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PHP - Coding To Display Data From Mysql
Morning all,
and Evening everyone else. Im using the following Query to display information from a table Code: [Select] <? include "config.php"; $query1="Select *,DATE_FORMAT(date_posted,'%W,%d %b %Y') as thedate FROM article WHERE DATE_SUB(CURDATE(),INTERVAL 30 DAY) ORDER BY date_posted DESC LIMIT 1 "; $blogarticles = mysql_query($query1) or die(mysql_error()); $num = mysql_num_rows($blogarticles); ?> What I am wanting to find out, is as the blog post itself could be pages long at times, is the anyway that further down the page where I actually call the data up onto the page itself I can limit how much of it is displayed? Ive succesfully added a limit to how many records are shown with the "DES LIMIT 1", is there something I can add to my query, or further down in my displaying table (which I will code just below here) to limit the lines/characters on display until the full article is opened? Display table: Code: [Select] <table width="75%" border="0" cellspacing="1" align="left"> <tr> <td> </td> </tr> <tr> <td> </td> </tr> <? if($num > 0){ while($row_articles = mysql_fetch_assoc($blogarticles)){ ?> <tr class="title"> <td><?=$row_articles['title'];?> </td> </tr> <tr> <td> </td> </tr> <tr class="tbody"> <td><?=$row_articles['comments'];?></td> </tr> <tr> <td> </td> </tr> <tr class="links"> <td>Date posted: <?=$row_articles['date_posted'];?> | <a href="comments.php?aid=<?=$row_articles['artid'];?>&cid=<?=$row_articles['categoryID'];?>">Comments(<? //echo $row_articles['artid']; //$thenum=row_articles['artid']; $getcomments = "SELECT * FROM article WHERE artchild='".$row_articles['artid']."'"; if(!$theResult=mysql_query($getcomments)){ echo mysql_error(); }else{ $num_comments=mysql_num_rows($theResult); echo $num_comments; } ?>) </a></td> </tr> <tr class="links"> <td> </td> </tr> <tr class="links"> <td> </td> </tr> <? } }else{ ?> <tr><td><p>There are no articles available at present</p></td></tr> <? } ?> <tr> </tr> </table> Just to clarify 'comments' is the table field that holds the blog data, it also holds comments on the blog, but the child ID is what differentiates them from one another. Thanks in advance for your help ladies & gents Tom Similar TutorialsI'm not sure what I'm doing wrong here... This is my index <html <head> <title>Admin applications</title> </head> <body text="#000000"> <center> <?php include("connect.php"); echo "<table cellpadding='3' cellspacing='2' summary='' border='3'>"; echo "<tr><td>Real name:<br><br>"; echo $row['real_name']; echo "</td><td>Age:<br><br>"; echo $row['age']; echo "</td><td>In-Game Name:<br><br>"; echo $row['game_name']; echo "</td><td>Steam ID:<br><br>"; echo $row['steamid']; echo "</td><td>Agreement:<br><br>"; echo $row['agreement']; echo "</td><td>Will use vent:<br><br>"; echo $row['vent']; echo "</td><td>Activity:<br><br>"; echo $row['activity']; echo "</td><td>Why this person wants to be an admin:<br><br>"; echo $row['why']; echo "</td></tr></table>"; ?> </center> </body> </html> and this is my database connect <?php $database="admin"; mysql_connect ("localhost", "root", "waygan914"); @mysql_select_db($database) or die( "Unable to select database"); mysql_query("SELECT * FROM applications"); ?> The database table "applications" has 8 fields, and 2 records, but when I view the page i get the table but no data: I am managing a shop website which is using php and mysql for data. The website has a section that will display the related products with the one that you are watching. Now my problem is that if there are more than 5 items it will continue to display all the items in one row with the consequent that the page will not display well. I need to find a way to begin a new row after the 5th item so it will display 5 items in each row. this is my current code that is responsible for showing the related items. There is also a picture attached with the problem. while ($row = mysql_fetch_array($retd)) { $code = $row["code"]; $name = $row["name"]; echo("<td width=150 align=center>"); echo ("<a href=../products/info.php?scode=$code><img src=pictures/$code.gif border=0 alt=Item $name</a>"); echo ("<br><a href=../products/info.php?scode=$code><span class=fs13>$name</span></a>"); echo("</td>"); } Does anyone know how can I do this? Hi guys I need your help. I would like the data from the column "user_login" being seen as an email link. How can I do that? Thousand Thanks Code: [Select] while($rows=mysql_fetch_array($result)){ ?> <tr> <td><span class="style10"><? echo $rows['user_id']; ?></span></td> <td><span class="style10"><? echo $rows['user_first_name']; ?></span></td> <td><span class="style10"><? echo $rows['user_surname']; ?></span></td> <td><span class="style10"><? echo $rows['user_login']; ?></span></td> <td align="center"><a href="update.php?id=<? echo $rows['user_id']; ?>" class="style10">update</a></td> </tr> <?php } I have got connection to the the mysql database, how do I get the data from the database to display on the webpage Hi all, I am trying to get data from MySQL to display in a html table in TCPDF but it is only displaying the ID.
$accom = '<h3>Accommodation:</h3>
<table cellpadding="1" cellspacing="1" border="1" style="text-align:center;">
<tr>
<th><strong>Organisation</strong></th>
<th><strong>Contact</strong></th>
<th><strong>Phone</strong></th>
</tr>
<tbody>
<tr>'.
$id = $_GET['id'];
$location = $row['location'];
$sql = "SELECT * FROM tours WHERE id = $id";
$result = $con->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
'<td>'.$location.'</td>
<td>David</td>
<td>0412345678</td>
</tbody>';
}
}
'</tr>
</table>';
Anyone got any ideas? im making a game and i need to show a users money but i dont know how help? I have created 5 websites under 5 different domains. Contents of this websites' are similar and using a same template for each one. Now I need to create an admin panel to control these websites. PHP and MySql I will use for this backend.
My problem is how can I manage these five website with one backend? Reason is I will use different domain for my backend. My all 5 client will use this same backend system to manage their own website.
So can I know from the professionals here, is it possible to display mysql data on these 5 website. If it is possible then how?
Any links to article or tutorials would be welcome and appreciated.
NOTE: I checked
mysql federated storage enginebut no idea? Is it the way where do I need to go? Thank you. Hey in my edit page i have 2 radio buttons in my form and i need to make sure the same value is still selected how can i do that? thanks i have a mysql table which contains name like mid mname 101 AAA 102 BBB 103 CCC now i have to print this name in a html table like AAA, BBB, CCC i am getting this by while loop in a variable but when loop changes then value also change so please tell me how i get this only in one variable & print I had this working, but when I try and get fancy and use AJAX the data doesn't display. I think this is a PHP problem though. My code for the select form including AJAX code Code: [Select] <!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-GB"> <head> <title>AJAX Example</title> <link rel="stylesheet" type="text/css" href="Form.css" media="screen" /> <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js"></script> <script type="text/javascript"> $(document).ready(function(){ $("tr:odd").addClass("odd"); }); </script> <script type="text/javascript"> function showPlayers(str) { var xmlhttp; if (str=="") { document.getElementById("DataDisplay").innerHTML=""; return; } if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp=new XMLHttpRequest(); else {// code for IE6, IE5 } xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); } xmlhttp.onreadystatechange=function() { if (xmlhttp.readyState==4 && xmlhttp.status==200) { document.getElementById("DataDisplay").innerHTML=xmlhttp.responseText; } } xmlhttp.open("GET","Query.php?category_id="+str,true); xmlhttp.send(); } </script> </head> <body> <h1">AJAX Example</h1> <?php #connect to MySQL $conn = @mysql_connect( "localhost","username","pw") or die( "You did not successfully connect to the DB!" ); #select the specified database $rs = @mysql_SELECT_DB ("MyDB", $conn ) or die ( "Error connecting to the database test!"); ?> <form name="sports" id="sports"> <legend>Select a Sport</legend> <select name="category_id" onChange="showPlayers(this.value)"> <option value="">Select a Sport:</option> <?php $sql = "SELECT category_id, sport FROM sports ". "ORDER BY sport"; $rs = mysql_query($sql); while($row = mysql_fetch_array($rs)) { echo "<option value=\"".$row['category_id']."\">".$row['sport']."</option>\n "; } ?> </select> </form> <br /> <div id="DataDisplay"></div> </body> </html> Query.php <?php #get the id $id=$_GET["category_id"]; #connect to MySQL $conn = @mysql_connect( "localhost","username","pw") or die( "Error connecting to MySQL" ); #select the specified database $rs = @mysql_SELECT_DB ("MyDB", $conn ) or die ( "Could not select that particular Database"); #$id="category_id"; #create the query $sql ="SELECT * FROM sports INNER JOIN players ON sports.category_id = players.category_id WHERE players.category_id = '".$id."'"; echo $sql; #execute the query $rs = mysql_query($sql,$conn); #start the table code echo "<table><tr><th>Category ID</th><th>Sport</th><th>First Name</th><th>Last Name</th></tr>"; #write the data while( $row = mysql_fetch_array( $rs) ) { echo ("<tr><td>"); echo ($row["category_id"] ); echo ("</td>"); echo ("<td>"); echo ($row["sport"]); echo ("</td>"); echo ("<td>"); echo ($row["first_name"]); echo ("</td>"); echo ("<td>"); echo ($row["last_name"]); echo ("</td></tr>"); } echo "</tr></table>"; mysql_close($conn); ?> I think the problem is either with this part in the AJAX Code: [Select] xmlhttp.open("GET","Query.php?category_id="+str,true); or most likely in my Query.php code when I wasn't using AJAX and using POST it worked fine, but adding the AJAX stuff and GET it doesn't work. When I echo out the SQL the result is Code: [Select] SELECT * FROM sports INNER JOIN players ON sports.category_id = players.category_id WHERE players.category_id = '' so the category_id is not being selected properly and that is the primary key/foreign key in the MySQL table which connects the JOIN. 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 need to create a web page that stores airline surveys in a MYSQL database. Include fields for the data and time of the flight and other fields. Also include group of radio buttons that allow the user to rate the airline. It should include a view past survey result button on the main survey page that displays a list of past results. I have made two php pasges and a Survey html main page but they don't work. Can anyone please help me?? I need to know what's wrong with my coding!!! THANKS 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. This could be PHP or MySql so putting it in PHP forum for now... I have code below (last code listed) which processes a dynamically created Form which could have anywhere from 0 to 6 fields. So I clean all fields whether they were posted or not and then I update the mySQL table. The problem with this code below is that if, say, $cextra was not posted (i.e. it wasnt on the dynamically created form), then this code would enter a blank into the table for $cextra (i.e. if there was already a value in the table for $cextra, it gets overwritten, which is bad). What is the best way to handle this? I'm thinking i have to break my SQL query into a bunch of if/else statements like this... Code: [Select] $sql = "UPDATE cluesanswers SET "; if (isset($_POST['ctext'])){ echo "ctext='$ctext',"; } else { //do nothing } and so on 5 more times.... That seems horribly hackish/inefficient. Is there a better way? Code: [Select] if (isset($_POST['hidden']) && $_POST['hidden'] == "edit") { $cimage=trim(mysql_prep($_POST['cimage'])); $ctext=trim(mysql_prep($_POST['ctext'])); $cextra=trim(mysql_prep($_POST['cextra'])); $atext=trim(mysql_prep($_POST['atext'])); $aextra=trim(mysql_prep($_POST['aextra'])); $aimage=trim(mysql_prep($_POST['aimage'])); //update the answer edits $sql = "UPDATE cluesanswers SET ctext='$ctext', cextra='$cextra', cimage='$cimage', atext='$atext', aextra='$aextra', aimage='$aimage'"; $result = mysql_query($sql, $connection); if (!$result) { die("Database query failed: " . mysql_error()); } else { } Hi. I have a database named 'Forums' with a table named 'forumlist'. The fields in the table are 'id', 'name', 'link', and 'keys'. I am trying to search the database in either the 'name' or 'keys' field and display the matches. I am getting this error: Parse error: syntax error, unexpected T_STRING, expecting ']' in C:\xampp\htdocs\search3.php on line 51 (The define fields) <html> <head> <title>My Forum Search</title> <style type="text/css"> table { background-color: #CCC } th { width: 150px; text-align: left; } </style> </head> <body> <h1>Forum Search</h1> <form method="post" action="search3.php"> <input type="hidden" name="submitted" value="true" /> <label>Search Category: <select name="category"> <option value="keys">Keyword</option> <option value="name">Name</option> </select> </label> <label><input type="text" name="criteria" /></label> <input type="submit" /> </form> <?php if (isset($_POST['submitted])) { DEFINE ('DB_USER', 'root'); DEFINE ('DB_PSWD', '******'); DEFINE ('DB_HOST', 'localhost'); DEFINE ('DB_NAME', 'Forums'); $dbcon = mysqli_connect(DB_HOST, DB_USER, DB_PSWD, DB_NAME); $category = $_POST['category']; $criteria = $_POST['criteria']; $query = "SELECT * FROM forumlist WHERE $category LIKE '$criteria'"; $result = mysqli_query($dbcon, $query) or die ('Error retrieving data') echo "<table>"; echo "<tr> <th>Name</th><th>Link</th> </tr>"; while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) { echo "<tr><td>"; echo $row['name']; echo "</td><td>"; echo $row['link']; echo "</td></tr>"; } echo "</table>"; } // end of main if state ?> </body> </html> Thanks in advance Hey there is no error with this code it works just fine but I would love to know if there is unnecessary coding in it for example do i have to do the global variables? is there a better way to do mysql editing page? thanks <?php include "../configdb.php"; $id = $_GET['id']; if(isset($_POST['submit'])) { //global variables $name = $_POST['name']; $footer = $_POST['footer']; //run the query which adds the data gathered from the form into the database $result = mysql_query("UPDATE pages SET name='$name', footer='$footer' WHERE id='$id' ",$connect); echo "<b>Your Page have been edited successfully"; } elseif($id) { $result = mysql_query("SELECT * FROM pages WHERE id='$id' ",$connect); while($row = mysql_fetch_assoc($result)) { $name = $row['name']; $footer = $row['footer']; ?> <h3>::Edit Page</h3> <form method="post" action="<?php echo $_SERVER['PHP_SELF'] ?>?id=<?php echo $row['id']?>"> <input type="hidden" name="id" value="<?php echo $row['id']?>"> <input name="name" size="40" maxlength="255" value="<?php echo $name; ?>"> <input name="footer" size="40" maxlength="255" value="<?php echo $footer; ?>"> <input type="submit" name="submit" value="Submit"> <?php } } Hi I am trying to select and order data/numbers from a colum in a mysql data base however i run the code and it returns no value just a blank page no errors or any thing so i think the code is working right but then it returns no result? Please help thanks Here is the code: <?php $host= "XXXXXX"; $mysql_user = "XXXXXX"; $mysql_password = "XXXXXX"; $mysql_database = "XXXXXXX"; $connection = mysql_connect("$host","$mysql_user","$mysql_password") or die ("Unable to connect to MySQL server."); mysql_select_db($mysql_database) or die ("Unable to select requested database."); $row = mysql_fetch_assoc( mysql_query( "SELECT XP FROM Game ORDER BY number DESC LIMIT 1" ) ); $number = mysql_result(mysql_query("SELECT XP FROM Game ORDER BY number DESC LIMIT 1"), 0); echo "The the highest XP is $number"; ?> hi guys, im new to this forum I'm new also to php, I need help from you guys: I want to display personal information from a certain person (the data is on the mysql database) using his name as a link: example: (index.php) names 1. Bill Gates 2. Mr. nice Guy i want to click Bill Gates (output.php) Name: Bill Gates Country:xxxx Age: xx etc. How can i make this or how to learn this? Hello. What I have created is a menu with links. The links will display content on the page from a database. I have the links with id displayed, but nothing is displayed when the link is clicks. I have seen isset() and $_GET used but not a form and I don't know how to change the code to allow the display of data using ID. My site is local, and I am using XAMPP. The database connections work. This is my code:
// query
$query = "SELECT * FROM topmenu ORDER BY id ASC";
$row = $PDO->query($query);
....
<table class="topmenu">
<tr>
<td>
<h1 class="siteName">site title</h1>
</td>
<?php
foreach($row as $data) {
?>
<td><a href="index.php?id=<?php echo $data['id']; ?>">
<?php echo $data['menuheader']; ?>
</a></td>
<?php
}
?>
</tr>
</table>
As a beginner, I would appreciate any help, no criticism of my code please! Also, is there a way of hiding the id in the url? Thanks in advance. Hi all. The following is a function to get and display data. However, I've hit a wall in my progress. It's only printing out one field. I need to have it print out every retrieved field. Any help is appreciated. function get_data($table_name){ $result=mysql_query("SELECT * FROM $table_name"); $num=mysql_num_rows($result); $i=0; while($i < $num){ $fields=mysql_fetch_field($result,$i); $display_fields=mysql_result($result,$i,$fields->name); echo $display_fields; $i++; }; }; I realized that (I think) I need the $display_fields var to be concatenated with the $i variable so there's a unique var name for each field retrieved from the mysql_fetch_field line. Any ideas? |
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