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PHP - How Do I Use Mysql_insert_id() In Conjuntion With Sprintf() To Link Tables?
Set up
Windows Vista * XAMPP 1.7.3, including: * Apache 2.2.14 (IPv6 enabled) + OpenSSL 0.9.8l * MySQL 5.1.41 + PBXT engine * PHP 5.3.1 * phpMyAdmin Ultimate objective: I want to insert session data into multiple tables whilst ensuring that the data is in the appropriate column. Problem: I managed to insert the data into the correct tables and column, but after reading various forums I have been given the impression that if I was to have multiple site users the data's columns could get muddled up if they execute the script at the same time (hope I'm making sense, say if I'm not). The suggested method was mysql_insert_id() but I do not know how to make this work with in conjunction with sprintf(). As I said the script worked before I added the code with the star by it in an attempt to reach my ultimate objective. <?php //let's start our session, so we have access to stored data session_start(); session_register('membership_type'); session_register('terms_and_conditions'); include 'db.inc.php'; $db = mysql_connect('localhost', 'root', '') or die ('Unable to connect. Check your connection parameters.'); mysql_select_db('ourgallery', $db) or die(mysql_error($db)); //let's create the query $query = sprintf("INSERT INTO subscriptions ( name, email_address, membership_type) VALUES ('%s','%s','%s')", mysql_real_escape_string($_SESSION['name']), mysql_real_escape_string($_SESSION['email_address']), mysql_real_escape_string($_SESSION['membership_type'])); //let's run the query $result = mysql_query($query, $db) or die(mysql_error($db)); *mysql_real_escape_string($_SESSION['name']) = mysql_insert_id($db);* $query = sprintf("INSERT INTO site_user_info ( terms_and_conditions, name_on_card, credit_card_number ) VALUES ('%s','%s','%s')", *mysql_real_escape_string($_SESSION['name']),* mysql_real_escape_string($_SESSION['terms_and_conditions']), mysql_real_escape_string($_POST['name_on_card']), mysql_real_escape_string($_POST['credit_card_number']), mysql_real_escape_string($_POST['credit_card_expiration_data'])); //let's run the query $result = mysql_query($query, $db) or die(mysql_error($db)); *mysql_real_escape_string($_SESSION['credit_card_number']) = mysql_insert_id($db);* $query = sprintf("INSERT INTO card_numbers ( credit_card_expiration_data) VALUES ('%s')", *mysql_real_escape_string($_POST['credit_card_number']),* mysql_real_escape_string($_POST['credit_card_expiration_data'])); $result = mysql_query($query, $db) or die(mysql_error($db)); echo '$result'; ?> All the other facets of the script work but I get the following error message with the above script: Fatal error: Can't use function return value in write context in C:\x\xampp\htdocs\form_process.php on line 27 But it's not so much the error message its the ultimate objective. Any help appreciated. Similar TutorialsThis looks perfect but for some reason it's not functioning. Could someone help fixing this. $resultF =sprintf("SELECT * FROM xy_db WHERE clubname ='%s'", mysql_real_escape_string($data3)); $resultFin = mysql_query($resultF); if (mysql_num_rows ($resultf) > 0){ $register = "Retry."; echo($register); } else { $go =sprintf("INSERT INTO ab_db (surnamez, firstnamez) VALUES ('%s', '%s'", mysql_real_escape_string($data1), mysql_real_escape_string($data2)); $result = mysql_query($go); Thank you. Ruth. \-) Hi guys, I had my code working fine as a login page untill I added sprintf and mysql_real_escape_string and since then when i test the form to login, server keep loading and then come up with this msg Fatal error: Maximum execution time of 30 seconds exceeded in ../Dashboard/index.php on line 35 which is (Line 35) Quote while($row=mysql_fetch_array(mysql_query($getpin))){ I have my code below, can u please help me what is wrong? im coding in dreamweaver and it doesnt have any error in there. Code: [Select] <?php include ('includes/db/db.php'); ?> <!DOCTYPE html> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-type" content="text/html; charset=utf-8" /> <title></title> <link rel="stylesheet" href="./css/reset.css" type="text/css" media="screen" title="no title" /> <link rel="stylesheet" href="./css/text.css" type="text/css" media="screen" title="no title" /> <link rel="stylesheet" href="./css/form.css" type="text/css" media="screen" title="no title" /> <link rel="stylesheet" href="./css/buttons.css" type="text/css" media="screen" title="no title" /> <link rel="stylesheet" href="./css/login.css" type="text/css" media="screen" title="no title" /> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /></head> <body> <div id="login"> <h1>Dashboard</h1> <?php if (isset($_POST['login']) && $_POST['login']){ $email=addslashes(strip_tags($_POST['email'])); $in_password=addslashes(strip_tags($_POST['password'])); $pin=addslashes(strip_tags($_POST['pin'])); $password=md5($in_password); if (!$email || !$in_password || !$pin) echo "<div class='error'>Please fill all required fields</div>"; else{ $getpin=sprintf("SELECT * FROM users WHERE UserEmail='%s' AND UserPassword='%s'", mysql_real_escape_string($email) , mysql_real_escape_string($password)); while($row=mysql_fetch_array(mysql_query($getpin))){ $pin_email=$row['UserEmail']; $pin_id=$row['UserId']; $pin_company_id=$row['company_id']; $pass=$row['UserPassword']; } $get=sprintf("SELECT pin FROM company WHERE company_id='%s' AND active='%s'", mysql_real_escape_string($pin_company_id), mysql_real_escape_string(1)) ; while($row=mysql_fetch_array(mysql_query($get))){ $pin_num= $row['pin']; } if($password==$pass && $pin_num==$pin && $email==$pin_email) { echo"success"; } else { echo "<div class='error'>Login Failed, Login details are incorrect!</div>"; } } } ?> <div id="login_panel"> <form action="" method="post" accept-charset="utf-8" /> <div class="login_fields"> <div class="field"> <label for="email">Email</label> <input type="text" name="email" value="" id="email" tabindex="1" placeholder="email@example.com" /> </div> <div class="field"> <label for="password">Password <small><a href="forgotpassword.php">Forgot Password?</a></small></label> <input type="password" name="password" value="" id="password" tabindex="2" placeholder="password" /> <div class="field"> <label for="pin">Pin Number</small></label><input type="password" name="pin" value="" id="password" tabindex="2" placeholder="pin"/> </div> </div> </div> <!-- .login_fields --> <div class="login_actions"> <input type="submit" name="login" value="Login" class="btn btn-grey"/> </div> </form> </div> <!-- #login_panel --> </div> <!-- #login --> </body> </html> thanks you all in advance. Hey everyone, I've come around this problem quite a few times now, so I'm looking for a cleaner way to do what I always resort to. Here's my scenario: The following line is fairly straightforward: $sql_where = " WHERE SUBSTRING(name,1,1) = '".$c."' "; But sometimes I need to explode on some values, rising a need for multiple instances of the above, as in MYSQL's WHERE bla = 1 || WHERE bla = 2 || WHERE bla = 3 Of course, I could always write it out manually, like this: foreach($c as $v) { $sql_str .= "WHERE SUBSTRING(name,1,1) = '".$v."' "; } But I want my code to be more modular. The only solution I've been using so far is something like this: $where_start = "WHERE SUBSTRING(name,1,1) = '"; $where_end = "' "; foreach($c as $v) { $sql_str .= $where_start . $v . $where_end; } I find this technique a bit ridiculous and I was wondering if there was a similar method as say when you want to sprintf like this: $v = 'Bla bla %s bla'; sprintf($str, $v); So my question is basically this: is there any way to use a sprintf-like function that will concatenate variables inside variables? Thanks for any suggestions or input. Hello dear friends, say i've database table with (id,name) and want to add more informations by insert new names within an file for example has Code: [Select] $q1 = "INSERT INTO `mytable` VALUES (???,name1); mysql_query($q1) or die(mysql_error()." at row ".__LINE__); $q2 = "INSERT INTO `mytable` VALUES (???,name2); mysql_query($q2) or die(mysql_error()." at row ".__LINE__); ect..... how then it automatic detect the last id to go on after it ? some says use mysql_insert_id but i didn't understand how to apply it in this way also the example at php.net is bad not explain much please how can i use it thanks How can you debug to find out why it's not getting the mysql_insert_id number? I have it echoed all the queries and all with all the correct values from my form but the only problem is that its not getting the insert id number of the id. $query1 = "INSERT INTO `efed_bio` (charactername,username,posername,style_id,gender,status_id,division_id,alignment_id,sortorder) VALUES ('".$charactername."','".$username."','".$posername."','".$style."','".$gender."','".$status."','".$division."','".$alignment."','".$sort."')"; mysql_query($query1); $query1_id = mysql_insert_id(); echo $query1; echo $query1_id; $query2 = "INSERT INTO `efed_bio_allies` (bio_id) VALUES (".$query1_id.")"; mysql_query($query2); echo $query2; $query3 = "INSERT INTO `efed_bio_rivals` (bio_id) VALUES (".$query1_id.")"; mysql_query($query3); echo $query3; $query5 = "INSERT INTO `efed_bio_singles` (bio_id) VALUES (".$query1_id.")"; mysql_query($query5); echo $query5; Hello there, well basically what this function does is look at all the entry's to find a unused port in the database but when I run the function the first time if the table is empty mysql_insert_id works fine, but then if I run it again the mysql_insert_id keeps returning 0, I am using it immediately after the query but its not working properly, can anybody shed some ideas into the matter, anything is appreciated!, thanks for your time!: function addServerToBuildPool($ServerGame, $ServerOwner, $ServerSlots, $ServerBox) { $Server = mysql_fetch_array(mysql_query("SELECT * FROM xhost_boxs WHERE box_id = '".mysql_real_escape_string($ServerBox)."'")); $AddServer = mysql_query("INSERT INTO xhost_servers (server_game, server_slots, server_owner, server_ip, server_is_setup) VALUES('".mysql_real_escape_string($ServerGame)."', '".mysql_real_escape_string($ServerSlots)."', '".mysql_real_escape_string($ServerOwner)."', '".mysql_real_escape_string($Server['box_ip'])."', 'No')"); $ServerID = mysql_insert_id(); $FindPort = mysql_query("SELECT server_port FROM xhost_servers ORDER BY server_id ASC"); $Port = 0; while($row = mysql_fetch_assoc($FindPort)) { $Port = ($Port == 0)? $row['server_port'] : $Port; if($row['server_port'] != $Port) { break; } $Port++; } mysql_query("UPDATE xhost_servers SET server_port = '".$Port."' AND server_username = 'server".$ServerID."' AND server_password = '".rand(5000000, 900000000)."' WHERE server_id = '".$ServerID."'") or die(mysql_error()); } hello, im asking for your help again. i have a problem with this mysql_insert_id() function im using. it's used to get the values of id's and insert it to a field of another table right?but in my case where i have 4 tables and using the function to get the unique id,it no longer gets the id on the third table which is supposed to be doing so that this id will be inserted to the field of the final table. these are dynamic textboxes im working on by the way..here's my php code: html Code: [Select] <html> <head> <script language="JavaScript"> function AddTextBox() { document.getElementById('container').innerHTML+='<input type="text" size="15" maxlength="15" name=block[]><br>'; } function AddTextBox2() { document.getElementById('container2').innerHTML+='<input type="text" size="15" maxlength="15" name=room[]><br>'; } </script> </head> <body> <form name="form1" method="post" action="adnew.php"> <input type="hidden" name="cid"> Course:<input type="text" name="course"> <input type="hidden" name="yid"> Year: <select name="year"> <option value="1">1</option> <option value="2">2</option> <option value="3">3</option> <option value="4">4</option> </select> <table width="23%" border="0"> <tr> <td width="37%" height="1" colspan="2"> <input type="hidden" name="block_id"> Block: <input name="button" type="button" onClick="AddTextBox();" value="Add textbox"></td> </tr> <tr> <td height="21"><div id="container"></div></td> </tr> </table> <table width="23%" border="0"> <tr> <td width="64%">Room: <input name="button2" type="button" onClick="AddTextBox2();" value="Add textbox"></td> </tr> <tr> <td><div id="container2"></div></td> </tr> </table> <br><br><input type="Submit" name="submit" value=" Add "> </form> </body> </html> php Code: [Select] <?php include("dbcon.php"); ?> <?php $id_c=$_POST['cid']; $block=$_POST['block']; $block_id=$_POST['block_id']; $room=$_POST['room']; $intblock=0; $introom=0; $sql=mysql_query("INSERT INTO course VALUES ('$id_c','$_POST[course]')") or die (mysql_error()); $id_c = mysql_insert_id(); $sql=mysql_query("INSERT INTO year VALUES ('$_POST[yid]','$id_c','$_POST[year]')") or die (mysql_error()); $_POST['yid'] = mysql_insert_id(); while(count($block)>$intblock) { if (($block[$intblock]<>"")){ $sql=mysql_query("INSERT INTO block VALUES ('$block_id', '$_POST[yid]', '".$block[$intblock]."')") or die (mysql_error()); mysql_query($sql); } else{ echo "Block ".($intblock+1)." is missing values and cannot be inserted."; } $intblock=($intblock+1); } $block_id = mysql_insert_id(); while (count($room)>$introom) { if (($room[$introom]<>"")){ $sql=mysql_query("INSERT INTO room VALUES ('$block_id', '".$room[$introom]."')") or die (mysql_error()); // this is the 4th table.. mysql_query($sql); } else{ echo "Room ".($introom+1)." is missing values and cannot be inserted."; } $introom=($introom + 1); } echo "Successfully added."; echo "<br><a href='index.php'>Add another</a>"; ?> Quote $block_id = mysql_insert_id(); this is what i'm having problems with.. Quote $sql=mysql_query("INSERT INTO room VALUES ('$block_id', '".$room[$introom]."')") or die (mysql_error()); and this is for my 4th table.. i have found out that you have to put the function after an INSERT command,but in my case, i have a switch statement and after i tried putting it inside the switch, i get a message that says "duplicate entry 1 for b_id...etc.." any suggestions?advise? Hi all, i'm a bit more than a newbie. Here's the scenario. I have a page with tabbed navigation. On each tab there's a form (diving centers, teachers, and so on). Each form has a submit button with a unique name. Every submit button is processed by a series of if statements in an external included php file. As you can see in my code below what the file does is to process the if statements based on the submit pressed. It' important to notice that a JS script checks that the first form is filled with all the infos. If not is not possible to proceed to complete all the others. On the first form the user can register general information. Obviously there's an Id field (auto increment) that i grab with mysql_nsert_id. Also obvious is the fact that, for query reasons, i want to store the grabbed id in an id field present in all the tables of my database. The tables get the data from the various forms displayed on the tabbed navigation. Here's the code of the included file (notice that i use to start with a simple coding to test everything 's working fine) <?php require_once('Connections/Scubadiving.php'); if (!empty($_POST['theButton'])) { $nome=$_POST["nome"]; $indirizzo=$_POST["indirizzo"]; $insertSQL = "INSERT INTO centrisub (nome, indirizzo) VALUES ('$nome' , '$indirizzo')"; mysql_select_db($database_Scubadiving, $Scubadiving); $Result1 = mysql_query($insertSQL, $Scubadiving) or die(mysql_error()); $last_id = mysql_insert_id ($Scubadiving); } if (!empty($_POST['istruttori'])) { echo $_SESSION["$last_id"]; $insertSQL = "INSERT INTO istruttori (nome, data_nascita, curriulum, altre_info, idcentrisub) VALUES ('$_POST[nome]','$_POST[data_nascita]','$_POST[curriculum]','$_POST[altre_info]','$last_id')"; mysql_select_db($database_Scubadiving, $Scubadiving); $Result1 = mysql_query($insertSQL, $Scubadiving) or die(mysql_error()); } ?> What i don't understand is why $last_id is not passed form the first if statment (processed when the user submit the generel infos form) to the other if statement (in this case the teachers form). I've tried also with $_SESSION (trying to assign $last_id) but no success. (Probably because i don't know exactly how to use it) Hope everything's clear. What i'm missing ? Thanks in advance for your help. When I run this Prepared Statement... // Build query. $q2 = "INSERT INTO member(email, activation_code, salt, hash, first_name, username, register_ip, register_hostname, location, created_on) VALUES(?, ?, ?, ?, ?, ?, ?, ?, ?, NOW())"; // Prepare statement. $stmt2 = mysqli_prepare($dbc, $q2); // Bind variables to query. mysqli_stmt_bind_param($stmt2, 'sssssssss', $email, $activationCode, $salt, $hash, $firstName, $username, $ip, $hostName, $location); // Execute query. mysqli_stmt_execute($stmt2); // Capture New ID. $_SESSION['memberID'] = mysql_insert_id(); I am getting this error... Quote Warning: mysql_insert_id() [function.mysql-insert-id]: A link to the server could not be established in /Users/user1/Documents/DEV/++htdocs/05_Debbie/members/create_account.php on line 269 What am I doing wrong? Debbie I have two tables. The first with an auto-increment field of id and the second with a article_id field. I want to get the value of the auto-increment field in the first table and insert it into the article_id field of the second table (so they match). I am using the mysql_insert_id() command for the first time and I am wondering if I can run a query like this and turn it into a variable or if I need to use a SELECT query from the the mysql_insert_id() field from the first table before inserting it into the second table? Any feedback is appreciated. Thanks, kaiman Here is what I have so far (untested): // insert data into blog database $sql1="INSERT INTO $tbl_name1(author, title, content, date)VALUES('$author', '$title', '$content', NOW()) LIMIT 1"; $result1=mysql_query($sql1) or trigger_error("A mysql error has occurred!"); $article_id = mysql_insert_id (); // if successfully inserted data into database, redirect user if($result1){ header( "Location: http://www.mydomain.com/blog/add/success/" ); } else { header( "Location: http://www.mydomain.com/blog/add/error/" ); exit; } // insert data into blog categories database $sql2="INSERT INTO $tbl_name2(article_id, category)VALUES('$article_id', '$category' LIMIT 1"; $result2=mysql_query($sql2) or trigger_error("A mysql error has occurred!"); // if successfully inserted data into database, redirect user if($result2){ header( "Location: http://www.mydomain.com/blog/add/success/" ); } else { header( "Location: http://www.mydomain.com/blog/add/error/" ); exit; } I get mysql_insert_id problem. It returns 0. I do not know how to fix it. Please tell me. Thank you very much. <?php session_id(); session_start();?> <!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=iso-8859-1" /> <title>Untitled Document</title> </head> <body> <?php include("connection.php"); $subtotal=$_POST['subtotal']; $tax=$_POST['tax']; $total=$_POST['total']; $today=date("Y-m-d"); $email=$_SESSION['email']; $od=mysql_insert_id(); $number=$_POST['number']; $name=$_POST['name']; $month=$_POST['month']; $year=$_POST['year']; $code=$_POST['code']; $method=$_POST['type']; $add1="select * from customer where email='$email'"; $add2=mysql_query($add1); $add3=mysql_fetch_array($add2); extract($add3); $qiu1="INSERT INTO test (oid) VALUES ('$email')"; $qiuzhen=mysql_query($qiu1); $orderid=mysql_insert_id(); $query44="INSERT INTO income (subtotal,tax,total,time,email,address,credit,name,month,year,code,method,custid) VALUES ('$subtotal','$tax','$total','$today','$email','$add1',$number,'$name','$month','$year','$code','$method','$orderid')"; $result=mysql_query($query44); echo "successful!"; $orderid=mysql_insert_id(); echo "$email"; echo "$total"; $sessid=session_id(); $qu="select * from carttemp where sess='$sessid'"; $result=mysql_query($qu); while($w=mysql_fetch_array($result)){ extract($w); $query7="INSERT INTO try (prodnum,quan,custnum) VALUES ('$prodnum','$quan','$custid')"; $iii=mysql_query($query7) or (mysql_error()); } ?> </body> </html> Hi guys Long time no post. Firstly is this possible, and can anyone point me in the right direction. I have two tables as below and I want to display only the results that are active against the master table. Table1 - Awards awardid award code active 1 Apple AFC 1 2 Orange OOC 1 3 Lemon LSO 1 Table 2 - Log userid AFC OOC LSO 1 1 0 0 2 1 1 1 The problem i'm facing is I can get the results from table 1 where active = 1 which is great and then return each row. however I only want to return the row where the user has that award for example user 1 only has the AFC. Any guidance as always appreciated. Many thanks If I am running a query like this: INSERT INTO `cam_locations` (`id`, `datacenter`, `address1`, `address2`, `city`, `state`, `zip`, `country`, `phone`) VALUES (1, 'Austin Data Center', '', '', 'Austin', 'Texas', '', 'United States', ''), (2, 'Sunnyvale Date Center', '', '', 'Sunnvale', 'California', '', 'United States', ''), (4, 'BoxBorough Data Center', '', '', 'Boston', 'Massachusetts', '', 'United States', '') It it possible to use mysql_insert_id() and get all the id's inserted? Maybe like an array of them or something? Thanks after inserting data i get 0 sometimes i created a mysql_connect for it Code: [Select] <?php $maincon = mysql_connect('local', "root", "root"); mysql_select_db("root"); then //insert and $success = mysql_query($query); echo $ref_id = mysql_insert_id($maincon); ?> $ref_id is returned as 0, although the connection is there var_dumping the query returns the values This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=354388.0 This portion is kind of stumping me. Basically, I have a two tables in this DB: users and users_access_level (Separated for DB normalization) users: id / username / password / realname / access_level users_access_level: access_level / access_name What I'm trying to do, is echo the data onto an HTML table that displays users.username in one table data and then uses the users.access_level to find users_access_level.access_name and echo into the following table data, I would prefer not to use multiple queries if possible or nested queries. Example row for users: 1234 / tmac / password / tmac / 99 Example row for users_access_level: 99 / Admin Using the examples above, I would want the output to appear as such: Username: Access Name: Tmac Admin I am not 100% sure where to start with this, but I pick up quickly, I just need a nudge in the right direction. The code I attempted to create just shows my lack of knowledge of joining tables, but I'll post it if you want to see that I did at least make an effort to code this myself. Thanks for reading! Im doing a search system and Im having some problems.
I need to search in two tables (news and pages), I already had sucess doing my search system for just one table, but for two tables its not easy to do.
I already use a select statment with two tables using UNION because I want to show number of search results, that is number of returned rows of my first sql statment.
But now I need to do a select statment that allows me to acess all fields of my news table and all fields of my pages table.
I need to acess in my news table this fields: id, title, content, link, date, nViews
I need to acess in my pages table this fields: id, title, content, link
Im trying to do this also with UNION, but in this case Im not having any row returning.
Do you see what I have wrong in my code?
<?php //first I get my $search keyword $search = $url[1]; $pdo = connecting(); //then I want to show number of returned rows for keyword searched $readALL = $pdo->prepare("SELECT title,content FROM news WHERE title LIKE ? OR content LIKE ? UNION SELECT title,content FROM pages WHERE title LIKE ? OR content like ?"); $readALL->bindValue(1,"%$search%", PDO::PARAM_STR); $readALL->bindValue(2,"%$search%", PDO::PARAM_STR); $readALL->bindValue(3,"%$search%", PDO::PARAM_STR); $readALL->bindValue(4,"%$search%", PDO::PARAM_STR); $readALL->execute(); //I show number of returned rows echo '<p>Your search keyword returned <strong>'.$readALL->rowCount().'</strong> results!</p>'; //If dont return any rows I show a error message if($readALL->rowCount() <=0){ echo 'Sorry but we didnt found any result for your keyword search.'; } else{ //If return rows I want to show, if it is a page result I want to show title and link that I have in my page table //if it is a news result I want to show title and link that I have in my news table and also date of news echo '<ul class="searchlist">'; $readALL2 = $pdo->prepare("SELECT * FROM news WHERE status = ? AND title LIKE ? OR content LIKE ? LIMIT 0,4 UNION SELECT * FROM pages where title LIKE ? OR content LIKE ? LIMIT 0,4"); $readALL2->bindValue(1, '1'); $readALL2->bindValue(2, "%$search%", PDO::PARAM_STR); $readALL2->bindValue(3, "%$search%", PDO::PARAM_STR); $readALL2->bindValue(4, "%$search%", PDO::PARAM_STR); $readALL2->execute(); while ($result = $readALL2->fetch(PDO::FETCH_ASSOC)){ echo '<li>'; echo '<img src="'.BASE.'/uploads/news/'.$result['thumb'].'"/>'; echo '<a href="'.BASE.'/news/'.$result['id_news'].'">'.$result['title'].'</a>'; //if it is a news result I also want to show date on my list //echo '<span id="date">'.$result['date'].'</span>'; echo '</li>'; } echo ' </ul>'; //but how can I do my select statement to have access to my news table fields and my page table fields?? } ?> I am trying to write some data from multiple SQL tables to a page. In the first table is a list of places. I then have more tables that are named after the different places. For example, say my first place in the list is called Place1. I have a table named Place1 with data that corresponds to place1. The data contained in the table named Place1 is a list of things to do in this place. It has 21 columns and each one is something to do in the morning, afternoon, and at night for each day of the week in the place Place1. What I am trying to do is display a sort of weekly calendar as a table on a webpage that lists all of the places in one column and then lists seven days of the week as 7 more columns. Then in each data cell I would want to list the things to do in the morning, afternoon and at night for the certain day of the week and for the place. The problem is that I am creating a CMS to allow other users with no coding knowledge to update events for other places, so I have to display data that could have been altered. The only solution I know of is to do a while loop that gets all of the place names and while there are still place names, write the place names to the page and set a variable equal to the place name. Inside the while loop I would create another while loop that each time the first while loop is executed uses the variable set in the first while loop to know which table to reference and then make a call to that table pulling out the 21 columns and writing them to the page. Each time the outer while loop executes, it would (hopefully) write the place name, and then set the variable as the current place name so that the inner while loop uses the variable to write the rest of the information about that place. I don't know if that would even work and if it did, I know it cannot be the best way to do this. I am pretty stuck here and don't really have a good solution so if anyone proposes a solution that is radically different to anything I have done, I am open to everything. Thank you! Hi I've got this database I created with fields ProductId ProductName Image I've managed to get it to list the ID,productname, and Image urls in a list. My next step is to have the image field actually display an image and make it clickable: heres what I've done so far: Code: [Select] <?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("productfeed", $con); $result = mysql_query("SELECT * FROM productfeeds"); echo "<table border='0'> <tr> <th>Firstname</th> <th>Lastname</th> <th>Image</th> </tr>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['ProductID'] . "</td>"; echo "<td>" . $row['ProductName'] . "</td>"; echo "<td>" . $row['ImageURL'] . "</td>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> Heres what I want to do: Code: [Select] while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['ProductID'] . "</td>"; echo "<td>" . $row['ProductName'] . "</td>"; // my changes beneath echo "<td>" . <a href="<?php echo $row['ImageURL'];?>"> <img src="<?php echo $row['LinkURL']; ?>"> </a>. "</td>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> Can you guys point me in the right direction? Many thanks |
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