understand MYSQL, PHP, SQL first! Instead of just finding any script that may/nearly produces the required results. and pasting into your own, and then get us to unravel third party code for you, which unrelated to your current code. Message me! well sorry, that's not going to happen. Link to comment Share on other sites More sharing options...
skyhighweb Posted May 20, 2017 Author Share Posted May 20, 2017 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> Link to comment Share on other sites More sharing options...
justsomeguy Posted May 22, 2017 Share Posted May 22, 2017 You're not printing any of those options into the select list. This is the problem with not understanding the code you are working with. We are here to help people learn how to program. We expect people to start on their own, and make an effort. If you're not going to go through the tutorials to learn the basics about how this stuff works, how can we help you? Just doing everything for you does not teach you anything except that when you have a problem you come here to get the answer. That's not programming. Link to comment Share on other sites More sharing options...
skyhighweb Posted May 23, 2017 Author Share Posted May 23, 2017 19 hours ago, justsomeguy said: You're not printing any of those options into the select list. This is the problem with not understanding the code you are working with. We are here to help people learn how to program. We expect people to start on their own, and make an effort. If you're not going to go through the tutorials to learn the basics about how this stuff works, how can we help you? Just doing everything for you does not teach you anything except that when you have a problem you come here to get the answer. That's not programming. yeah true, but fixed anyway. Link to comment Share on other sites More sharing options...
skyhighweb Posted May 26, 2017 Author Share Posted May 26, 2017 On 5/20/2017 at 9:39 AM, dsonesuk said: IS $willlose = mysql_real_escape_string($_POST['willlose']); returning a value? Just make sure it returns a value first, and this </select></div> should be UNDER the php outputted options. 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> Link to comment Share on other sites More sharing options...
skyhighweb Posted May 26, 2017 Author Share Posted May 26, 2017 <?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>"; } ?> Link to comment Share on other sites More sharing options...
dsonesuk Posted May 26, 2017 Share Posted May 26, 2017 IF you are using MYSQLI then this is mysql_real_escape_string($_POST['willlose']) IS WRONG! it should be mysqli_real_escape_string($_POST['willlose']) Notice 'i' at end of 'mysql', most if not all of these specific mysqli functions begin with 'mysqli'. Link to comment Share on other sites More sharing options...
skyhighweb Posted May 26, 2017 Author Share Posted May 26, 2017 4 minutes ago, dsonesuk said: IF you are using MYSQLI then this is mysql_real_escape_string($_POST['willlose']) IS WRONG! it should be mysqli_real_escape_string($_POST['willlose']) Notice 'i' at end of 'mysql', most if not all of these specific mysqli functions begin with 'mysqli'. tried all that, aam dropping the code. Link to comment Share on other sites More sharing options...
justsomeguy Posted May 26, 2017 Share Posted May 26, 2017 tried that didnt work This is the single worst response you can probably give. If you did it right, then it would have worked. If it didn't work, then you didn't do it right, but you didn't show any code to allow anyone to point out what you did wrong. It's not like people suggest all of these PHP functions that just don't work, PHP works fine, you're just not using it right, and you're also not showing anyone else what you did so that we can point out your errors. Incidentally, this is also why I stopped responding to your threads. I'm not here to do peoples' work for them, I'm trying to teach people. Link to comment Share on other sites More sharing options...
skyhighweb Posted May 26, 2017 Author Share Posted May 26, 2017 3 minutes ago, justsomeguy said: This is the single worst response you can probably give. If you did it right, then it would have worked. If it didn't work, then you didn't do it right, but you didn't show any code to allow anyone to point out what you did wrong. It's not like people suggest all of these PHP functions that just don't work, PHP works fine, you're just not using it right, and you're also not showing anyone else what you did so that we can point out your errors. Incidentally, this is also why I stopped responding to your threads. I'm not here to do peoples' work for them, I'm trying to teach people. 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. Link to comment Share on other sites More sharing options...
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