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PHP - Moved: Table Results With Rank
This topic has been moved to MySQL Help.
http://www.phpfreaks.com/forums/index.php?topic=312470.0 Similar TutorialsEver since I switched over to pagination, the ranks for the sites have been off. I use to have an $i++ increment that kept the ranks, but now with pagination, it just ranks those on the current page. Is there a way to bypass this? public function grabSites() { //amount of sites $amount = mysql_num_rows(mysql_query("SELECT * FROM sites")); if($amount > 0) { //amount per page $p_page = 13; //track page $start = $_GET['page']; //max_pages $max_pages = $amount / $p_page; //set default if(!$start) $start = 0; //new query $query = mysql_query("SELECT id,title,description,votes,date,outl,site_url FROM sites ORDER BY votes DESC LIMIT $start, $p_page"); while($fetch = mysql_fetch_assoc($query)) { ++$i; $i = $i + $p_page; echo " <div id='post-1' class='post'> <h2 class='title'><font color='white'>#". $i." ".$fetch['title'] ."</font></h2> <div class='entry'> <b>". $fetch['title'] ." currently has ". $fetch['votes'] ." votes. <a href='vote.php?id=". $fetch['id'] ."'>Vote now</a>.</b> <br/><br/> <p>". nl2br(stripslashes($fetch['description'])) ."</p> </div> <p class='meta'><a href='". $fetch['site_url'] ."'>Visit</a></p> </div> "; } //set back and forth variables $previous = $start - $p_page; $next = $start + $p_page; ?> <hr> <center> <?php if(!($start<=0)) echo "<a href='index.php?page=". $previous ."'><<</a>"; $i = 1; for($x = 0; $x < $amount; $x = $x + $p_page) { echo "<a href='index.php?page=". $x ."'> ". $i ." </a>"; $i++; } if(!($start>=$amount-$p_page)) echo "<a href='index.php?page=". $next ."'>>></a></center>"; ?> </center> <?php } else { echo " <div id='post-1' class='post'> <h2 class='title'><a href='#'>Oh NOEZ!</a></h2> <h3 class='date'>". date("M-d-Y") ."</h3> <div class='entry'> <p>There are currently no sites to display!</p> </div> <p class='meta'>View</p> </div> "; } } I am having trouble on user post = rank This is the rank table: ----------------- post | rank ----------------- 0 | Noob 10 | Member 50 | Metal 100 | Silver 200 | Gold $user_post="120"; $query_rank=mysql_query("SELECT * FROM rank WHERE post<='$user_post' OR post>='$user_post'") or die ("Error"); while ($row=mysql_fetch_assoc($query_rank)){ $rank=$row['rank']; } echo "User Rank: $rank"; The above code output: User Rank: Gold The right output should be: User Rank: Silver Please help It will be better if someone willing to share a better ranking code. Hi all, the other I coded a simular script to this which didnt work, so I re coded it but im having a problem with this one aswell, what its meant to do is just change the users rank to what is posted, but every time I hit submit on the form it locates me to a different place. <?php session_start(); include ("../includes/db_connect.php"); include ("../includes/functions.php"); logincheck(); $username = $_SESSION['username']; $rankget = mysql_query ("SELECT * FROM users WHERE username='$username'"); $newrank = mysql_fetch_object($rankget); if ($_POST['change']){ if ($newrank->userlevel >= "2"){ $time = now(); $updateuser = $_POST['usernameone']; // User Posted $updaterank = $_POST['rankone']; // Rank posted if ($updateuser == NULL && $updaterank == NULL){ $message = "You must enter a user and rank"; }elseif ($updateuser == NULL){ $message = "You must enter a username"; }elseif ($updaterank == NULL){ $message = "You must enter a rank"; } mysql_query ("UPDATE users SET rank = '$updaterank' WHERE username='$updateuser'") or die (mysql_error()); mysql_query ("INSERT INTO logs (`id` , `who`, `action`, `time`) VALUES ('', '$username' , '$username Updated rank; $updateuser to $updaterank' , '$time')") or die (mysql_error()); mysql_query ("INSERT INTO inbox (`id` , `to` , `from` , `message` , `date` , `read` , `saved` , `event_id`) VALUES ('', '$updateuser', 'System', 'Your rank has been updated to $updaterank!', '$date', '0', '0', '')") or die (mysql_error()); echo ("Rank Changed!"); }} ?> <html> <head> <title>Change Rank</title> <link rel="stylesheet" href="../include/in.css" type="text/css"> <style type="text/css"> .infobg { font-family: Arial; font-weight:normal; font-size:12px; border-top: 1px solid #000000; border-right: 1px solid #000000; border-bottom: 1px solid #000000; border-left: 1px solid #000000; background: URL(textbg1.png); font-weight:300; } .button { font-size: 12px; background:url(button.png); vertical-align: middle; border-top: 1px solid #000000; border-right: 1px solid #000000; border-bottom: 1px solid #000000; border-left: 1px solid #000000; color: #FFFFCC; height:23px; font-weight:300; border-radius: 10px; padding-bottom:2px; } </style> </head> <body> <form action='' name='form1' id='form1' method='POST'> <table width='50%' cellpadding='0' align='center' cellspacing='0' border='1' bordercolor='#000000' bgcolor='#808080' style='border-collapse: collapse'> <tr> <td><?php echo ("$message"); ?></td> </tr> <tr> <td background='../header.jpg' colspan='2' align='center'>Change Rank</td> <tr> <td>Username:</td><td><input type='text' name='usernameone'></td> </tr> <tr> <td>Rank:</td><td><input type='text' name='rankone'></td> </tr> <tr> <td> </td><td><input type='submit' name='change' value='Change Rank!'></td> </tr> </table> </form> </body> </html> Anyone see why it locates me else? Thanks Hello everyone,
I'm create an application where I have member, admin, and superadmin
What I'm try to do is superadmin update member to admin, so they can have more privilege. The name of the table is member where I have member, admin account, and i have row call 'rank'. Rank 1 is for member they have access basic stuff and rank 2 they have more privilege. Want the superadmin to type the name of the member and change to rank 2.
This is what happening, if I dont put nothing on the text boxes, it said "Please check if you have put the right information." , if i put member and the rank i want to change is says ""User has upgrade to admin", but than when i check my database the member is still a member when i wanted to change to admin.
Can anyone see where I'm going home, or what should I change.
<form> </form>
<form action='<?php htmlentities($_SERVER['PHP_SELF']); ?>' method='POST' enctype='multipart/form-data '>
Username: <input class="fontsize" name='user' type='text'/>
Rank: <input class="fontsize" name='rank' type='text' /> <br><br>
<input class="button" type='submit' name='get' value='Upgrade' />
</form>
</h2>
<?php
if(isset($_POST['get']))
{
$username1 = $_POST['user'];
$sql_user = mysqli_real_escape_string ($connection, $username1);
$rank = $_POST['rank'];
$rank = mysqli_real_escape_string ($connection, $rank);
$query1 = mysqli_query($connection, "SELECT rank FROM member WHERE username='$sql_user'");
if ($row = mysqli_fetch_array($query1))
{
if ($rank == '1')
{
$data1 = mysqli_query($connection, "UPDATE member SET rank= '1' WHERE username='sql_user'") or die (mysql_error($connection));
echo "User has downgrade to member";
}
elseif ($rank == '2')
{
$data2 = mysqli_query($connection, "UPDATE member SET rank= '2' WHERE username ='sql_user'") or die (mysql_error($connection));
echo "User has upgrade to admin";
}
else
{
echo "Please check if you have put the right information.";
}
}
}
?>
Consider the following code; Code: [Select] <div class="blockrow"> <table cellpadding="10" align="center" border="0"> <tr> <td> <center> <font size=6>Donation Hall of Fame</font><br><br> <font size=2><b>Anonymous Donations are not Counted</b></font><br><br> <font size=3><a href="/donate">Get on This List</a></font> <br> <br> <?php //database connection $connectdb = mysql_connect ("localhost", "user", "pass") or die ('I cannot connect to the database because: ' . mysql_error()); mysql_select_db ("thatsact_donations"); //query setup $sql = "SELECT * FROM totals ORDER BY donation_total DESC LIMIT 100"; $result = mysql_query($sql) or die(mysql_error()); //initialize container $container = array(); //get entries from db while ($get = mysql_fetch_assoc($result)){ if(!$get[userid]=="0") { $username = "<b><a style=\"color:#00CC00\" href=/member.php/$get[userid]>$get[username]</a></b> donated <b>$$get[donation_total]</b> total so far!"; } $container[] = "<font size=3>$username</font>"; } //combine collected entries $output = implode("<br /><hr>",$container); echo $output; mysql_close($connectdb); ?> </center> </td> </tr> </table> </div> I would like to add a rank number to the container starting with one and going 23456 as each piece is echoed. How might I do this because I am very new to php. Example: 1. Username donated $5...... 2. Username donated $4........ 3. Username donated $3....... Thanks in advance. Hello everyone... My following code is supposed to order rank by from Greatest covertaction but for some reason it is ranking the greatest last eg. 900,000 Rank 1 900,001 Rank 2 900,002 Rank 3 Here is the code: Code: [Select] <?php $q = "SELECT * FROM `accountinfo_db` ORDER BY `covertaction`, `anticovertaction` ASC"; $res = mysql_query($q) or die(mysql_error()); $i = 1; while($player=mysql_fetch_array($res)){ $id = securevar($player['id']); $time = time(); $user = securevar($player['username']); $q = "UPDATE `accountinfo_db` SET `covertrank` = '$i', `lastTurnTime` = '$time' WHERE `id` = '$id'"; if(mysql_query($q)){ echo "Covert & Anti Covert Rank set to $user as ".number_format($i)."!<br />"; } else { // query failed with an error // put your error reporting/logging code here... } $i++; } ?> Any help will be appreciated! Brian hi i have a table like this rank xp 1 0 2 50 3 100 4 200 say my username has 75xp how would a select which rank i am not sure how i can do this? :S cheers matt This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=329323.0 This topic has been moved to Application Design. http://www.phpfreaks.com/forums/index.php?topic=347295.0 This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=321088.0 when I do a "SELECT * FROM system_disks WHERE system_id = 'aNumber'" and their is more than one result with that system_id, I get more than one result obviously. I essentially need two results in one row. I've took a screen shot of what it is doing and what I need it to do. The blurred row is a different system_id This is what the while ($row = mysql_fetch_array($query)) is currently doing: This is what I would like it to do: Sorry if this is a dumb question, I've been coding the past 48 hours and my brain is fried This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=330251.0 This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=354388.0 This topic has been blindfolded and driven across the boarder to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=344623.0 (That was a pun, not a typo.) This topic has been moved to Third Party PHP Scripts. http://www.phpfreaks.com/forums/index.php?topic=359102.0 This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=347091.0 Ive gotten some results user selcts check box on first page The php page will say which brackets it falls between example price is between 100-300 say I dont seem to be able to populate a table with the data in the database:S Code: [Select] $row_number= 0; while ($row = mysql_fetch_array($result3)) { If (($row["price"] = $price_low) && ($row["price"] <= $price_high)) //price If (($row["storage"] = $storage_low) && ($row["storage"] <= $storage_high)) //storage if (($row["Processor "] = $processor_low) && ($row["Processor"] <= $processor_high)) //Processor { $row_number++; ?> <tr> <td align="center"><?php print $row["Computer_Price"]; ?> </td> <td align="center"><?php print $row["Computer_Storage"]; ?> </td> <td align="center"><?php print $row["Computer_ProcessorSpeed"]; ?> </td> </tr> </table> Thats wat im usign atm the price_low and price_high are what sets the low and high price for the search Hi guys, I am working with an old script at the moment, there is one page which just will not populate the table results. I have tried running multiple debugging commands but the only one it flags is the line displaying Quote last; saying it's not a used function. If I comment out this line, no errors are produced but the results do not enter the table. Can anyone shed some light on this please, I've spent hours and hours and banging my head against a brick wall would probably be more constructive right now. Many thanks indeed for any help or advice. <?php mysql_connect("localhost", "$db_user","$db_upwd") or die ("Error - Could not connect: " . mysql_error()); mysql_select_db("$database"); $query="select host,count(*) from badc_mis_prog group by host"; $result = mysql_query($query) or die ("Error - Query: $query" . mysql_error()); $count=0; $hosts=array(); while ($row = mysql_fetch_array($result, MYSQL_NUM)) { $html_hlname=$row[0]; $html_hlname=preg_replace("/</","<",$html_hlname); $html_hlname=preg_replace("/>/",">",$html_hlname); array_push($hosts, $html_hlname,$row[1],0); $count++; } $query="select host,count(*) from badc_mis_prog where reported=1 group by host"; $result = mysql_query($query) or die ("Error - Query: $query" . mysql_error()); while ($row = mysql_fetch_array($result, MYSQL_NUM)) { $html_hlname=$row[0]; $html_hlname=preg_replace("/</","<",$html_hlname); $html_hlname=preg_replace("/>/",">",$html_hlname); for ($i=0; $i<($count*3); $i+=3) { if ($hosts[$i] == $html_hlname) { $hosts[($i+2)]=$row[1]; last; } } } for ($i=0 ; $i<(($count-1)*3); $i+=3){ for ($j=$i+3 ; $j<($count*3); $j+=3){ if ($hosts[($i+1)] < $hosts[($j+1)]){ $temp=array(); $temp[0]=$hosts[$i]; $temp[1]=$hosts[($i+1)]; $temp[2]=$hosts[($i+2)]; $hosts[$i]=$hosts[$j]; $hosts[($i+1)]=$hosts[($j+1)]; $hosts[($i+2)]=$hosts[($j+2)]; $hosts[$j]=$temp[0]; $hosts[($j+1)]=$temp[1]; $hosts[($j+2)]=$temp[2]; } } } print "<br><br><br><center><table border=\"1\">\n"; print "<tr><td>Host Name</td><td>Hosted</td><td>Reported</td><td>Ratio H/R</td></tr>\n"; for ($i=0; $i<($count*3); $i+=3) { if ($hosts[($i+1)]<15){ break;} printf ("<tr><td> %s </td><td> %d </td><td> %d </td><td>%.1f %%</td></tr>\n",$hosts[$i],$hosts[($i+1)],$hosts[($i+2)],(($hosts[($i+2)]/$hosts[($i+1)])*100)); } print "</table></center>\n"; ?> My db is called localDB, and the table in that db is called MonthlySales. I want to display the table in a table in php. But to be able to filter the table by the column year, so if 2000 is selected only the values with 2000 are shown. Is there also a better way to populate the drop down menu? My script looks like: Code: [Select] <form action='<?php echo $_SERVER['PHP_SELF']; ?>' method='post' name='form_filter' > <select name="value"> <option value="all">All</option> <option value="2000">2000</option> <option value="2001">2001</option> </select> <input type='submit' value = 'Filter'> </form> <?php $link = mysql_connect('localhost', 'root', 'root'); if (!$link) { die('Could not connect: ' . mysql_error()); } $db_selected = mysql_select_db('localDB', $link); if (!$db_selected) { die (mysql_error()); } // process form when posted if(isset($_POST['value'])) { if($_POST['value'] == '2000') { $query = "SELECT * FROM MonthlySales WHERE Year='2000'"; } elseif($_POST['value'] == '2001') { $query = "SELECT * FROM MonthlySales WHERE Year='2001'"; } else { $query = "SELECT * FROM MonthlySales"; } $sql = mysql_query($query); while ($row = mysql_fetch_array($query)) { $Id = $row["Id"]; $ProductCode = $row["ProductCode"]; $Month = $row["Month"]; $Year = $row["Year"]; $SalesVolume = $row["SalesVolume"]; echo "<tr>"; echo "<td>" . $row['Id'] . "</td>"; echo "<td>" . $row['ProductCode'] . "</td>"; echo "<td>" . $row['Month'] . "</td>"; echo "<td>" . $row['Year'] . "</td>"; echo "<td>" . $row['SalesVoulme'] . "</td>"; echo "</tr>"; } mysql_close($con); } ?> Hi, I'm quite new to this and I'm trying to get this to line up in a table with 3 columns (unlimited rows) I have searched and tried but I'm not having any luck. Can any one help? Thank you Code: [Select] <?php echo '<div class="resultados_sub_cat">'; foreach($this->subcats as $key => $subcat) { $subcat->link = JRoute::_('index.php?option=classcliff&view=list&catid='.$subcat->id."&Itemid=".$this->Itemid); if ($key != 0) echo ' - '; echo '<a href="'.$subcat->link.'">'.$subcat->name.'</a>'; } ?> |
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