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PHP - Displaying Mysql Contents In 4x2 Table
I want to display my pictures I stored in Mysql in a 4 column, 2 row table WITH pagination.
Here's the code I use to display the data currently: Code: [Select] //your username $username = "username"; //your password $password = "password"; //your mySQL server $host = "host"; //The name of the database your table is in $database = "database"; //connect, but if there is a problem, display an error message telling why there is a problem $conn = mysql_connect($host,$username,$password) or die("Error connecting to Database!<br>" . mysql_error()); //Choose the database from your mySQL server, but if there is a problem, display an error telling why $db = mysql_select_db($database) or die("Cannot select database!<br>" . mysql_error()); // find out how many rows are in the table $sql = "SELECT COUNT(*) FROM myphotos"; $result = mysql_query($sql, $conn) or trigger_error("SQL", E_USER_ERROR); $r = mysql_fetch_row($result); $numrows = $r[0]; // number of rows to show per page $rowsperpage = 4; // find out total pages $totalpages = ceil($numrows / $rowsperpage); // get the current page or set a default if (isset($_GET['currentpage']) && is_numeric($_GET['currentpage'])) { // cast var as int $currentpage = (int) $_GET['currentpage']; } else { // default page num $currentpage = 1; } // end if // if current page is greater than total pages... if ($currentpage > $totalpages) { // set current page to last page $currentpage = $totalpages; } // end if // if current page is less than first page... if ($currentpage < 1) { // set current page to first page $currentpage = 1; } // end if // the offset of the list, based on current page $offset = ($currentpage - 1) * $rowsperpage; // get the info from the db $sql = "SELECT * FROM myphotos ORDER BY id ASC LIMIT $offset, $rowsperpage"; $result = mysql_query($sql, $conn) or trigger_error("SQL", E_USER_ERROR); /****** build the pagination links ******/ // range of num links to show $range = 3; // if not on page 1, don't show back links if ($currentpage > 1) { // show << link to go back to page 1 echo "<span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=1'>First</a></span> "; // get previous page num $prevpage = $currentpage - 1; // show < link to go back to 1 page echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$prevpage'>Previous</a></span> "; } // end if // loop to show links to range of pages around current page for ($x = ($currentpage - $range); $x < (($currentpage + $range) + 1); $x++) { // if it's a valid page number... if (($x > 0) && ($x <= $totalpages)) { // if we're on current page... if ($x == $currentpage) { // 'highlight' it but don't make a link echo " <span class=\"paginationDown\"><b>$x</b></span> "; // if not current page... } else { // make it a link echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$x'>$x</a></span> "; } // end else } // end if } // end for // if not on last page, show forward and last page links if ($currentpage != $totalpages) { // get next page $nextpage = $currentpage + 1; // echo forward link for next page echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$nextpage'>Next</a></span> "; // echo forward link for lastpage echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$totalpages'>Last</a></span><br> "; } // end if /****** end build pagination links ******/ // while there are rows to be fetched... while ($list = mysql_fetch_assoc($result)) { extract ($list); // echo data $url = $list['url'];; $title = $list['title']; $description = $list['description']; echo("$title<br><a rel=\"example_group\" title=\"$description\" href=\"$url\"><img src=\"$url\" alt=\"\" width=\"\" height=\"\" class=\"gallery_images\" /></a>"); } // end while /****** build the pagination links ******/ // range of num links to show $range = 3; // if not on page 1, don't show back links if ($currentpage > 1) { // show << link to go back to page 1 echo "<span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=1'>First</a></span> "; // get previous page num $prevpage = $currentpage - 1; // show < link to go back to 1 page echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$prevpage'>Previous</a></span> "; } // end if // loop to show links to range of pages around current page for ($x = ($currentpage - $range); $x < (($currentpage + $range) + 1); $x++) { // if it's a valid page number... if (($x > 0) && ($x <= $totalpages)) { // if we're on current page... if ($x == $currentpage) { // 'highlight' it but don't make a link echo " <span class=\"paginationDown\"><b>$x</b></span> "; // if not current page... } else { // make it a link echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$x'>$x</a></span> "; } // end else } // end if } // end for // if not on last page, show forward and last page links if ($currentpage != $totalpages) { // get next page $nextpage = $currentpage + 1; // echo forward link for next page echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$nextpage'>Next</a></span> "; // echo forward link for lastpage echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$totalpages'>Last</a></span><br>"; } // end if /****** end build pagination links ******/ The above code just displays the pictures vertically. Here's this code live: http://www.djsmiley.net/gallery/albums/my_photos.php (you can't view the page in IE) Now how would I make it so that the first 8 images in the database display in a 4x2 table, etc.? Similar TutorialsGreetings,
My current code logs into a database, opens a table named randomproverb, randomly selects 1 proverb phrase, and then SHOULD display the proverb in the footer of my web page.
As of right now, the best I can do is get it to display "Array", but not the text proverb... this code below actually causes my whole footer to not even show up.
Please help!
<?php
include("inc_connect.php"); //Connects to the database, does work properly, already tested
$Proverb = "randomproverb";
$SQLproverb = "SELECT * FROM $Proverb ORDER BY RAND() LIMIT 1";
$QueryResult = @mysql_query($SQLproverb, $DBConnect);
while (($Row = mysql_fetch_assoc($QueryResult)) !== FALSE) {
echo "<p style = 'text-align:center'>" . {$Row[proverb]} . "</p>\n";
}
$SQLString = "UPDATE randomproverb SET display_count = display_count + 1 WHERE proverb = $QueryResult[]";
$QueryResult = @mysql_query($SQLstring, $DBConnect);
...
?>
Hi, I am trying to array a mysql tables data into a php table. Not having luck... <?php include('dbconnect.php') ?> <?php // Make a MySQL Connection $query = "SELECT * FROM cars"; $result = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_array($result)){ echo $row['CarName']. " - ". $row['CarTitle']; echo "<br />"; echo "<TABLE CELLPADDING=0 CELLSPACING=0 WIDTH=100%>"; echo "<TR />"; echo "<TD />"; echo $row['CarName'].; echo "</TD>"; echo "<TD />"; echo $row['CarTitle'].; echo "</TD>"; echo "</TR>"; echo "</TABLE>"; } ?> Error says: Parse error: syntax error, unexpected ';' in /home/wormste1/public_html/tilburywebdesign/shop/FTPServers/barryottley/viewcars.php on line 69 echo </TD /> is line 69. Please help Ian So I have a jobs database with the following columns: id, jobtext, jobdate, and id. This is how it looks right now: http://prahan.com/jobs/display.html.php I have another table called author. In the authorid column in need the results of this query, SELECT name FROM author WHERE id = (SELECT authorid FROM job) , to be displayed for each row. I also want to be able to customize the header title for each column. Thanks in advance! Not really sure how to get the images I have stored in MySQL into a html form. I can call-up the text fields from the database but it cannot seem to find the index for the images. Here is my code:- <?php session_start(); mysql_connect("localhost","root","abc") or die ("Error! Cannot connect to database"); mysql_select_db("theimageworks") or die ("Cannot find database"); $query = "SELECT * FROM jobs"; $result = mysql_query($query) or die (mysql_error()); ?> <?php //display data in html table echo "<table>"; echo "<tr><td>Username</td><td align='center'>Message</td><td>Product Image</td></tr>"; while($row = mysql_fetch_array($result)) { echo "</td><td>"; echo $row['username']; echo "</td><td>"; echo $row['message']; echo "</td></tr>"; echo $row['image']; } echo "</table>"; ?> The error message I get is "Notice: Undefined index: image in....." Thanks in advance! Hi, I'm trying to make a dynamic html table to contain the mysql data that is generated via php. I'm trying to display a user's friends in a table of two columns and however many rows, but can't seem to figure out what is needed to make this work. Here's my code as it stands: Code: [Select] <?php //Begin mysql query $sql = "SELECT * FROM friends WHERE username = '{$_GET['username']}' AND status = 'Active' ORDER BY friends_with ASC"; $result = mysql_query($sql); $count = mysql_num_rows($result); $sql_2 = "SELECT * FROM friends WHERE friends_with = '{$_GET['username']}' AND status = 'Active' ORDER BY username ASC"; $result_2 = mysql_query($sql_2); $count_2 = mysql_num_rows($result_2); while ($row = mysql_fetch_array($result)) { echo $row["friendswith"] . "<br>"; } while ($row_2 = mysql_fetch_array($result_2)) { echo $row_2["username"] . "<br>"; } ?> The above simply outputs all records of a user's friends (their usernames) in alphabetical order. The question of how I'd generate a new row each time a certain amount of columns have been met, however, is beyond me. Anyone know of any helpful resources that may solve my problem? Thanks in advance =) So, I'm trying to get this read my text document "news.txt" and display the contents. My code: Code: [Select] div.newscontent{ padding:0px; margin-top:60px; margin-bottom:0px; margin-left:400px; margin-right:0px; position:absolute; width:457px; height:330px; text-align:center; color:#FFFFFF; } Code: [Select] <div class='newscontent'> <?PHP $filename = "news.txt"; $arr = file($filename); foreach($arr as $line){ print $line . "<br/>"; } ?> </div> This is the output I get: Code: [Select] "; } ?> Whats wrong? need a little help guys! I use the script below to display profile images, trouble is it shows 1 on top of the other, and i need it to double up 2 profile images on top of 2 profile images ect any ideas how i can do this. require("./include/mysqldb.php"); $con = mysql_connect("$dbhost","$dbuser","$dbpass"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("$dbame", $con); $result = mysql_query("SELECT * FROM Search_profiles_up WHERE upgrade_one ='1' ORDER BY RAND() LIMIT 40"); print "<table width=\"293\" height=\"111\" border=\"0\"> <tr>\n"; while($row = mysql_fetch_array($result)) { print "<td width=\"142\"><img src=" . $row['search_small_image'] . " width=\"144\" height=\"169\" /></td>\n"; print " </tr>\n"; print " <tr> \n"; print "<td>" . $row['star'] . "</td>\n"; print " </tr>\n"; print " <tr>\n"; print "<td>" . $row['username_search'] . "</td>\n"; print " </tr>\n"; print " <tr> \n"; print "<td>" . $row['phone_search'] . "</td>\n"; print " </tr> \n"; } print "</table>"; mysql_close($con); ?> Hi i want to make a form LIST that automaticly gets the content "orderid" to choose for editing. can i please get a pointing pin on which function(s) to use ? i have tried with: Code: [Select] $result = mysql_query("SELECT * FROM {$table} WHERE 'ordreid'"); if (!$result) { die("Query to show fields from table failed"); } echo($result); $i = 0; while($i<mysql_num_fields($result)) { $meta=mysql_fetch_field($result,$i); echo $i.".".$meta->name."<br />"; $i++; } like i want to get this part: <option>first item in field orderid</option> <option>2'nd item in field orderid</option> and so on...i know its easy but i have struggled wit this enough now. Hello all, I'm sure something like this has been brought up before but I think it's quite a unique problem and I didn't get anywhere searching... so here's my first post. Yay. Right, so, basically I have a database table that looks like this: Code: [Select] +----------+-------+ | name | value | +----------+-------+ | settingA | true | | settingB | false | | settingC | false | +----------+-------+ And I have been racking my brains trying to figure out how to get that data into an array formatted like so: Code: [Select] Array ( [settingA] => true [settingB] => false [settingC] => false ) The best I have been able to do, using nested foreach() statements, is this: Code: [Select] Array ( [0] => Array ( [name] => settingA [value] => true ) [1] => Array ( [name] => settingB [value] => false ) [2] => Array ( [name] => settingC [value] => false ) ) ...which is not very practical, or at least it isn't for what I want to do. Any ideas on what the best method might be for achieving this? I am using ADOdb Lite but even an example using PHP's native MySQL functions could help me figure out what to do. Thanks! Hi Everyone, I have a problem displaying some information from an array (selected from my database). The array is below. Array ( [0] => Array ( [ssc_skill_categories] => Web [sc_skill_categories] => Programming ) [1] => Array ( [ssc_skill_categories] => Actionscript [sc_skill_categories] => Programming ) [2] => Array ( [ssc_skill_categories] => C# [sc_skill_categories] => Programming ) [3] => Array ( [ssc_skill_categories] => CSS [sc_skill_categories] => Programming ) [4] => Array ( [ssc_skill_categories] => Graphic [sc_skill_categories] => Designers ) [5] => Array ( [ssc_skill_categories] => Logo [sc_skill_categories] => Designers ) [6] => Array ( [ssc_skill_categories] => Illistration [sc_skill_categories] => Designers ) [7] => Array ( [ssc_skill_categories] => Animation [sc_skill_categories] => Designers ) ) What i would like to to is display this information in a table like so: <html> <body> <table> <tr> <td>Programming</td><td>Web</td><td>Actionscript</td><td>C#</td><td>CSS</td> <tr> <tr> <td>Designers</td><td>Graphic</td><td>Logo</td><td>Illistration</td><td>Animation</td> <tr> <table> </body> </html> I have been trying and failing all day to do this. Posting my "progress" will clog up the thread, so for now i wont post it. Does anyone have an idea how i would achieve this? Regards, -Ben So i pull some records out of a mysql table and i want to display them in 5 even columns. I'm not entirely sure how to do the math & logic to accomplish this. The pull is simple $qry = "SELECT DIST_PART_NUM FROM $tablename"; $sql = mysql_query($qry) or die(mysql_error()); while($res = mysql_fetch_assoc($sql)) { // CREATE 5 even columns here. } so let's say i just retrieved 5,000 part numbers, i'd like to display then in a table of 5 columns with 1000 records per column. This is easy math, but i need the script to automatically figure out the #'s. Also the tricky part is that i dont want to display the part numbers like so 11111 22222 33333 44444 55555 66666 77777 88888 99999 00000 but rather 11111 44444 77777 22222 55555 88888 33333 66666 99999 00000 the remainder if there is one can go in the last column or whatever is easier. I'd tried googling this, but it's not easy to phrase what i'm looking for. Thanks for the help. PS: I'm not looking to copy and paste code, if possible please explain your way so that i can learn the logic. Guys, i need your help,i have one table employee with empno=number,image=blob, images uploaded sucessfully and insert into database but when i am trying to retrieve records image are not displaying instead of it some encrypted form shows please help me Ok... I need some help - I want to show a players balance in a game beside there name (Balance is in mysql database)I can do that but... - I also want to show if there online or offline at the same time( This is stored in a different database) I have the code which says whether they are online or offline <?PHP // Conect to the Mysql Server $connect = mysql_connect("localhost","scswccla_bukkit","********"); //connect to the database mysql_select_db("scswccla_bukkit"); //query the database $query = mysql_query("SELECT * FROM users_online WHERE online = 1"); //title echo "<font size='100'>NovaCraft Users</font><br>"; // fetch the results / convert into an array WHILE($rows = mysql_fetch_array($query)): $users = $rows['name']; echo "<font color='black'>|Online|<br><font color='green'>$users</font></font><br>"; endwhile; //query the database $query = mysql_query("SELECT * FROM users_online WHERE online = 0"); //title echo "<font color='black'>|Offline|</font><br>"; // fetch the results / convert into an array WHILE($rows = mysql_fetch_array($query)): $users = $rows['name']; echo "<font color='red'>$users</font><br>"; endwhile; ?> Here is the page: www.scswc.com/Offline_Users.php displaying that But I want to Create something like this: Nocvacraft Players |Online| Name:Player Balance:$20 |Offline| Name:Player Balance:$15 Here is what I have tried: <?PHP // Conect to the Mysql Server $connect = mysql_connect("localhost","scswccla_bukkit","**********"); //connect to the database mysql_select_db("scswccla_bukkit"); //query the database $query = mysql_query("SELECT * FROM users_online WHERE online = 1"); $query2 =mysql_query("SELECT * FROM iBalances WHERE player = $users"); //title echo "<font size='100'>NovaCraft Users</font><br>"; // fetch the results / convert into an array WHILE($rows = mysql_fetch_array($query) $rows2 = mysql_fetch_array($query2)): $users = $rows['name']; $balance = $rows2['balance']; echo "<font color='black'>|Online|<br><font color='green'>Name:$usersBalance:$balance</font></font><br>"; endwhile; //query the database $query = mysql_query("SELECT * FROM users_online WHERE online = 0"); //title echo "<font color='black'>|Offline|</font><br>"; // fetch the results / convert into an array WHILE($rows = mysql_fetch_array($query)): $users = $rows['name']; echo "<font color='red'>$users</font><br>"; endwhile; ?> I know I am trying to use a variable before it is been set - but if I don't how I have tried this as well... <?PHP // Conect to the Mysql Server $connect = mysql_connect("localhost","scswccla_bukkit","**********"); //connect to the database mysql_select_db("scswccla_bukkit"); //query the database $query = mysql_query("SELECT * FROM users_online WHERE online = 1"); //title echo "<font size='100'>NovaCraft Users</font><br>"; // fetch the results / convert into an array WHILE($rows = mysql_fetch_array($query)): $users = $rows['name']; echo "<font color='black'>|Online|<br><font color='green'>Name:$users</font></font><br>"; endwhile; //query the database $query = mysql_query("SELECT * FROM users_online WHERE online = 0"); //title echo "<font color='black'>|Offline|</font><br>"; // fetch the results / convert into an array WHILE($rows = mysql_fetch_array($query)): $users = $rows['name']; echo "<font color='red'>$users</font><br>"; endwhile; $query = mysql_query("SELECT * FROM iBalances WHERE player = $users"); WHILE($rows = mysql_fetch_array($query)): $balance = $rows['balance']; echo "<font color='red'>$users $balance</font><br>"; endwhile; // ?> Can you use variables in mysql_query()?Is that why it isn't working? This is my first php script so if I need to give you more information for you to help me just tell me Thanks Here is database pictures iBalances users_online Hey guys have been trying to get this script to work for a while now, i am new to php and mysql so i am sure i am missing something simple. I have DB setup and need to pull data based on the key item code and get the following I want to get the fields item_code description allergy_statement useable_units region_availability order_lead_time ingredients for item_code 12-100 LITERALLY 12-100, no range, but like i said before i am really new to php and mysql. I have 1187 items that when a user clicks a link in search results it takes them to the product details page for that item code All that data is in my database just can't figure out how to get it out of the database. Is this even the right script to achieve that result. here is the code to get the data from database Code: [Select] <?php require_once('includes/mysql_connect_nfacts_ro.php'); $query = "SELECT item_code, description, allergy_statement, useable_units, region_availability, order_lead_time, ingredients " . "FROM products " . "WHERE item_code = '12-100' "; $resuts = mysql_query($query) or die(mysql_error()); ?> And need to display the data like so : Code: [Select] <td width="715" align="center" valign="top"> <h1>Product Details</h1> <h3>DISPLAY description HERE</h3> <table width="420" border="0"> <td class="ingreg"> </td> </table> <h5>Item Number</h5> <table width="420" border="0"> <tr> <td class="ingreg">DISPLAY ITEM_CODE HERE</td> </tr> </table> <h3>Ingredients:</h3> <table width="420" border="0"> <tr> <td class="ingreg">DISPLAY INGREDIENTS HERE</td> </tr> </table> <h4>Allergy Statement:</h4> <table width="420" border="0"> <tr> <td class="ingreg">DISPLAY Allergy Statement HERE</td> </tr> </table> <h4>Useable Units Per Package:</h4> <table width="420" border="0"> <tr> <td class="ingreg">DISPLAY Useable Units Per Package HERE</td> </tr> </table> <h4>Region Availability: </h4> <table width="420" border="0"> <tr> <td class="ingreg">&DISPLAY ITEM_CODE HERE</td> </tr> </table> <h4>Order Lead Time:</h4> <table width="420" border="0"> <tr> <td class="ingreg">&DISPLAY order lead time HERE</td> </tr> </table> <p> </p> <div align="right"></div></td> </tr> </table> how do i get data in database to display where i need it to? Can any one shine some light on this Hello everyone
Hoping someone could lend me a hand. I have a form that takes some end-user's details and adds the date and time into a MySQL table of when the form was submitted. I wish to display that date/time + 4 days ahead using PHP.
I believe the MySLQ DATE_ADD should do the trick quite nicely. In fact plumbing the following statement into phpMyAdmin gives me exactly the results I requi
SELECT DATE_ADD(`datetime`,INTERVAL 4 DAY) FROM `faults` WHERE fault_id = '51';However, just having a pig of a time getting this displayed using PHP. I'm sure this is elementary so forgive me. Here's what I had in mind but is no working. Can someone please point me in the right direction:
<?php
$date_query = mysqli_query($con, "SELECT DATE_ADD(`datetime`,INTERVAL 4 DAY) FROM `faults` WHERE fault_id = '51'");
while($row = mysqli_fetch_assoc($date_query)){
echo $row ['datetime'];
}
?>
Many thanks for your help and advice.Hi Guys, I am a complete novice as you will soon notice. Can anyone suggest what I am doing wrong with this code. When I run the query in phpmyadmin it produces the correct answer. However when I try to output on my site with php it returns the result "Array". I am guessing I have oversimplified somewhere, aint got a clue how though Code: [Select] <?php include("configure.php"); // To grab the DB info $dbh = mysql_connect ("localhost", DB_SERVER_USERNAME, DB_SERVER_PASSWORD) or die ('<BR> - Could not connect to the database because: '.mysql_error()); mysql_select_db (DB_DATABASE, $dbh) or die(mysql_error( )); $query = "SELECT `options_values_price` FROM `rain_products_attributes` WHERE `products_id` = 526 AND `options_id` = 3 AND `options_values_id` = 3"; $result = mysql_query($query); if (!$result) { $message = "Error! Invalid Query: ".mysql_error()."\n Original Query: ".$query; die($message); } while($row = mysql_fetch_array($result)) { echo $row; } mysql_close(); ?> hi guys I am having trouble displaying a table in a mySQL database i get the error message Warning: mysqli_fetch_row() expects parameter 1 to be mysqli_result, boolean given in /home/students/accounts/s7188633/hit3323/www/htdocs/Assiment2V2/main.php on line 47 Warning: mysqli_fetch_row() expects parameter 1 to be mysqli_result, boolean given in /home/students/accounts/s7188633/hit3323/www/htdocs/Assiment2V2/main.php on line 53 these are the two line that it realtes to row 47: $Row = mysqli_fetch_row($result); row 53: $Row = mysqli_fetch_row($result); below is all the code if you wanted to look at it, thanks <?php $choice = addslashes ($_POST["selection"]); { $DBConnect = @mysqli_connect("neptune.it.swin.edu.au", "*****", "***") Or die("<p>Unable to connect to the database server.</p>" . "<p>Error code " . mysqli_connect_errno() . ": " . mysqli_connect_error()) . "</p>"; $DBName = "*****_db"; if (!@mysqli_select_db($DBConnect, $DBName)) echo "<p>The database is not available.</p>"; $SQLstring = "SELECT * FROM Books";// WHERE category = 'Programing'"; $QueryResult = @mysqli_query($DBConnect, $SQLstring) Or die("<p>Unable to execute the query.</p>" . "<p>Error code " . mysqli_errno($DBConnect) . ": " . mysqli_error($DBConnect)) . "</p>"; $NumRows = mysqli_num_rows($QueryResult); if ($NumRows == 0) echo "<p>No records returned.</p>"; else { mysqli_select_db($DBConnect, $DBName); $SQLstring = "SELECT * FROM Books"; $result = @mysql_query($DBConnect, $SQLstring); $Row = mysqli_fetch_row($result); do { echo "<tr><td>{$Row[0]}</td>"; echo "<td>{$Row[1]}</td>"; echo "<td align='right'>{$Row[2]}</td>"; echo "<td align='right'>{$Row[3]}</td></tr>"; $Row = mysqli_fetch_row($result); } while ($Row); } } mysqli_close($DBConnect); ?>
create table mimi (mimiId int(11) not null, mimiBody varchar(255) );
<?php
//connecting to database
include_once ('conn.php');
$sql ="SELECT mimiId, mimiBody FROM mimi";
$result = mysqli_query($conn, $sql );
$mimi = mysqli_fetch_assoc($result);
$mimiId ='<span>No: '.$mimi['mimiId'].'</span>';
$mimiBody ='<p class="leading text-justify">'.$mimi['mimiBody'].'</p>';
?>
//what is next?
i want to download pdf or text document after clicking button or link how to do that I am querying my database to show the visit statistics for a particular week and it shows the number of visits for the countries, but does not display the country name.
I have proved that the MySQL works by going into phpMyAdmin and pasting the query into SQL query tab, replacing the POST with 1, for week 1.
I can't see why it is not displying the country.
Here is the code:
<?php
include('connect_visits.php');
doDB7();
$WVisit_data="SELECT WeekNo15.WNo, WeekNo15.WCom, Countries.Country, Countries.CID, ctryvisits15.CVisits
FROM ctryvisits15
LEFT JOIN Countries
ON ctryvisits15.country = Countries.CID
LEFT JOIN WeekNo15
ON ctryvisits15.WNo = WeekNo15.WNo
WHERE ctryvisits15.WNo = '{$_POST['WeekNo']}'
ORDER BY ctryvisits15.CVisits DESC";
$WVisit_data_res = mysqli_query($mysqli, $WVisit_data) or die(mysqli_error($mysqli));
$display_block ="
<table width=\"20%\" cellpadding=\"3\" cellspacing=\"1\" border=\"1\" BGCOLOR=\"white\" >
<tr>
<th>Country</th>
<th>Visits</>
</tr>";
while ($WV_info = mysqli_fetch_array($WVisit_data_res)){
$Ctry = $WV_info['country'];
$Visits = $WV_info['CVisits'];
//add to display
$display_block .="
<tr>
<td width=\"10%\" valign=\"top\">".$Ctry."<br/></td>
<td width=\"5%\" valign=\"top\">".$Visits."<br/></td>
";
}
mysqli_free_result($WVisit_data_res);
mysqli_close($mysqli);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
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Version : 1.0
Released : 20091106
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<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta name="keywords" content="" />
<meta name="description" content="" />
<meta http-equiv="content-type" content="text/html; charset=utf-8" />
<title>1066 Cards 4U - Stats for country</title>
<link href="style.css" rel="stylesheet" type="text/css" media="screen" />
</head>
<body>
<div id="wrapper">
<div id="menu">
<ul>
<li class="current_page_item"><a href="index.php">Home</a></li>
<li><a href="Links.html">Links</a></li>
<li><a href="Verse_Menu.html">Verses</a></li>
<li><a href="Techniques.html">Techniques</a></li>
<li><a href="blog.php">Blog</a></li>
<li><a href="Gallery.html">Gallery</a></li>
<li><a href="contact.html">Contact</a></li>
<li><a href="AboutUs.html">About Us</a></li>
<li><a href="stats1.html">Stats</a></li>
</ul>
</div><!-- end #menu -->
<div id="header">
<div id="logo">
<h1><a href="http://www.1066cards4u.co.uk">1066 Cards 4U</a></h1>
</div><!-- end #wrapper -->
</div><!-- end #header -->
<div id="page">
<div id="page-bgtop">
<div id="page-bgbtm">
<div id="content">
<h3>Statistics for Week Commencing <? echo $WkCom; ?> in 2015</h3>
<div id="table">
<?php echo $display_block; ?></div>
</div><!-- end #content -->
</body>
</html>
Can you help please?
Edited by rocky48, 07 January 2015 - 07:33 AM. |
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