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PHP - Problems With Mysql_insert_id
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()); } Similar TutorialsHow 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 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 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($result2){ header( "Location: http://www.mydomain.com/blog/add/success/" ); } else { header( "Location: http://www.mydomain.com/blog/add/error/" ); exit; } Hi all, i'm a bit more than a newbie. Here's the scenario. I have a page with tabbed navigation. On each tab there's a form (diving centers, teachers, and so on). Each form has a submit button with a unique name. Every submit button is processed by a series of if statements in an external included php file. As you can see in my code below what the file does is to process the if statements based on the submit pressed. It' important to notice that a JS script checks that the first form is filled with all the infos. If not is not possible to proceed to complete all the others. On the first form the user can register general information. Obviously there's an Id field (auto increment) that i grab with mysql_nsert_id. Also obvious is the fact that, for query reasons, i want to store the grabbed id in an id field present in all the tables of my database. The tables get the data from the various forms displayed on the tabbed navigation. Here's the code of the included file (notice that i use to start with a simple coding to test everything 's working fine) <?php require_once('Connections/Scubadiving.php'); if (!empty($_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. 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> 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? 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 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. class curl2{ private $curl_init; private $CURLOPT_URL; public function connect(){ $this->curl_init = curl_init(); } public function debug(){ curl_setopt($this->curl_init, CURLOPT_VERBOSE, TRUE); $fp = fopen("curl2.txt", "w"); curl_setopt($this->curl_init, CURLOPT_STDERR, $fp); curl_setopt($this->curl_init, CURLOPT_RETURNTRANSFER, TRUE); } public function setUrl($url = null){ $this->CURLOPT_URL = $url; curl_setopt($this->curl_init, CURLOPT_URL, $this->CURLOPT_URL); } public function execute(){ $out = curl_exec($this->curl_init); curl_close($this->curl_init); return $out; } } $curl2 = new curl2; $curl2->connect(); $curl2->setUrl("http://www.linuxformat.co.uk"); $curl2->debug(); echo $curl2->execute(); It display a blank page like attachment result1.jpg, but if I move the $fp = fopen("curl2.txt", "w"); curl_setopt($this->curl_init, CURLOPT_STDERR, $fp); curl_setopt($this->curl_init, CURLOPT_RETURNTRANSFER, TRUE); from function debug() and join it with function execute() like this: public function execute(){ $fp = fopen("curl2.txt", "w"); curl_setopt($this->curl_init, CURLOPT_STDERR, $fp); curl_setopt($this->curl_init, CURLOPT_RETURNTRANSFER, TRUE); $out = curl_exec($this->curl_init); curl_close($this->curl_init); return $out; } it return me Linuxformat content ( expected result ) like result2.jpg below is the working code : class curl2{ private $curl_init; private $CURLOPT_URL; public function connect(){ $this->curl_init = curl_init(); } public function debug(){ curl_setopt($this->curl_init, CURLOPT_VERBOSE, TRUE); } public function setUrl($url = null){ $this->CURLOPT_URL = $url; curl_setopt($this->curl_init, CURLOPT_URL, $this->CURLOPT_URL); } public function execute(){ $fp = fopen("curl2.txt", "w"); curl_setopt($this->curl_init, CURLOPT_STDERR, $fp); curl_setopt($this->curl_init, CURLOPT_RETURNTRANSFER, TRUE); $out = curl_exec($this->curl_init); curl_close($this->curl_init); return $out; } } $curl2 = new curl2; $curl2->connect(); $curl2->setUrl("http://www.linuxformat.co.uk"); $curl2->debug(); echo $curl2->execute(); Why I couldn't split "CURLOPT_STDERR, CURLOPT_RETURNTRANSFER" with "curl_exec" Hi I have plans in developing a connect function for remote login to my web side. I can't find any useful on Google. Some idees on how to code a API connect button? Something similiar to Facebook connect, Twitter connect etc. BUT this should not rely on facebook api. I'm going to make my own stand alone api. I know I need to use REST in backend, but I'm missing the knowledge to know how to send / recive the login data, and how to know when a user are online or not. I'm greatfull for any help. Also links. And also tips on how to make a developer plattform for apps, much like the way Facebook have it. Hello, Alright I am having a few issues here. Let me explain what I am doing...this is a script that allows a customer to split an order..he can split certain items out of one order and it will create another order with the items he split added to this order, and then subtracting the items from the old order. The products are stored in one field called products, they are stored in a delimited list like this qty, product name, price each, product id | qty, product name, price each, product id |etc. Now, before you say why are you storing your products like that?? The reason I have to store my products like so is because the customer does not want the products stored in the order table to be dependent off the products table like they were before. In that case if he modified a product it would modify all of the orders with that product stored in them. Make sense? I am aware a delimited list probably wasn't the best way to do this but it was the best way I could think of. Ok here is my list of issues. 1) Everytime you split an order, it is not creating a new order to store the items that were split into. It created an order the first time I did it, and now all it does is it keeps updating the products field in that order, rather than creating a new order. 2) Next I do not know how I can update the old products (stored in the $prod array). It needs to update the products qtys after the order was split, and then if all of the qty for that product were moved to the new order it just needs to remove that product from the $prod array rather than updating the quantity. The code is below. Please let me know if I can provide any more information to help! Thanks everyone! // Get Old Order $get_order = @mysql_query("SELECT * FROM orders WHERE order_id = {$_POST['order_id']}"); $order = @mysql_fetch_assoc($get_order); // Get Old Order Items $products = $order['products']; //breaking products text down for display $prod = array(); $_products = explode('|', $products); foreach ($_products AS $p) $prod[] = explode(',', $p); /* $get_items = @mysql_query("SELECT product_id, qty FROM order_items WHERE order_id = {$order['order_id']}"); $items = array(); while(($row = @mysql_fetch_assoc($get_items)) !== false) { $items[] = $row; } */ if(empty($prod)) { header("Location: tracking.php"); die(); } // Create New Order mysql_query("INSERT INTO orders SET customer_id = {$order['customer_id']}, order_status = {$order['order_status']}, order_date = '{$order['order_date']}', order_date_paid = '{$order['order_date_paid']}', order_shipping = '{$order['order_shipping']}', order_shipping_fee = '{$order['order_shipping_fee']}', order_insurance = '{$order['order_insurance']}', order_insurance_fee = '{$order['order_insurance_fee']}', order_insurance_total = '{$order['order_insurance_total']}', order_grand_total = '{$order['order_grand_total']}', order_date = '{$order['order_date']}', order_filled = '{$order['order_filled']}', order_ship_date = '{$order['ship_date']}'"); $get_new_order = @mysql_query("SELECT MAX(order_id) AS order_id FROM orders") or die(mysql_error()); $new_order_id = @mysql_result($get_new_order, 'order_id', 0); // Add Items to New Order & Remove Items from Old Order $new_items = array(); $_new_items = ''; foreach($prod as $p2) { for($i = 0; $i < $p2[0]; $i++) { if(!empty($_POST[trim($p2[3]).'_'.$i])) { $new_items[trim($p2[3])]++; } } } //construct new static products list foreach($new_items as $id=>$qty) { $get_product = mysql_query("SELECT name, price FROM products WHERE product_id = '{$id}'"); $got_product = mysql_fetch_assoc($get_product); $_new_items .= $qty.','.$got_product['name'].','.$got_product['price'].','.$id.'|'; //echo $id.' - '.$qty.'<br>'; } //remove last character in products text before going into DB $_new_items = substr_replace($_new_items ,"",-1); //update products field in new order mysql_query("UPDATE orders SET products = '{$_new_items}' WHERE order_id = '{$new_order_id}'"); the image in my comments box wont show up it gives me an error Code: [Select] Notice: Undefined index: avatar in /home/ecabrera/public_html/profile.php on line 277 and i dont whats wrong with it Code: [Select] // display comments $perpage = 10; $start=0; if(@$_GET['s']) $start = $_GET['s']; $query = mysql_query("SELECT * FROM profile_comments WHERE profile_id='$getid' ORDER BY id DESC LIMIT $start, $perpage"); $numrows = mysql_num_rows($query); if ($numrows > 0){ $next = $start + $perpage; $prev = $start - $perpage; while($row = mysql_fetch_assoc($query)){ $user_id = $row['user_id']; $user_name = $row['user_name']; $comment = nl2br($row['comment']); $date = $row['date']; $avatar = $row['avatar']; echo "<img src='avatars/$avatar'></img><a href='$site/profile?id=$getid'></a> <b> on $date</b><br />"; echo "<div style='margin-left: 10px;'>$comment</div><hr>"; } } else echo "This user has no profile comments.<br />"; // end diplay comment area I'm trying to either update my database if some data matches the if statement, or insert the data in a new row if it doesn't match (if possible). The code will update the database just fine if the data matches, but if not, it won't insert the new data. (When I leave off the if statement it will insert just fine). I fear I'm not using the if...else correctly. Thanks for any help. <?php header("Location: admin_schedule.php"); include("opendatabase.php"); $date=("$_POST[gamedate]"); $week=("$_POST[Week]"); $game=("$_POST[Game]"); $hometeam=("$_POST[team_name_1]"); $awayteam=("$_POST[team_name_2]"); $check = mysql_query("SELECT week_id,game_id FROM schedule"); $row[] = mysql_fetch_array($check); if ($row[week_id]='$week' && $row[game_id]='$game') { mysql_query("UPDATE schedule SET week_id = '$week', game_id = '$game', date = '$date', H_team = '$hometeam', A_team = '$awayteam' WHERE week_id = '$week' AND game_id = '$game'"); } else { mysql_query("INSERT INTO schedule (week_id,game_id,date,H_team,A_team) VALUES('$week','$game','$date','$hometeam','$awayteam')"); } mysql_close($con); Hi, I hope someone can help me. I am trying to generate a script for our Intranet that tells me which birthdays = today. My field in my table = d/m/y, eg 09/05/2011 (9 May 2011). How do I get the script to just look at the day & month, not the year? My script looks like this: Code: [Select] <?php echo date("d/m/Y") . "<br />"; $myDate = date('d/m/Y'); $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("dbname", $con); $sql = "SELECT * FROM detail WHERE dob='$mydate'"; $result=mysql_query($sql); echo mysql_num_rows($result); ?> I want to gettwitter feeds, this gives me a xml http://twitter.com/statuses/user_timeline.rss?screen_name=fleuragemapvv&count=3 based on this: http://www.w3schools.com/PHP/php_xml_simplexml.asp i try to get the xml with: <?php$xml = simplexml_load_file("http://twitter.com/statuses/user_timeline.rss?screen_name=fleuragemapvv&count=3");?>() i get this error: Quote Warning: simplexml_load_file(http://twitter.com/statuses/user_timeline.rss?screen_name=fleuragemapvv&count=3): failed to open stream: HTTP request failed! HTTP/1.1 400 Bad Request in /home/vhosting/x/vhost0033377/domains/doekewartena.nl/htdocs/www/temp/twitter03.php on line 2 Warning: simplexml_load_file(): I/O warning : failed to load external entity "http://twitter.com/statuses/user_timeline.rss?screen_name=fleuragemapvv&count=3" in /home/vhosting/x/vhost0033377/domains/doekewartena.nl/htdocs/www/temp/twitter03.php on line 2 i have no clue, i'm very new to php can i even use simplexml_load_file cause it has no .xml file type extention. First Since I first started learning PHP, I've been writing code for forms like this: $name = $_POST["name"]; if ($name == "") { echo "You didn't type anything"; } I've written scripts like this and tested them on a GOdaddy server, and they always worked fine. Now, after testing scripts like that on my own server (WAMP2 [PHP5, Mysql, Apache]), I get an error saying "Unidentified index: name", and so I have to nest that if statement AS WELL AS the variable assignment inside another if statement like this: if (isset ($_POST["name"])) { $name = $_POST["name"]; if ($name == "") { echo "you didn't type anything. } } The error doesn't actually stop the script from running, but a medium sized box appears in the browser with the error message contained that's after the element in the document of which is getting an invalid error. Which way is truly the correct way? Because now that I have to use the second way in order for my scripts to run correctly on my own server, I have to write more code and it all gets disorganized and more complicated to read. So...is this a configuration thing that I can tweak in WAMP2 to stop messages like that from popping up? Or is the second way the way you're supposed to do it? Also, I was wondering why ereg got deprecated and what the replacement of it is in PHP5? I've asked before and even googled it but I simply get taken to a "Pearl" function on PHP.net. I'm not sure what it is, or how to use it, and haven't got any hints from either PHP.net or anywhere else. Any suggestions? why cant i login in i have everything in my db this is the code Code: [Select] <?php session_start(); $email= $_SESSION['email']; ?> <?php if(loginbtn){ $email = $_POST['email']; $password = $_POST['password']; if($email && $password){ //connection to db require 'scripts/connect.php'; $query = mysql_query("SELECT * FROM users WHERE email='$email'"); $numrows = mysql_num_rows($query); if($numrows == 0){ $rows = mysql_fetch_assoc($query); $dbemail = $row['email']; $dbpassword = $row['password']; if($password==$dbpassword){ $_SESSION['email'] = $dbemail; echo "You have been logged in as $email"."<br/>"; }else echo "You enter a incorrect password"; }else echo "This person does not exsit"; }else echo "You did not fill in all the fields"; }else echo "$form"; $form = "<form action='login.php' method='POST'> <table> <tr> <td>Email</td> <td><input type='email' name='email'></td> </tr> <tr> <td>Password</td> <td><input type='password' name='password'></td> </tr> <tr> <td></td> <td><input type='submit' name='loginbtn' value='Login'></td> </tr> </table> </form>"; echo $form; ?> It seems like every new language I learn, I get this exact same issue. I want to make a clickable link that will go to "viewpost?postid=" and then add the value of a variable. The variable is an array. Here's what I'm trying. I'm not really sure why it doesn't work: <?php $poststr = "<a href=\'viewpost.php?"+$postid[$num]+"\'>View This Post</a>"; echo $poststr; ?> $postid[$num] in this example should be an integer. But what it outputs is really weird. It's not "View This Post" as a link, which is what I want. It just outputs the value of the variable $postid[$num] with no link. Can someone tell me what my problem is? |
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