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PHP - Linking Mysql Output Result
Hi,
I'm trying to make a mysql output to a link so the name will be a link so when you hit this link you will get the full information of this mysql input. Can someone point me in the correct direction? Here is how i get the output from my mysql database into my table. Code: [Select] <td>"; echo $row['name']; echo "</td> Similar TutorialsHi there, i've had a good look around the forums and on various other sites and can't get my head around a silly little php issue i'm having and was wondering if any of the php guru's on here could possibly provide me with a simple answer/solution please, i've not even been able to get results for what i'm trying to do, as I haven't known the best text to search for this little thing. I'm retrieving data from mysql and one of the cells that I parse the results for always has either a Y or an N for that field. I'm having no issues getting it to display the Y or the N correctly. However... Rather than have it display just an N or a Y, i've been trying to get PHP to make it display either a YES or a NO instead, purely for cosmetic purposes. The php code i've been working with was part of a pre built script I'm currently customising and this bit of functionality never worked in it: Code: [Select] <td>" . ($result[7] ? "Yes" : "No") . "</td> $result[7] being the field that contains either a y or a n for every result. Unfortunately, all that my output text does is displays a Yes regardless, even if there is a n in the dbase field. Could someone please help me, so I can get this to display a YES if the field contains a y and display a NO if the field contains a n I know this is probably a bit of a nOObie question, but I am learning this stuff still as I go along... Thanks in advance and sorry for the rather non-technical way i've worded this. Okay so I have 2 tables in my database. One called user and one called messages. A user logs in to the message board and leaves a message (eg nice website). They write in the author name and the message then after the message is posted it says "Nice website" Posted by (author) on (date). All is good so far. It works. However if you look at my code you will see I have a session started. This session is storing the username of the logged in user. From the column username in the users table. (This table has has an id for each user). Ive played around with the code trying to make it so the user doesnt have to fill in the author box. I want rid of that box So the logged in user just leaves a message then it says "posted by (username) on (date). Im missing something from my code. Can anyone tell me what? Please? <?php session_start(); mysql_connect("*************", "*****************", "***************"); mysql_select_db("***********************"); $time = time(); //this checks to see if the $_SESSION variable has been not set //or if the $_SESSION variable has been not set to true //and if one or the other is not set then the user gets //sent to the login page if (!isset($_SESSION['username'])) { header('Location: http://***************.com/login.php'); } $query = "INSERT INTO messages VALUES( NULL, '". mysql_real_escape_string($_POST['message']) ."', '". mysql_real_escape_string($_POST['username']) ."', '$time' )";if( $result = mysql_query($query) ) { if(mysql_affected_rows() > 0 ) { echo "Message Posted.<br><a href='messageboard.php'>Return</a>"; } else { echo 'There was an error posting your message. Please try again later.'; } } else { echo "There was a database error."; // comment out next line for live site. echo "<br>Query string: $query<br>Returned error: " . mysql_error() . '<br>'; } ; hi all, I am trying to link 2 mysql tables and display some information from each of them. I have a list of all the possible items for sale in table1 and I am trying to count the number of rows in the other table2 where the items exist. E.g. 'table1' manufacturer model man1 item1 man1 item2 man1 item3 'table2' id model 1 item3 2 item3 3 item2 And the result would show: item1(0) item2(1) item3(2) It would list all the items from table1 and show next to it how many rows are related to that item from table2. I have inserted a quote where I have tried many times to enter something similar to that show in the note below - but I can not get it to work - it just shows the total number of models in table1 for a given manufacturer. The php I have made so far is: <?php case 'manufacturer': $query = " SELECT * FROM table1"; $query .= " WHERE manufacturer = '".$data."' "; $query .= " ORDER BY model "; $result = mysqli_query($cxn,$query); $returnData[''] = "Select a Model..."; while($row = mysqli_fetch_assoc($result)){ // I THINK I NEED TO INSERT SOMETHING LIKE $query2 = "SELECT * FROM table2 WHERE model = table1.model"; $k=$row['model']; $k2=$row2['model']; $counter[$k]+=1; $returnData[$k]=$k; } foreach($counter as $k => $row) { $returnData[$k] .= " ($row)"; } break; ?> Hi everyone, I hope I explain my problem well enough. I have created a cms with the help of a tutorial for my website, it allows me to click on the page from a menu and shows me the results. The menu list is taken from my mysql database, so for example i have homepage and recent news in my mysql table and these are what are shown in the menu. What I want is for the menu to to be images that can be clicked to take you to the correct page. I have attached a print screen to show you what it looks like at the minute and what I want the menu to look like. I have no idea if this is even possible can someone please help me out? Hopefully I will hear a reply, I will send my code if needed when I know whether it is possible or not. Thanks in hope Hi, Currently on my website I have a section where you can select an article held within a database, I still want this to happen but I wouldn't a different word to be linked to the article. Currently it reads 'Click here to view this entry' underneath the title of the article, I want he title of all the different articles to link to the right article is there anyway to do this? My code currently reads: Code: [Select] <?php $blog_postnumber = 5; if(!isset($_GET['page'])) { $page = 1; } else { $page = (int)$_GET['page']; } $from = (($page * $blog_postnumber) - $blog_postnumber); $sql = "SELECT * FROM cms_article ORDER BY timestamp DESC LIMIT $from, $blog_postnumber"; $result = mysql_query($sql) or print ("Can't select entries from table cms_article.<br />" . $sql . "<br />" . mysql_error()); while($row = mysql_fetch_array($result)) { $date = date("l F d Y", $row['timestamp']); $title = stripslashes($row['title']); $entry = stripslashes($row['entry']); $id = $row['id']; if (strlen($entry) > 0) { $entry = substr($entry, 0, 0); $entry = "$entry<a href=\"journal.php?id=" . $id . "\">Click here to view this entry.</a>"; } ?> im finding it hard to link PHP, MySQL and dreamweaver together. when php document is alaunched within the localhost all the data in the database is present how ever when the php coding is entered into the html file via dreamweaver no data is present which i think connection to the mysql database cannot be establish why is this can someone help me please, the php document is attached with all the coding [attachment deleted by admin] I have a few tables in a project and i'm linking pages together via PK/FK's. i have a main table with client names, ID, CLIENT NAME, ETC i have a table with communications COMMID, ID, CONTACT NAMES, ETC and another of communication history. COMMHISTORYID, COMMID, DATE, TIME, ETC I'm not sure how to write the query so that i can view/update information from all 3 tables on the same webpage page using php/mysql. I'd appreciate any assistance you may provide. thanks. How can I check if a returned mysql value is equal to '' i.e. nothing? I keep getting an error where the page won't load because the returned value is '' so i need to check for it hello I'm using this code: Code: [Select] $query="SELECT * FROM `second_content` WHERE CHANGED =0 limit 0,1";to query first row data when I want see it and echo it I recived : Code: [Select] Resource id #2can anyone help me ? thank you Im not sure where to post this but since it includes php il post it here instead of in the mysql forum. ok so, i have a table and i get the values using while($row = mysql_fetch_array($result)){ and then echo them in rows. that works fine but i need to add a class to the last row of my table. I would need somehow to fetch the last row of the array and make it echo something different. Any help is appreciated Thank you Hello, i created this script for a client and have ran into an annoying error with the results displaying on a new line for each result instead of side by side, any help is welcome Cheers Code: [Select] <?php $subcat = mysql_real_escape_string(strip_tags(htmlspecialchars(protect($_GET['subcat'])))); $cat = mysql_real_escape_string(strip_tags(htmlspecialchars(protect($_GET['cat'])))); $sql = @mysql_query("SELECT * FROM cakes WHERE category =\"$cat\" AND sub_cat=\"$subcat\" ORDER BY id DESC"); while ($row = mysql_fetch_array($sql)) { $reference = $row['reference']; $image = $row['image']; echo ("<p><img src='./images/cakes/$image' height='289px' width='177px' alt='IMAGE OF CAKE'></img><br />"); echo ("<b>Reference:</b>$reference"."</p>"); } if (!$reference) { echo 'There are no cakes in this category yet.'; } ?> I know the errors only going to be something small i'm missing, but i've been coding all day Hi, I don't know anything about php and Mysql but I found a tutorial for a shopping cart and everything is working. The only thing is they don't have the email part in the totorial so I'm kind of stuck with my file. Anyone know how to email the resul to me via email with this code? <? include("includes/db.php"); include("includes/functions.php"); if($_REQUEST['command']=='update'){ $name=$_REQUEST['name']; $email=$_REQUEST['email']; $address=$_REQUEST['address']; $phone=$_REQUEST['phone']; $result=mysql_query("insert into customers values('','$name','$email','$address','$phone')"); $customerid=mysql_insert_id(); $date=date('Y-m-d'); $result=mysql_query("insert into orders values('','$date','$customerid')"); $orderid=mysql_insert_id(); $max=count($_SESSION['cart']); for($i=0;$i<$max;$i++){ $pid=$_SESSION['cart'][$i]['productid']; $q=$_SESSION['cart'][$i]['qty']; $price=get_price($pid); mysql_query("insert into order_detail values ($orderid,$pid,$q,$price)"); } die('Thank You! your order has been placed!'); } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Billing Info</title> <script language="javascript"> function validate(){ var f=document.form1; if(f.name.value==''){ alert('Your name is required'); f.name.focus(); return false; } f.command.value='update'; f.submit(); } </script> </head> <body> <form name="form1" onsubmit="return validate()"> <input type="hidden" name="command" /> <div align="center"> <h1 align="center">Billing information</h1> <table border="0" cellpadding="2px"> <tr><td>Total:</td><td><?=get_order_total()?></td></tr> <tr><td>Name :</td><td><input type="text" name="name" /></td></tr> <tr><td>Address :</td><td><input type="text" name="address" /></td></tr> <tr><td>Email :</td><td><input type="text" name="email" /></td></tr> <tr><td>Phone :</td><td><input type="text" name="phone" /></td></tr> <tr><td> </td><td><input type="submit" value="Place Order" /></td></tr> </table> </div> </form> </body> </html> I am seeking to learn more about the noted subject, how to use PHP to allow a user to enter search terms and search a database. I have experimented with this with little results save for errors. Please see code listed below: search.php <? //// filename = search.php <form method="post" action="result.php3"> <select name="metode" size="1"> <option value="row_name1">metode1</option> <option value="row_name2">metode2</option> </select> <input type="text" name="search" size="25"> <input type="submit" value="Begin Searching!!"> </form> ?> results.php //// filename = result.php3 <? $hostname = "mysql7.000webhost.com"; // Usually localhost. $username = "a4542527_root"; // If you have no username, leave this space empty. $password = "*******"; // The same applies here. $usertable = "people"; // This is the table you made. $dbName = "a4542527_test1"; // This is the main database you connect to. MYSQL_CONNECT($hostname, $username, $password) OR DIE("Unable to connect to database"); @mysql_select_db( "$dbName") or die( "Unable to select database"); ?> <? //error message (not found message) $XX = "No Record Found"; $query = mysql_query("SELECT * FROM $usertable WHERE $metode LIKE '%$search%' LIMIT 0, 30 "); while ($row = mysql_fetch_array($query)) { $variable1=$row["row_name1"]; $variable2=$row["row_name2"]; $variable3=$row["row_name3"]; print ("this is for $variable1, and this print the variable2 end so on..."); } //below this is the function for no record!! if (!$variable1) { print ("$XX"); } //end ?> Upon viewing search.php I receive the error message: Parse error: syntax error, unexpected '<' in /home/a4542527/public_html/search.php on line 3 I believe I may be missing something and am a bit lost. Thank-you in in advance for any help or suggestions. ~Matty Hello Everyone, I have a quick question for you all, I think its fairly simple... I have created a database and I am using PHP to grab the data: $usera = $_SESSION['username']; $query2 = "SELECT * FROM tracker WHERE id = '$usera', hidden = yes"; mysql_query($query2) or die('Error, query failed : ' . mysql_error()); This hopefully will return multiple rows which look like this in the database. id username date reps hidden 1 supremebeing 2011-01-02 30 yes 4 supremebeing 2011-04-02 46 yes How would i turn each result into a variable eg: $date1 = 2011-01-02; $date2 = 2011-04-02; $reps1 = 30; $reps2 = 46; I think i have explained that well enough for you to understand, please reply if not though and i will provide more information. Thanks in Advance Hey guys, Currently Im using: $row = mysql_fetch_array($result) or die(mysql_error()); echo $row['user_family']. " - ". $row['user_registered']; $row['user_family'] = $fam; $_SESSION['family'] = $fam; to take data from a mysql table & set it as SESSION family. However, I cant seem to get this to set. The information IS being taken from mysql because its being echo'd earlier up in the code, but its just not passing to the session. Any ideas? im getting an error: Warning: mysql_fetch_array() [function.mysql-fetch-array]: The result type should be either MYSQL_NUM, MYSQL_ASSOC or MYSQL_BOTH. in C:\www\library\mysql.class.php on line 353 $result_type = "MYSQL_ASSOC"; $row = mysql_fetch_array($result, $result_type) but if i use (below) it works fine...is the constant() function the right way to what im doing? $result_type = "MYSQL_ASSOC"; $row = mysql_fetch_array($result, constant($result_type)) thank you Hi, I am new to php but not to programming. I created a script on a windows platform which connects to the mysql database and returns the results of a table. A very basic script which I wrote to simply test my connection worked. The script works fine on my windows machine but not on my new mac. On the mac it simply does not display any records at all. I know that the database connection has been established because there is no error but I can not see why the result set is not being displayed on screen, as I said it worked fine on my windows machine. The Mac has mysql (with data) and apache running for php. Please could someone help as I have no idea what to do now? Script below: Code: [Select] $dbhost = 'localhost'; $dbuser = 'root'; $dbpass = 'root'; $conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql'); $dbname = 'test'; mysql_select_db($dbname); mysql_select_db("test", $conn); $result = mysql_query("SELECT * FROM new_table"); while($row = mysql_fetch_array($result)) { echo $row['test1'] . " " . $row['test2'] . " " . $row['test3']; echo "<br />"; } mysql_close($con); hey guys, i need some help with my php/mysql iplogger. My code: <?php //finds out ip $ip = $_SERVER['REMOTE_ADDR']; //conects to the mysql server $connection = mysql_connect('localhost', 'root', ''); //sellects the database mysql_select_db('iplog', $connection); //looks for duplacute ips $dup = mysql_query("SELECT COUNT(number) FROM logged_ips WHERE ip_address = '$ip'",$connection); $count = mysql_result($dup, 0); //checks to see if there is a duplecate name if ($count == 0){ //inserts the ip in to the database $string = 'INSERT INTO `logged_ips` (`aid`, `ip_address`, `ip_visits`) VALUES (\'' . null . '\', \'' . $ip . '\', \'0\')'; mysql_query($string, $connection); }else{ //adds a visit to the database $string2 = "UPDATE `logged_ips` SET `ip_visits` = '++1' WHERE `ip_address` = $ip LIMIT 0,1"; mysql_query($string2, $connection); } //outputs the ip echo $ip; ?> error: Warning: mysql_result() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\randoms\logger\test.php on line 11 127.0.0.1 It also doesnt put anything in to the mysql database. Please help. Thanks jragon |
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