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PHP - Display Data That Has Been Written To The Database On The Same Page Load
Hi guys,
Should be a simple 1. If i have the following at the top of the page: $page_views = $row['page_views'] + 1; mysql_query("UPDATE table SET page_views='$page_views'"); and then the following at the bottom of the page: echo $row['page_views']; Should I see the page views as 1 the first time the page is visited, 2 the second time the page is visited, and so on......? At the moment im seeing 0 on the first page visit, 1 on the second page visit, 2 on the third..... I had this problem before on another page i was working, and i simply solved it by displaying the mysql query above the echo similar to the code above. However now it does not seem to be working. Am i missing something really simple? lol Thanks Similar TutorialsHi I am trying to display data from the table "event" in my database, I use the code below but it will not work and I cannot figure out why. CAn anyone help? CODE: <?php $host="localhost"; // Host name $username="root"; // Mysql username $password=""; // Mysql password $db_name="test"; // Database name $tbl_name="event"; // Table name mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $result = my_sql_query("SELECT * FROM event WHERE eventid = '1'"); while($row = mysql_fetch_array($result)) { $eventname= $row["eventname"]; $eventdate= $row["eventdate"]; echo "<b><u>Event Name:</b></u> $eventname" echo "<b><u>Event Date:</b></u> $eventdate<br>"; } ?> DISPLAY: Event Name: $eventname echo "Event Date: $eventdate "; } ?> hiii all, i want to show the database field (alise name) as the url after index.php when page or site loads first time. how i can achieve it? please help.... thanks in advance im making a game and i need to show a users money but i dont know how help? how i want to display data from database to look like this : <table width="633" height="224" border="1"> <tr bgcolor="#999900"> <td width="45">Bil</td> <td width="121">Course_name</td> <td width="83">session</td> <td width="83">start_date</td> <td width="83">end_date</td> <td width="83">notes</td> <td width="89">pre-req</td> </tr> <tr bgcolor="#6A7AEA"> <td rowspan="2">1.</td> <td rowspan="2" bgcolor="#6A7AEA">Math</td> <td>1st session </td> <td>1 jan 11 </td> <td>6 jan 11 </td> <td rowspan="2"> </td> <td rowspan="2"><image icon that will link to the oter site> </td> </tr> <tr> <td bgcolor="#6A7AEA">2nd session </td> <td bgcolor="#6A7AEA">8 jan 11 </td> <td bgcolor="#6A7AEA">15 jan 11 </td> </tr> <tr> <td bgcolor="#0066CC">2.</td> <td bgcolor="#0066CC">English</td> <td bgcolor="#0066CC">1st session </td> <td bgcolor="#0066CC">1 feb 11 </td> <td bgcolor="#0066CC">6 feb 11 </td> <td bgcolor="#0066CC"> </td> <td bgcolor="#0066CC"><image icon that will link to the oter site></td> </tr> <tr> <td rowspan="2" bgcolor="#6A7AEA">3.</td> <td rowspan="2" bgcolor="#6A7AEA">Science</td> <td height="29" bgcolor="#6A7AEA">1st session </td> <td bgcolor="#6A7AEA">8 march 11 </td> <td bgcolor="#6A7AEA">15 march 11 </td> <td rowspan="2" bgcolor="#6A7AEA"> </td> <td rowspan="2" bgcolor="#6A7AEA"><image icon that will link to the oter site></td> </tr> <tr> <td bgcolor="#6A7AEA">2nd session</td> <td bgcolor="#6A7AEA">16 march 11 </td> <td bgcolor="#6A7AEA">21 march 11 </td> </tr> </table> ** all the view data is called from database including the icon image thanks... I have got connection to the the mysql database, how do I get the data from the database to display on the webpage hey guys so im trying to display data into text boxes that are fetched from database according to checkbox with value id. processing is located before <!DOCTYPE html>:
if(isset($_POST['edit_event']) && isset($_POST['check']))
{
require "connection.php";
foreach ($_POST['check'] as $edit_id)
{
$edit_id = intval($GET['event_id']); //i tried (int)$edit_id;
$sqls = "SELECT event_name,start_date,start_time,end_date,end_time,event_venue FROM event WHERE event_id IN $edit_id ";
$sqlsr = mysqli_query($con, $sqls);
$z = mysqli_fetch_array($sqlsr);
{
}
button and form opens:
<form method="post" action="event.php"> <input type="submit" name="edit_event" value="Edit Event">this is the html where the data will be echoed:
<div id="doverlay" class="doverlay"></div>
<div id="ddialog" class="ddialog">
<table class="cevent">
<thead><tr><th>Update Event</th></tr></thead>
<tbody>
<tr>
<td>
<input type="text" name="en_" value="<?php echo $z['event_name']; ?>">
</td>
</tr>
<tr>
<td>
<input type="text" name="dates_" value="<?php echo $z['start_date']; ?>">
<input type="text" name="times_" value="<?php echo $z['start_time']; ?>">
</td>
</tr>
<tr>
<td><input type="text" name="datee_" value="<?php echo $z['end_date']; ?>">
<input type="text" name="time_" value="<?php echo $z['end_time']; ?>">
</td>
</tr>
<tr>
<td><input type="text" name="ev_" value="<?php echo $z['event_venue']; ?>">
</td>
</tr>
<tr>
<td><input type="submit" name="update" value="Update Event" id="update">
<input type="submit" id="cancelupdate" name="cancel" value="Cancel" >
</td>
</tr>
</tbody>
</table>
</div>
this is the part which is populated by data from database where isset($_POST['check']) gets the 'check' from:
echo
"<tr>
<td><input type='checkbox' name='check[]' value='$id'>$name
</td>
</tr>";
</form>
thanks in advance!
Edited by noobdood, 19 May 2014 - 10:42 PM. This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=321339.0 This is actually a page to edit the drama details. Database- 3 tables 1. drama dramaID drama_title 1 friends 2. drama_genre drama_genreID dramaID genreID 1 1 2 2 1 1 3 1 3 3. genre genreID genre 1 comedy 2 romance 3 family 4 suspense 5 war 6 horror <?php $dbhost = 'localhost'; $dbuser = 'root'; $dbpass = ''; $dbname = 'drama'; $link = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname) or trigger_error('Error connecting to mysql'); $id = $_GET['dramaID']; $link2 = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname) or trigger_error('Error connecting to mysql'); $sql2 = "SELECT * from drama , drama_genre , genre WHERE drama.dramaID='".$id."' AND drama_genre.dramaID='".$id."' AND drama.dramaID = drama_genre.dramaID AND drama_genre.genreID = genre.genreID"; $status2 = mysqli_query($link2,$sql2) or die (mysqli_error($link2)); $link3 = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname) or trigger_error('Error connecting to mysql'); $sql3 = "SELECT * from genre "; $status3 = mysqli_query($link3,$sql3) or die (mysqli_error($link3)); <form method='Post' action='doUpdate.php' enctype="multipart/form-data"> <?php while ($row2 = mysqli_fetch_assoc($status2)) { ?> <td height="1"></td> <td><select name="gdrop"> <option value="<?php echo $row2['genreID'];?>"><?php echo $row2['genre'];?></option> <?php while ($row3 = mysqli_fetch_assoc($status3)) { ?> <option value="<?php echo $row3['genreID'];?>"><?php echo $row3['genre'];?></option> <?php } mysqli_close($link3); ?> <input type='hidden' id='drama_genreID' name='drama_genreID' value = "<?php echo $row2['drama_genreID']; ?>"/> </select> <input type="hidden" id="dramaID" name="dramaID" value = "<?php echo $row['dramaID'];?>"/> <input type="submit" Value="Update"/> The result was there were 3 dropdown menus but only the first dropdown menu has all 6 genres from my database and also the genre that belongs to the drama. I'm also wondering how I can bring all the drama_genreIDs to my save(doupdate.php) page and update all 3 of them because it seems like only the last dropdown menu's data is saved. And also how can I display only 6 genres instead of 7 with the genre that belongs to the drama , being set as the default selection. As a complete newbie to php and webdesigning i have a following problem.I would like to retrieve the data from database and display it in a drop down menu.Then i should allow the user to select the values from drop down list along with other details,in other words i have to embed the drop down output as the form input for the user and store the form data in another table.I am running a xampp server and i am using php 5.4 version.Please help.My code is as follows.In this case project_name is displayed as the drop down output.but how do i use the same drop down output as a input in the form. <html> <head></head> <body> <?php error_reporting(E_ALL ^ E_DEPRECATED); include 'connect.php' ; $tbl_name="projects"; $sql="SELECT project_name FROM $tbl_name "; $result=mysql_query($sql); if($result === FALSE) { die(mysql_error()); } ?> <form name="resources" action="hourssubmit.php" method="post" > <?php echo "<select name='project_name'>"; while ($row = mysql_fetch_array($result)) { echo "<option value='" . $row['project_name'] ."'>" . $row['project_name'] ."</option>"; } echo "</select>"; ?> </form> </body> </html> Hi there, Im working on my little project and I would appreciate your help. I have only basic knowledge of php, mostly I just copy some scripts that could be useful for me. Im trying to find some simple script that allows me to see the written text on web page no. 1 on webpage no. 2.. Something like send the form to email, except I dont want to send it on email, but different webpage. Something like different way of eshop, where you get your order shipped to email, but I wanna send this information to webpage. Is there such a script like Im describing? Probably is but I dont know how to search for this..
Thank You
Edited by Radim, 21 October 2014 - 07:15 AM. i have 8 division (div), i want to display 4 rows in 4 division and the remain 4 rows in the next 4 division here is my code structure for carousel
<div class="nyie-outer"> second row third row
fourth row fifth row sixth row seven throw eighth row
</div><!--/.second four rows here-->
sql code
CREATE TABLE product( php code
<?php how can i echo that result in those rows
I have been playing around with this. Thought I had it nailed, and then found that It was fubar when I started to add more entries to the database. I basically want to display 2 tables from my database on the page. One is called "players" the other is called "editlog". The output with my current script came out as this: and here is the dreaded code: <html> <body> <u><h3>Performance Point Monitor (PPM): Knights of Shadow WoW Officers.</h3></u> <?php include('/home/a3269923/public_html/ppm/admin/config.php'); include('/home/a3269923/public_html/ppm/admin/dbopen.php'); $query="SELECT * FROM players"; $result=mysql_query($query); $num=mysql_numrows($result); ?> <table border="1" style="position:absolute;width:500;height:10;left:0;top:70"> <tr> <th><font face="Arial, Helvetica, sans-serif">PLAYER</font></th> <th><font face="Arial, Helvetica, sans-serif">POINTS</font></th> <th><font face="Arial, Helvetica, sans-serif">DATE</font></th> </tr> <?php $i=0; while ($i < $num) { $f2=mysql_result($result,$i,"PLAYER_NAME"); $f3=mysql_result($result,$i,"ND_POINTS"); $f4=mysql_result($result,$i,"DATE ADDED"); ?> <tr> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f2; ?></font></td> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f3; ?></font></td> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f4; ?></font></td> </tr> </table> <?php $i++; } ?> <?php include('/home/a3269923/public_html/ppm/admin/config.php'); include('/home/a3269923/public_html/ppm/admin/dbopen.php'); //USER LOG $query="SELECT * FROM editlog"; $result=mysql_query($query); $num=mysql_numrows($result); ?> <table border="1" style="position:absolute;width:600;height:10;left:510;top:70"> <tr> <th><font face="Arial, Helvetica, sans-serif">USER</font></th> <th><font face="Arial, Helvetica, sans-serif">TYPE</font></th> <th><font face="Arial, Helvetica, sans-serif">POINTS</font></th> <th><font face="Arial, Helvetica, sans-serif">NOTES</font></th> <th><font face="Arial, Helvetica, sans-serif">DATE</font></th> </tr> <?php $i=0; while ($i < $num) { $f1=mysql_result($result,$i,"USER"); $f2=mysql_result($result,$i,"TYPE"); $f3=mysql_result($result,$i,"POINTS"); $f4=mysql_result($result,$i,"NOTE"); $f5=mysql_result($result,$i,"TIMESTAMP"); ?> <tr> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f1; ?></font></td> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f2; ?></font></td> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f3; ?></font></td> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f4; ?></font></td> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f5; ?></font></td> </tr> </table> <?php $i++; } ?> What am I doing wrong? Halp! Hello guy , how do I write PHP page that have table to list all the data I entered to my system.
I used to be good at this but I changed servers and everything is different... Heres my code so far: Code: [Select] <?php $rated=$_REQUEST['rated']; echo $rated; $rating=$_REQUEST['rating']; echo $rating; // Make a MySQL Connection mysql_connect("localhost", "********", "********") or die(mysql_error()); mysql_select_db("*********") or die(mysql_error()); $result = mysql_query("SELECT * FROM main WHERE username = '$rated'") or die(mysql_error()); $row = mysql_fetch_array( $result ); $votes = $db_field['$rating']; $newvotes = $votes + 1; echo $newvotes; mysql_query("UPDATE main SET $rating = '$newvotes' WHERE username = '$rated'"); ?> Whats going on here is the colomb that I want to update comes as a variable $rated (That works) and then the database selects the row to update with $username (That works) and gets the variable $newvotes by taking the original value of the data its about to update and add 1 to it (That works) Then it updates the field to $newvotes.... I don't know why the update won't go through... there are no errors.... Is there anyway of saving data to phpmyadmin, and linking it to another page with one button. I can do it separately but I can't seem to do it using one button. Does anyone has any ideas! I'm having a problem and need an answer to why its happening and how to prevent it. Scenario: I begin load my home page which starts with a session_start(); .... Before it FULLY completes loading I try to navigate to another page and BOOM, that page will not load and any other page that begins with session_start(); will not load unless I close and restart the entire browser or wait about 10 minutes.... I will note my website makes ajax calls every 5 seconds or so, but I use setTimeout for them. Any help??? Thanks ahead! This is the code im using: Code: [Select] <?php include "config.php"; if (isset($_GET['id'])) { $id = mysql_real_escape_string($_GET['id']); $sql = "SELECT email FROM gusers WHERE id = $id LIMIT 1"; if ($result = mysql_query($sql)) { if (mysql_num_rows($result)) { $row = mysql_fetch_assoc($result); echo "Email: " . $row['email']; } } } I am trying to get more informations than just the e-mail address, but everytime it seems to fail, i am a completly newb to php but ill try my best.... Could someone tell how this is done properly (like loading name, info, and so on from the database and showing it aswell) Hello to all, I have problem figuring out how to properly display data fetched from MySQL database in a HTML table. In the below example I am using two while loops, where the second one is nested inside first one, that check two different expressions fetching data from tables found in a MySQL database. The second expression compares the two tables IDs and after their match it displays the email of the account holder in each column in the HTML table. The main problem is that the 'email' row is displayed properly while its while expression is not nested and alone(meaning the other data is omitted or commented out), but either nested or neighbored to the first while loop, it is displayed horizontally and the other data ('validity', 'valid_from', 'valid_to') is not displayed.'
Can someone help me on this, I guess the problem lies in the while loop? <thead> <tr> <th data-column-id="id" data-type="numeric">ID</th> <th data-column-id="email">Subscriber's Email</th> <th data-column-id="validity">Validity</th> <th data-column-id="valid_from">Valid From</th> <th data-column-id="valid_to">Valid To</th> </tr> </thead> Here is part of the PHP code:
<?php
while($row = $stmt->fetch(PDO::FETCH_ASSOC))
{
echo '
<tr>
<td>'.$row["id"].'</td>
';
while ($row1 = $stmt1->fetch(PDO::FETCH_ASSOC)) {
echo '
<td>'.$row1["email"].'</td>
';
}
if($row["validity"] == 1) {
echo '<td>'.$row["validity"].' month</td>';
}else{
echo '<td>'.$row["validity"].' months</td>';
}
echo ' <td>'.$row["valid_from"].'</td>
<td>'.$row["valid_to"].'</td>
</tr>';
}
?>
Thank you. So just to preface this, I have been part of two operations (one as developer, one with a 3rd party company developing) where the business was forced to cease due to difficulties in database load balancing and lots of people lost lots of money. I am talking about big data and high performance needed at the same time.
So for my new project, I am going to try and design it around having a forever expanding infrastructure of servers but that means setting it correctly from the beginning. I have put together some ideas for possible ways to split the database load across multiple servers. Any input to which idea(s) are best would be great so I know which to explore further. Also any relevant info on this type of thing would be helpful as this is the first time I am personally doing this.
Thanks!
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