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PHP - Is This The Best Way To Use Mysql_insert_id() Or Do I Need To Select It?
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; } Similar TutorialsHello 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; 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 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 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 $last_id) but no success. (Probably because i don't know exactly how to use it) Hope everything's clear. What i'm missing ? Thanks in advance for your help. Hello there, well basically what this function does is look at all the entry's to find a unused port in the database but when I run the function the first time if the table is empty mysql_insert_id works fine, but then if I run it again the mysql_insert_id keeps returning 0, I am using it immediately after the query but its not working properly, can anybody shed some ideas into the matter, anything is appreciated!, thanks for your time!: function addServerToBuildPool($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? 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 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 '$result'; ?> All the other facets of the script work but I get the following error message with the above script: Fatal error: Can't use function return value in write context in C:\x\xampp\htdocs\form_process.php on line 27 But it's not so much the error message its the ultimate objective. Any help appreciated. hirealimo.com.au/code1.php this works as i want it: Quote SELECT * FROM price INNER JOIN vehicle USING (vehicleID) WHERE vehicle.passengers >= 1 AND price.townID = 1 AND price.eventID = 1 but apparelty selecting * is not a good thing???? but if I do this: Quote SELECT priceID, price FROM price INNER JOIN vehicle....etc it works but i lose the info from the vehicle table. but how do i make this work: Quote SELECT priceID, price, type, description, passengers FROM price INNER JOIN vehicle....etc so that i am specifiying which colums from which tables to query?? thanks I have 2 queries that I want to join together to make one row
Dear All, I wish to have 2 drop down boxes, Country Select Box and Locality Select Box. The locality select box will be affected by the value chosen in the country select box. All is working fine except that the locality select box is not being populated. I know that the problem is in the sql statement WHERE country_id='$co' because i am having an error that $co is an undefined variable. All the rest works fine because i have replaced the $co variable directly with a number (say 98) for a particular country id and it worked fine. In what way can i define this variable $co so that it is accepted by my sql statement? Thank you for your help in advance. MySQL Tables indicated below: CREATE TABLE countries( country_id INT(3) UNSIGNED NOT NULL AUTO_INCREMENT, country_name VARCHAR(30) NOT NULL, PRIMARY KEY(country_id), UNIQUE KEY(country_name), INDEX(country_id), INDEX(country_name)) ENGINE=MyISAM; CREATE TABLE localities( locality_id INT(10) UNSIGNED NOT NULL AUTO_INCREMENT, country_id INT(3) UNSIGNED NOT NULL, locality_name VARCHAR(50), PRIMARY KEY (locality_id), INDEX (country_id), INDEX (locality_name)) ENGINE=MyISAM; Extract PHP script included below: // connect to database require_once(MYSQL); if(isset($_POST['submitted'])) { // trim the incoming data /* this line runs every element in $_POST through the trim() function, and assigns the returned result to the new $trimmed array */ $trimmed=array_map('trim',$_POST); // clean the data $co=mysqli_real_escape_string($dbc,$trimmed['country']); $lc=mysqli_real_escape_string($dbc,$trimmed['locality']); } ?> <form action="form.php" method="post"> <p>Country <select name="country"> <option>Select Country</option> <?php $q="SELECT country_id, country_name FROM countries"; $r=mysqli_query($dbc,$q) or trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc)); while($row=mysqli_fetch_array($r)) { $country_id=$row[0]; $country_name=$row[1]; echo '<option value="' . $country_id . '"'; if(isset($trimmed['country']) && ($trimmed['country']==$country_id)) echo 'selected="selected"'; echo '>' . $country_name . '</option>\n'; } ?> </select> </p> <p>Locality <select name="locality"> <option>Select Locality</option> <?php $ql="SELECT locality_id, country_id, locality_name FROM localities WHERE country_id='$co' ORDER BY locality_name"; $rl=mysqli_query($dbc,$ql) or trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc)); while($row=mysqli_fetch_array($rl)) { $locality_id=$row[0]; $country_id=$row[1]; $locality_name=$row[2]; echo '<option value="' . $locality_id . '"'; if(isset($trimmed['locality']) && ($trimmed['locality']==$locality_id)) echo 'selected="selected"'; echo '>' . $locality_name . '</option>\n'; } // close database connection mysqli_close($dbc); ?> </select> </p> <p><input type="submit" name="submit" value="Submit" /></p> <input type="hidden" name="submitted" value="TRUE" /> </form> I'm using a PHP helpdesk script, but when a customer submits a ticket, they are able to assign a category. I want to be able to have the category function so I can move tickets around, but only allow customers to assign to one. (Basically I want to hide the Select Box on the new ticket page) This is the code which displays the select box: Quote <td style="text-align:right" width="150"><?php echo $hesklang['category']; ?>: <font class="important">*</font></td> <td width="80%"><select name="category"> Can you tell me what I can do to "hide" the category select box without getting rid of it? (The page requires a variable for category) Thanks Hi guys, I'm trying to do make this code so that IF a user owns a property (in this code a bulletfactory) then the BF CP Shows up... Here's the code so far..... Code: [Select] <?php session_start(); include_once"includes/db_connect.php"; include_once"includes/functions.php"; logincheck(); $username=$_SESSION['username']; $query=mysql_query("SELECT * FROM users WHERE username='$username'"); $fetch=mysql_fetch_object($query); $query_bf=mysql_query("SELECT * FROM bf WHERE location='England, Japan, Colombia, USA, Russia, Italy, Turkey'"); $fetch_bf=mysql_fetch_object($query_bf); if (strtolower($fetch_bf->owner) == (strtolower($fetch->username))){ require_once"bulletCP.php"; exit(); } ?> hi. just started a website talkietaco.com and at the momment my code selects the first row and displays it. this is all well and good but when i add a new row to the table it gos to the bottom. I want to be able to select the last row and echo it out. Any ideas? would i need to add an id row or something? Heres the code and thank you in advance for any help people can offer. Code: [Select] <?php $con = mysql_connect("localhost","",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("mainbase", $con); $result = mysql_query("SELECT * FROM matteroffact"); echo "<table border='0'> <tr> <th></th> </tr>"; $row = mysql_fetch_array($result) or die(mysql_error()); echo "<tr>"; echo "<td><strong>" . $row['question']. "</strong> ". $row['answer']; "</td>"; echo "<tr>"; echo "</table>"; ?> Hi, I'm trying to make a 3 page registration form. On the 3rd page I want it to echo back all the options they have selected. I can get the input type to work but having problems with the selects. First Page Code: [Select] <form action="choose.password.php" method="post"> eMail: <input type="text" name="eMail" /><br /> Forename: <input type="text" name="fname" /><br /> Surname: <input type="text" name="lname" /><br /> Date of Birth: <input type="text" name="dob" /><br /> Gender: <select name="gender"> <option value="male">Male</option> <option value="female">Female</option> </select> <br /> I am not a lawyer: <input type="checkbox" name="lawyer" value="lawyer"> <br /> <br /> <input type="submit" value="Next" /> </form> Second Page Code: [Select] <form action="confirm.details.php" method="post"> <p>email address: <?php echo $_POST["eMail"]; ?></p> <p>password: <input type="password" name="eMail" /></p> <p>Confirm Password: <input type="password" name="eMail" /></p> <p>Hint: <input type="text" name="hint" /></p> <p>My email is correct, I am happy to be sent occassional emails, and i am not a family lawyer. <input type="checkbox" name="lawyer" value="lawyer"></p> <input type="hidden" name="fname" value="<?php echo ($_POST["fname"]); ?>" /> <input type="hidden" name="eMail" value="<?php echo ($_POST["eMail"]); ?>" /> <input type="hidden" name="lname" value="<?php echo ($_POST["lname"]); ?>" /> <input type="hidden" name="dob" value="<?php echo ($_POST["dob"]); ?>" /> <select style="display:none;" name="gender" <?php echo ($_POST["gender"]); ?> /> <p><input type="submit" value="Next" /></p> Third page Code: [Select] <p>Please confirm your details are correct to register.</p> <p>Email Address: <?php echo $_POST["eMail"]; ?> <p>Forename: <?php echo $_POST["fname"]; ?></p> <p>Surname: <?php echo $_POST["lname"]; ?></p> <p>Date of Birth: <?php echo $_POST["dob"]; ?></p> <p>Gender: <?php echo $_POST["gender"]; ?></p> How can I make the select work and echo it back on the third page? Thanks Hi, what I want to figure out is for instance a person has registered with their country e.g. England. Now if I echo the country in a select box giving the person an option to change their country and Showing the person which currently they have selected already. The select box shows two England`s to select from. Could some one tell me how can I have one of each country and echo their already selected country from the database. I don't know how to explain any better what I am after but just basically there are two Englands showing one which is already selected (echoing from the mysql database) and one is already in the select box. Any help is much appreciated thank you. ok here's my problem $Sql1 returns two values: 8 and 10 and these numbers get put into a <select>. So far so good. I assign a onchange to it. When i select 8 it makes the changes but when I select 10 nothing happends. I preciate some help. Code: [Select] <?php require("status.php"); require("id.php"); $Link = mysql_connect($Host, $User, $Password); mysql_select_db('sportsportal', $Link); $Sql1 = "select distinct week, (select max(week) from coupons where user='$User') max from coupons where user='$User'"; $Result1 = mysql_query($Sql1, $Link); print "<tr>"; print "<td align=left valign=top> </td>"; print "<td align=left valign=top>"; print "<select name=current_week onchange=window.location='coupon.php?curwk='+this.value>"; while($Row1 = mysql_fetch_array($Result1)){ //if($Row1[week] == $Row1[max]){ //print "<option value='$Row1[week]' selected>Vecka $Row1[week]</option>"; //} else { //print "<option value='$Row1[week]'>Vecka $Row1[week]</option>"; //} print "<option value='$Row1[week]'>Vecka $Row1[week]</option>"; } print "</select>"; print "<p></td>"; print "</tr>"; if(isset($_REQUEST['curwk'])){ $Curwk = $_REQUEST['curwk']; } else { $Curwk = 0; } $Sql = "select home, away, home_score, away_score, winner from coupons where user='$User' and week='$Curwk'"; $Result = mysql_query($Sql, $Link) or die(mysql_error()); while($Row = mysql_fetch_array($Result)){ if(@$Row[home] == @$Row[winner]){ print "<tr>"; print "<td align=left valign=top> </td>"; print "<td align=left valign=top><b>$Row[home]</b> - $Row[away] $Row[home_score]-$Row[away_score]</td>"; print "</tr>"; } else { print "<tr>"; print "<td align=left valign=top> </td>"; print "<td align=left valign=top>$Row[home] - <b>$Row[away]</b> $Row[home_score]-$Row[away_score]</td>"; print "</tr>"; } } mysql_close($Link); ?> hi guys ive just finished this task after hours of head scratching since svg is only really supported good in firefox and opera ive chosen firefox as my browser to view this url www.deansignori.com/phpsvgpie/index.php i do need more help with this task but a different problem (creating select box to call different stylesheet and to change from 2d - 3d i have the code set out so that i can explode any segment or change size of slices or change from 2d-3d but i have to do this manually in the code to render different piecharts im wanting to use 1 but change it using a select box and echo my variable into it basically im unsure of the syntax for this problem psuedo code for style maybe something like if select box value isset onchange stylecolour echo stylecolour if select box value isset onchange stylegrey echo stylegrey and for 3d-2d if select box value isset onchnage format3 echo format3 if select box value isset onchange format2 echo format2 this would be on my index page can anyone advise me please regards Dean |
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