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PHP - Mysql Database Display Info Separately
Hi there.
I have this simple code which displays 5 results. How can i grab each element separately instead of displaying all the results at once. Thanks:) Code: [Select] <?php $query = mysql_query("SELECT product_name, product_price FROM products WHERE product_type = 'laptop' LIMIT 5"); $numrows = mysql_num_rows($query); if ($numrows != 0) { while ($row = mysql_fetch_assoc($query)) { $product_name = $row['product_name']; $product_price = $row['product_price']; echo $product_name . '<br />'; echo $product_price . '<br /><br />'; } } ?> Similar TutorialsI'm currently trying...struggling....to teach myself PHP and I'm really bugged with this database stuff. I am slowly managing but I'm a tad bit stuck now. I want to show a specific piece of information from a table. Lets say my table is structured like so: id user email 1 Bob Bob@Name.com 2 Fred Fred@Name.com 3 Matt Matt@Name.com What would I need to do to display ONLY Freds user? One way I tried only displayed the first rows info (Bob) the second way I tried (with a while loop) only displayed the last rows info (Matt) Heres my current code: <html> <body> <?php include 'dbwire.php'; $query = mysql_query('SELECT * FROM user'); $row = mysql_fetch_array($query); while ($row = mysql_fetch_array($query)) { echo '<b>User:</b> ' . $row['user'] . '<br />'; } ?> </body> </html> Howdy, I'm trying to display text from a table in a database. It's a list of quotes, so I just need to pull out the quote and the author name. However, the quote and name are not fields in the same record; they are separate records. Example data: Code: [Select] quoteid name value 1 content You guys are great! Thanks for being awesome. 1 author John Jackson 2 content Gosh you're amazing! Always been so darn helpful! 2 author Peter Davis So, I just need to rip out the data from 'content' and 'author', and then group them together based on the quoteid. This is my code thus far: $testimonial_resource = mysql_query("SELECT name, value FROM quotes GROUP BY quoteid ORDER BY author ASC") or die(mysql_error()); while ($testimonial = mysql_fetch_assoc($testimonial_resource)) { echo '<p>'.$testimonial['value']'.<br /><strong>'.$testimonial['value'].'</strong></p>'; } Any help would be greatly appreciated for this novice. Cheers. I have a 'user' table and a display users page. How would I display them in alphabetical order by their username? Heres my current basic display page: <?php include 'dbwire.php'; include 'header.php'; $query = mysql_query('SELECT * FROM user'); while ($row = mysql_fetch_array($query)) { echo '<b>Username:</b> ' . $row['user'] . '<br />'; echo '<b>Real Name:</b> ' . $row['name'] . '<br />'; echo '<b>Email:</b> ' . $row['email'] . '<br />'; echo '<b>Location:</b> ' . $row['location'] . '<br /> <hr>'; } ?> Theres also an id row but since user wouldn't be added alphabetically I wouldn't be able to order them by id. Hi Im having some trouble implementing info into a database well I know the standard way of retrieving mysql data was through the following codes: Code: [Select] $query = "SELECT * FROM {$tablename} WHERE columnmame = '{$var}'"; $result = mysql_query($query); $row = mysql_fetch_array($result); This will return all properties inside a table row by an associative array indexed by column names. I am, however, wondering if there is an easier way to retrieve database info from more than one table. For now, what I am doing is: Code: [Select] $result = mysql_query( "SELECT * FROM {$tablename} WHERE columnmame = '{$var}'"); $row = mysql_fetch_array($result); $result2 = mysql_query( "SELECT * FROM {$tablename2} WHERE columnmame2 = '{$var2}'"); $row2 = mysql_fetch_array($result2); which is a bit tedious and can cause problems when two or more coders work on the same project(it will be difficult to tell what is $row1, $row2 and $row3...). Is there away to write a simpler code than the one above? I mean, if it is possible to run mysql_fetch_array only once and retrieve database info from multiple tables? Hi there, I am using this code to send the users email address to the database. That works fine, but i keep getting blank info added to the database. Does anyone know how i can stop this? <?php $con = mysql_connect("*","*","*"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("ogs_mailinglist1", $con); $sql="INSERT INTO mailinglist (email) VALUES ('$_POST[rec_email]')"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } mysql_close($con); ?> Thanks, Hi, I have a website where users can log on and edit their profile pic, name, biography etc. I was wondering about the correct way to:- Add data to the database through forms (Register.php) Display the data on a page Using mysql escape sting, however, the way I am currently using will display a '\' before any ' symbol. So it's >> it\'s ... Here is a snippet of the code I am using... Code: [Select] //insert data $about1 = mysql_real_escape_string($_POST['about']); //get $query = mysql_query("SELECT * FROM `staff` WHERE username='$username'"); $row = mysql_fetch_array($query); $about = $row['about']; echo $about; Hi I have tried and tried and tried again to get this to work
in simple terms I have very little knowledge with PHP and even less with mysql
I have a paid subscription and domain in order to learn more and I feel I have made ok progress so far
then I realised how unsafe my current work is;
here is my experience this far
I created a site for a group of voluntary online game hosts where they can posts points from their tournaments in a forum
and some info pages to go with this,
however what I did was create a base template and style sheet and then an admin dashboard linked to individual forms to allow the group admin to edit the info pages they go to my form and enter the desired info and submit this then sends through and action file which posts the text and <BR> to a .txt file,
then the connecting page reads the .txt file using the PHP code of " <? php include ( 'index.txt'); ?>
yes you are seeing this correctly I have allowed a direct edit of text in a .txt file rather silly of me but I didn't realise how unsafe this was until now I guess its a good job I trust that the admin has no knowledge or skills in coding
ok since all this I have created a DB in MySQL on my server,
My server uses PHPMyAdmin I have create a DB named " mnvbcou1_content1 " and a table named " home " with rows " ID " and " home "
what I am trying to do:
I want my page to display the content of the table row home and a form once submitted to send to the table row home
or if needed I can re make this DB if the names are not suitable
I have tried to create the needed coding to make this work but for some reason this just will not work I have already added 2 rows to my table to try and make the page to display the content but it just is not working I got an error every time
so I hope that someone out there is rather patient and is willing to help me learn how to do this correctly and safely,
also this is a closed group website the address to this site is only known by a handful of none programmers I am mainly trying to make this work for my own personal knowledge and server safety please help me
I am working on a project where I want a select form to display information from a MySQL table. The select values will be different sports (basketball,baseball,hockey,football) and the display will be various players from those sports. I have set up so far two tables in MySQL. One is called 'sports' and contains two columns. Once called 'category_id' and that is the primary key and auto increments. The other column is 'sports' and contains the various sports I mentioned. For my select menu I created the following code. <?php #connect to MySQL $conn = @mysql_connect( "localhost","uname","pw") or die( "You did not successfully connect to the DB!" ); #select the specified database $rs = @mysql_SELECT_DB ("test", $conn ) or die ( "Error connecting to the database test!"); ?> <html> <head>Display MySQL</head> <body> <form name="form2" id="form2"action="" > <select name="categoryID"> <?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> </body> </html> this works great. I also created another table called 'players' which contains the fields 'player_id' which is the primary key and auto increments, category_id' which is the foreign key for the sports table, sport, first_name, last_name. The code I am using the query and display the desired result is as follows <html> <head> <title>Get MySQL Data</title> </head> <body> <?php #connect to MySQL $conn = @mysql_connect( "localhost","uname","pw") or die( "Err:Db" ); #select the specified database $rs = @mysql_SELECT_DB ("test", $conn ) or die ( "Err:Db"); #create the query $sql ="SELECT * FROM sports INNER JOIN players ON sports.category_id = players.category_id WHERE players.sport = 'Basketball'"; #execute the query $rs = mysql_query($sql,$conn); #write the data while( $row = mysql_fetch_array( $rs) ) { echo ("<table border='1'><tr><td>"); echo ("Caetegory ID: " . $row["category_id"] ); echo ("</td>"); echo ("<td>"); echo ( "Sport: " .$row["sport"]); echo ("</td>"); echo ("<td>"); echo ( "first_name: " .$row["first_name"]); echo ("</td>"); echo ("<td>"); echo ( "last_name: " .$row["last_name"]); echo ("</td>"); echo ("</tr></table>"); } ?> </body> </html> this also works fine. All I need to do is tie the two together so that when a particular sport is selected, the query will display below in a table. I know I need to change my WHERE clause to a variable. This is what I need help with. thanks I want to make a "hint" feature for an app ive been working on. Its a guessing game, and I want to be able to give out hints. Ok so say there is an answer in the database. Say the value is "firetruck". I want to retrieve that value, and obscure it, displaying only one or two letters that are in the word. Or, for a number, say 178, I want to display only one digit from it. Is this possible? Im sure it is, anything seems possible with PHP & Mysql. If so, how could I implement this? Trust me, I read the manuals. The manuals for both PHP and Mysql are so vast and a little advanced for my level.. then again im not complete novice and know the basics plus more of both. I have got connection to the the mysql database, how do I get the data from the database to display on the webpage im making a game and i need to show a users money but i dont know how help? Hi, here's my problem: I am trying to make a simple online buying website and I want to display a table with all the fields for each item. So I got that part down which is to just use mysql_fetch_assoc("SELECT * FROM myTable") and use the html table tags stuff, but now I want to display my images in the table, so here's my code to display my mysql database table in html's table tag along w/ php: <html> <head> <title>My Online buying website project</title> </head> <body> <?php mysql_connect("localhost","root"); mysql_select_db("myTable"); $imagesArray=array("Apple_iPhone3GS.jpg","Apple_iPhone4.jpg","product3.jpg","product4.jpg","product5.jpg"); $result=mysql_query("SELECT Name, Manufacturer, Price, Description, SimSupport FROM myTable"); if(mysql_num_rows($result))//if there is at least one entry in bellProducts, make a table { print "<table border='border'>"; print "<tr> <th>Name</th> <th>Manufacturer</th> <th>Price</th> <th>Description</th> <th>SimSupport</th> </tr>"; //NB: now output each row of records while($row=mysql_fetch_assoc($result)) { extract($row); print "<tr> <td>$Name</td> <td>$Manufacturer</td> <td>$Price</td> <td>$Description</td><td>$SimSupport</td> </tr>"; }//END WHILE }//END IF ?> </table> </body> </html> *So how do I go about adding my images in this table? Say a user puts in a support request, and for every request it generates a unqiue string, and enters it into the database. Ok, now say there is a text field, when the user enters their unique string and it finds a match, it displays the data along with it. How can I accomplish this? Im kind of new to mysql, but I know basic SQL. Would be great if somebody could point me in the right direction! Thanks I have a standard form that displays users current data from a mysql database once logged in(code obtained from the internet). Users can then edit their data then submit it to page called editform.php that does the update. All works well except that the page does not display the updated info. Users have to first logout and login again to see the updated info. even refreshing the page does not show the new info. Please tell me where the problem is as i am new to php.
my form page test.php
<?PHP
require_once("./include/membersite_config.php");
if(!$fgmembersite->CheckLogin())
{
$fgmembersite->RedirectToURL("login.php");
exit;
}
?>
<form action="editform.php?id_user=<?= $fgmembersite->UserId() ?>" method="POST">
<input type="hidden" name="id_user" value="<?= $fgmembersite->UserId() ?>"><br>
Name:<br> <input type="text" name="name" size="40" value="<?= $fgmembersite->UserFullName() ?>"><br><br>
Email:<br> <input type="text" name="email" size="40" value="<?= $fgmembersite->UserEmail() ?> "><br><br>
Address:<br> <input type="text" name="address" size="40" value="<?= $fgmembersite->UserAddress() ?> "><br><br>
<button>Submit</button>
my editform.php
<?php
$con = mysqli_connect("localhost","root","user","pass");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_query($con,"UPDATE fgusers3 SET name = '".$_POST['name']."', email= '".$_POST['email']."', address= '".$_POST['address']."' WHERE id_user='".$_POST['id_user']."'");
header("Location: test.php");
?>
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
How do I do I prevent a broken image icon if there is no image? Here is my code: Code: [Select] <?php if ($_POST){ $county = $_POST['county']; } $con = mysql_connect("localhost","",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("", $con); $imageLocation = $row['imageurl1']; $result = mysql_query("SELECT * FROM places WHERE `county` = '".mysql_real_escape_string($county)."' order by `date_created` DESC"); if ( mysql_num_rows($result) > 0 ) { echo "<strong>Click Headers to Sort</strong>"; echo "<table border='0' align='center' bgcolor='#999969' cellpadding='3' bordercolor='#000000' table class='sortable' table id='results'> <tr> <th> Title </th> <th> Borough </th> <th> Town </th> <th> Phone </th> <th> Rooms </th> <th> Bath </th> <th> Fees </th> <th> Rent </th> <th> Image </th> </tr>"; while($row = mysql_fetch_array($result)) { echo "<tr> <td bgcolor='#FFFFFF' style='color: #000' align='center'> <a href='classified/places/index.php?id=".$row['id']."'>" . $row['title'] . "</a></td> <td bgcolor='#FFFFFF' style='color: #000' align='center'>" . $row['county'] . "</td> <td bgcolor='#FFFFFF' style='color: #000' align='center'>" . $row['town'] . "</td> <td bgcolor='#FFFFFF' style='color: #000' align='center'>" . $row['phone'] . "</td> <td bgcolor='#FFFFFF' style='color: #000' align='center'>" . $row['rooms'] . "</td> <td bgcolor='#FFFFFF' style='color: #000' align='center'>" . $row['bath'] . "</td> <td bgcolor='#FFFFFF' style='color: #000' align='center'>" . $row['feeornofee'] . "</td> <td bgcolor='#FFFFFF' style='color: #000' align='center'>" . $row['rent'] . "</td> <td bgcolor='#FFFFFF' style='color: #000' align='center'><img src=user/". $row['imageurl1'] ." width='50'></td> </tr>"; } echo "</table>"; print_r($apts); } else { echo "<p> </p><p> </p> No Results <br /><p> </p><FORM><INPUT TYPE='button' VALUE='Go Back' onClick='history.go(-1);return true;'></FORM> and Refine Your Search <p> </p><p> </p>"; } ?> Thanks in advance Hello. I am trying to display info from two tables with this code. Code: [Select] <?php $query = mysql_query("SELECT users.username, users2.username FROM users INNER JOIN users2 ON users.id = users2.id"); $numrows = mysql_num_rows($query); if ($numrows != 0) { while ($row = mysql_fetch_assoc($query)) { $username = $row['username']; echo $username . "<br />"; } } ?> However it displays info only from one db (users). I suppose there's something wrong with Code: [Select] <?php ... while ($row = mysql_fetch_assoc($query)) { $username = $row['username']; echo $username . "<br />"; } ... ?> Any help will be appreciated. Thank you. Hello all, I have a directory page that is populated with a foreach loop pulling info from a db table. I want to make a "Download Directory pdf" button that will print out the directory entries in a nicely formatted way. The thing is that the foreach loop uses an OO call to what every template is being used for the HTML display: foreach ( (array) $results as $row) { ... $out .= '<div class="cn-list-row' . $alternate . ' vcard ' . $template->slug . ' ' . $entry->getCategoryClass(TRUE) . '">' . "\n"; $out = apply_filters('cn_entry_before', $out, $entry); ob_start(); include($template->file); $out .= ob_get_contents(); ob_end_clean(); $out = apply_filters('cn_entry_after', $out, $entry); $out .= '</div>' . "\n"; } $out .= '<div class="clear"></div>' . "\n"; $out .= '</div>' . "\n"; $out = apply_filters('cn_list_after', $out, $results); return $out; } Any Ideas?? I have created a form that i can see the form input info in to the form boxes but once submitted it is not updating or adding new input to the database. I am trying to keep it as simple as possible. I am also new at PHP.
Here is my code:
<?php // define variables and set to empty values $amp_20_parts_idErr = $part_numberErr = $locationErr = $quantityErr = ""; $amp_20_parts_id = $part_number = $discription = $location = $quantity = ""; if ($_SERVER["REQUEST_METHOD"] == "POST") { if (empty($_POST["amp_20_parts_id"])) { $amp_20_parts_idErr = "ID is required."; } else { $amp_20_parts_id= test_input($_POST["amp_20_parts_id"]); } if (empty($_POST["part_number"])) { $part_numberErr = "Part number is required."; } else { $email = test_input($_POST["part_number"]); } if (empty($_POST["description"])) { $descriptionErr = ""; } else { $description = test_input($_POST["description"]); } if (empty($_POST["location"])) { $locationErr = "A location is required."; } else { $location = test_input($_POST["location"]); } if (empty($_POST["quantity"])) { $quantityErr = "Quantity is required"; } else { $quantity = test_input($_POST["quantity"]); } } function test_input($data) { $data = trim($data); $data = stripslashes($data); $data = htmlspecialchars($data); return $data; } ?> <div id="update_form"> Please follow instructions for updating the data base.<br>instructional text goes here.<br/><br/> <table><tr><form method="POST" action="new path"> <td>ID: <input name="amp_20_parts_id" type="text"><span><?php echo $amp_20_parts_idErr; ?></span><br/><br/></td> <td>Part Number: <input name="part_number" type="text"><span><?php echo $part_numberErr; ?></span><br/><br/></td> <td><label>Discription: <textarea name="description" rows="3" col="20"></textarea> <br/><br/></td> <td>Location: <input name="location" type="text"><span><?php echo $locationErr;?> </span><br/><br/></td> <td>Quantity: <input name="quantity" type="text"><span><?php echo $quantityErr; ?> </span><br/><br/></td><br/> <td><input type="submit"></td> </form></tr></table> <?php $con=mysqli_connect("server","user","password","db"); // Check connection if (mysqli_connect_errno()) { echo("Connect failed: %s\n", mysqli_connect_error()); // escape variables for security $amp_20_parts_id = mysqli_real_escape_string($con, $_POST['amp_20_parts_id']); $part_number = mysqli_real_escape_string($con, $_POST['part_number']); $discription = mysqli_real_escape_string($con, $_POST['description']); $location = mysqli_real_escape_string($con, $_POST['location']); $quantity = mysqli_real_escape_string($con, $_POST['quantity']); $sql="INSERT INTO amp_20 (amp_20_parts_id, part_number, description, location, quantity) VALUES ('$amp_20_parts_id', '$part_number', '$description', '$location', '$quantity')"; if (!mysqli_query($con,$sql)) { die('Error: ' . mysqli_error($con)); } echo "1 record added"; mysqli_close($con); } ?> |
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