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PHP - Please Help With Double $_post Function
Hello,
its a bit difficult to explain the issue that I have but I will try my best. I have this small code $message = $_POST['message1']; in my php code and message1 is the name of a textarea in my page. i also have another textarea called message2 in my page. what I need to do is to use the message1 and message2 in the $message = $_POST function. For example: $message = $_POST['message1'] + $_POST['message2'] or anything that is acceptable in PHP coding and works in PHP. I hope I havent confused you guys and you undersstood what I mean... Thanks Similar TutorialsHi, I was going to through a PHP tutorial, and came across a form validation user defined function . Below is an example.. Code: [Select] function check_required_fields($fields){ $field_errors = array(); foreach ($fields as $requried_fields) { if (!isset($_POST[$requried_fields]) || (empty($_POST[$requried_fields]) && $_POST[$requried_fields] != 0)) if (!isset($_POST[$fieldname]) || (empty($_POST[$fieldname]) && $_POST[$fieldname] != 0)) { $errors[] = $requried_fields; } } return $field_errors; } When the user defined function was called they added an $_POST and the end of the user defined function. Here is a the code snipped of how it looks Code: [Select] $errors = array(); // perform validations on the form data $required_fields = array('username', 'password'); $errors = array_merge($errors, check_required_fields($required_fields, $_POST)); If the validation can be preformed with out the $_POST, why bother inserting it? Thanks! Hello everyone, Can someone show me a way how to pass $_post variable from a form to a function? So, input username and input password to a function login($username, $password). I have a basic form for collecting data, I have a function for collecting the ip address of the visitors unfortunately I cannot get the ip collection working with the rest of the form once I put the if($_POST['emailaddress']) {.........} in. My aim is once the user has completed the form, data gone to MySQL database that the data will then be printed via the echo statements on to the form as the action sends it back to the same page. Currently my code is <?php /* This script is a form handler, each section is commented as to what it does. 1. connects to the database (see connect.php). 2. it gets the users ip address. 3. Strips tags from the data entered so that no malicious code can be entered and corrupt/cause problems with the database or site. 4. Inserts the relevant data into the database in this case it is the ip, haveemail, emailaddress, browser, otherbrowser, resolution, otherresolution. 5. Sends the data just entered by the user back to the screen so they can see what they entered */ // 1. connection to MySQL require ("php/connect.php"); // if this script is unavailable then the rest of the code is pointless as need a connection to the database. // if fields are completed if($_POST['emailaddress']) { // if this field has had data entered then process the data // 2. collect ip address //$ip = getRealIpAddr(); // 3. Strips tags and POST variables from the form $haveemail = strip_tags($_POST['haveemail']); $emailaddress = strip_tags($_POST['emailaddress']); $browser = strip_tags($_POST['browser']); $otherbrowser = strip_tags($_POST['otherbrowser']); $resolution = strip_tags($_POST['resolution']); $otherresolution = strip_tags($_POST['otherresolution']); // 4. insert data to dbase $query="INSERT INTO datacollection1 (id, ip, haveemail, emailaddress, browser, otherbrowser, resolution, otherresolution) VALUES ('Null', '$ip', '$haveemail', '$emailaddress', '$browser', '$otherbrowser', '$resolution', '$otherresolution')"; // Null is in the id field as this is added automatically in the database as it is set to auto increment upon an entry going in and is primary key. // message to say if database has been updated mysql_query($query) or die (mysql_error()); //echo "<b>Your IP address is: $ip</b> <br />"; /*function getRealIpAddr() { if (!empty($_SERVER['HTTP_CLIENT_IP'])) //check ip from share internet { $ip=$_SERVER['HTTP_CLIENT_IP']; } elseif (!empty($_SERVER['HTTP_X_FORWARDED_FOR'])) //to check ip is pass from proxy { $ip=$_SERVER['HTTP_X_FORWARDED_FOR']; } else { $ip=$_SERVER['REMOTE_ADDR']; } return $ip; } */ // 5. displays information on form to advise what the user just entered echo '<br/>'."The database has just been updated with the following information: ".'<br/><br/>'; // echo "Your ip address is ".$ip.'<br/>'; echo "You answered ".$haveemail." to having an email address.".'<br/>'; echo "Your email address is ".$emailaddress.'<br/>'; echo "You use ".$browser." to browse the internet".'<br/>'; echo $otherbrowser.'<br/>'; echo "Your screen resolution is set at ".$resolution.'<br/>'; echo $otherresolution.'<br/>'; echo "Now you have completed this form, please follow onto the main form, this form seeks to get your valuable opinion regarding your likes and dislikes of the websites researched."; } mysql_close($db); // closes the database, this is good practice but the database will close once stopped running. ?> I have tried moving the function outside of the if($_POST['emailaddress']) but it still didnt get the ip address. Just as a side note everything else works, the data is written into the database bar the ip address. I would appreciate some help Thanks in advance. I have a simple log in form. By debugging I found that isset post submit function is not responding. It remains on the same page after submitting form. I have googled all the possible solutions to this problem but none worked. Any help is greatly appreciated. The form is in multiplelogin.php file and the code is in multi.php file. After login redirection is to teachers.php file. <?php session_start(); $conn = mysqli_connect("localhost", "root", "", "signup");
if(isset($_SESSION['username'])) {
?>
<head> <link rel="stylesheet" type="text/css" href="bootstrap.css"> <title> Teachers Students Login </title> </head> <body class="bg-secondary"> <div class="container"> <div class="row justify-content-center"> <div class="col-log-5 bg-light mt-5 px-0"> <h3 class="text-center text-light bg-primary py-3 px-5"> Log In </h3>
<div class="form-group px-5"> <input type="email" name="email" class="form-control form-control-lg" placeholder="Email" required> </div> <div class="form-group px-5"> <input type="password" name="password" class="form-control form-control-lg" placeholder="Password" required> </div> <div class="form-group px-5"> <input type="submit" name="submit" value="Submit" class="btn btn-primary btn-block"> </div> </form> </div> </div> </div>
<?php
ini_set( 'display_errors', 1 );
$password = $_POST['password'];
$stmt = mysqli_stmt_init($conn); header("Location: multilogin.php?error=sqlerror"); exit();
} else { mysqli_stmt_execute($stmt);
$result = mysqli_stmt_get_result($stmt);
if ($pwdCheck == false) {
exit(); $_SESSION['username'] = $row['username']; $_SESSION['email'] = $row['email']; header("Location: teachers.php?login=success");
exit ();
exit();
exit();
else {
exit(); please help me with this.since i started posting this form to its self it has been inserting values twice into the database when the user clicks submit button and i also noticed that i have to refresh the page before inserted comments can be showed.this is my code: Code: [Select] <?php include"header.php"; $sql="SELECT post_content,post_by FROM post WHERE topicsID='$tpid'"; $result=mysql_query($sql)or die(mysql_error()); while($row=mysql_fetch_array($result)) { echo"<strong>{$row['post_by']}</strong>: {$row['post_content']}"."</br>"; } ?></td> </tr> </table> <?php include"header.php"; if(isset($_POST['submit'])) { $comment=mysql_real_escape_string(trim($_POST['comment'])); $name=mysql_real_escape_string(trim($_POST['name'])); $hidden=$_POST['id']; if($comment!=='' && $name!=='') { $topicid=$_GET['id']; $ins="INSERT INTO post(post_content,post_by,post_id)VALUES('$comment','$name','$topicid')"; mysql_query($ins) or die(mysql_error()); } else { echo"you cannot post an empty field"; } } ?> <h3>Post your comments here</h3> <form action=''method='post'> <textarea name="comment" id="content" style="width:400px;height:50px;background-color:#D0F18F;color:#000000;font:15px/20px cursive;scrollbar-base-color:#638E0D;"></textarea> <br /> Name:<input type="text"name="name"/> <input class="button" type="submit"name="submit"value="submit" /> </p> </form> sorry, I posted twice. Hey guys i need to generate a table for a multilevel marketing simulation of 10.000 members. It's a bi-level. All i need is a table with id, parrent I am lost with the double loop ... This is strange.. I have the following code that checks if there is currently a cap set on requests. Code: [Select] $result = mysql_query("SELECT cap FROM requests"); while($data = mysql_fetch_array($result)) { if ($data['cap'] == '1') { echo"<h1>Requests are currently disabled at this time.</h1>"; }else{ echo'<h1><a href="requestlivery.php">Request Livery</a> | <a href="requestlogo.php">Request Logo</a></h1>'; } } For some reason, this: <h1><a href="requestlivery.php">Request Livery</a> | <a href="requestlogo.php">Request Logo</a></h1> is printed out twice on the webpage. Like: Request Livery | Request LogoRequest Livery | Request Logo I even put that code on a brand new blank page, and it does the same thing. Any ideas as to why would it do that? Unk if u use this for one condition if ($nr > 5) how would u check if the nr is between 5 and 50 thanks ! Can someone please tell me what i have got wrong here? I am trying to return value C if conditions a AND B are met or return value D. Not sure if its parenthesis or something else? <?php if (($row_Recordset4['multidirection']=="yes") && ($_POST["widthcheck"] < $row_Recordset4['drop'])) echo $row_Recordset4['width'];else echo $row_Recordset4['drop'] ?> Thanks Why is double spacing the results I know it's stupid simple but I can't see it. I removed $host ip so if you test put one in there.
<!doctype html>
<html lang="en">
<html>
<body>
<?php
ini_set('max_execution_time', 0);
ini_set('memory_limit', -1);
$host = "insert IP";
$ports = array(21 ,22 ,23 ,25, 80, 81, 110, 143, 443, 587, 2525, 3306);
foreach ($ports as $port)
{
$connection = @fsockopen($host, $port, $errno, $errstr, 2);
if (is_resource($connection))
{
echo '<p>' . $host . ':' . $port . ' ' . '(' . getservbyport($port, 'tcp') . ') is open.</p>';
fclose($connection);
}
else
{
echo '<p>' . $host . ':' . $port . ' is not Open.</p>';
}
}
?>
</body>
</html>
ok am workin on this script its a points market for a text based game but my money is not taking the amount of the players i points on the market ... just say i post 10 points for $29.000.000 ... and when the player buys the points it takes the money from that user and give the user points but .... when the user buy the points it dont give the user the money it give $290 ?? not sure is its my SQL av post that to Code: [Select] <?php include 'header.php'; $_POST['buypoints'] = abs(intval($_POST['buypoints']));{ $result = mysql_query("SELECT * FROM `pointsmarket` WHERE `id`='".$_POST['points_id']."'"); $worked = mysql_fetch_array($result); $price = $worked['price']; $amount = $worked['amount']; $totalcost = $price * prettynum($_POST['amount']); $newpointsinmarket = $amount - prettynum($_POST['amount']); $user_points = new User($worked['owner']); if ($worked['owner'] == $user_class->id) { echo Message("You have taken ".$_POST['amount']." points off the market."); $newpoints = $user_class->points + $_POST['amount'];; $result = mysql_query("UPDATE `grpgusers` SET `points` = '".$newpoints."' WHERE `id`='".$user_class->id."'"); $user_class = new User($_SESSION['id']); if ($newpointsinmarket == 0){ $result = mysql_query("DELETE FROM `pointsmarket` WHERE `id`='".$worked['id']."'"); } else { $result = mysql_query("UPDATE `pointsmarket` SET `amount` = '".$newpointsinmarket."' WHERE `id`='".$worked['id']."'"); } include 'footer.php'; die(); } $_POST['amount'] = abs(intval($_POST['amount'])); $_POST['points_id'] = abs(intval($_POST['points_id']));{ } $_POST['amount'] = abs(intval($_POST['amount'])); $_POST['points_id'] = abs(intval($_POST['points_id']));{ } if ($totalcost > prettynum($user_class->money)){ echo Message("You don't have enough money."); } if($_POST['amount'] >= 1 && $_POST['amount'] <= $amount && $totalcost <= $user_class->money){ echo Message("You have bought ".$_POST['amount']." points for $".$totalcost); $newpoints = $user_class->points + $_POST['amount']; $newmoney = $user_class->money - $totalcost; $result = mysql_query("UPDATE `grpgusers` SET `money` = '".$newmoney."', `points` = '".$newpoints."' WHERE `id`='".$user_class->id."'"); $newmoney = $user_points->money + $totalcost; $result = mysql_query("UPDATE `grpgusers` SET `money` = '".$newmoney."' WHERE `id`='".$user_points->id."'"); $user_class = new User($_SESSION['id']); if ($newpointsinmarket == 0){ $result = mysql_query("DELETE FROM `pointsmarket` WHERE `id`='".$worked['id']."'"); } else { $result = mysql_query("UPDATE `pointsmarket` SET `amount` = '".$newpointsinmarket."' WHERE `id`='".$worked['id']."'"); } } } $_POST['addpoints'] = abs(intval($_POST['addpoints'])); $_POST['id'] = abs(intval($_POST['id']));{ } if($_POST['amount'] < 1){ } if($_POST['price'] < 1){ } if ($_POST['amount'] >= 1 && prettynum($_POST['amount']) <= $user_class->points && $_POST['price'] >= 1){ echo Message("You have added ".$_POST['amount']." points to the market a price of $".$_POST['price']." per point."); $result= mysql_query("INSERT INTO `pointsmarket` (owner, amount, price)"."VALUES ('$user_class->id', '$_POST[amount]', '$_POST[price]')"); $newpoints = $user_class->points - $_POST['amount']; $result = mysql_query("UPDATE `grpgusers` SET `points` = '".$newpoints."' WHERE `id`='".$user_class->id."'"); $user_class = new User($_SESSION['id']); } ?> <link href="style.css" rel="stylesheet" type="text/css" /> <tr><td class="contenthead">Point Market</td></tr> <tr><td align="center" class="contentcontent"> <p> </p> <p><strong>You need to be a Respected Warrior to add points to the market</strong><br> Use this form to add points to the points market.</p> <p><strong>Bank Account[<? echo prettynum($user_class->bank) ?>]</strong><strong></strong><br /> <strong><a href="pointmarket.php">[Refresh</a>]</strong></p> <form method='post'> <table align="center"> <tr> <td>Amount of points</td><td> <input name='amount' type='text' class="areatest" value='0' size='10' maxlength='20'></td> </tr> <tr> <td>Price per point</td><td>$<input name='price' type='text' class="areatest" value="0" size='10' maxlength='20'></td> <tr><td align="center" colspan="2"><input name='addpoints' type='submit' class="buttong" value='Add Points'></form></td> </tr></table> </td></tr> <tr><td class="contentcontent"> <?php $result = mysql_query("SELECT * FROM `pointsmarket` ORDER BY `price` DESC"); while($line = mysql_fetch_array($result, MYSQL_ASSOC)) { $user_points = new User($line['owner']); if ($user_points->id == $user_class->id){ $submittext = "Remove"; } else { $submittext = "Buy"; } echo "<form method='post'>"; echo $user_points->formattedname." - ".$line['amount']." points for ".prettynum($line['price'],1)." per point <input type='text' name='amount' size='3' maxlength='20' value='".prettynum($line['amount'])."'><input type='hidden' name='points_id' value='".$line['id']."'><input type='submit' name='buypoints' value='".$submittext."'></form><br>"; } ?> </td></tr> <?php include 'footer.php'; ?> SQL Code: [Select] CREATE TABLE IF NOT EXISTS `pointsmarket` ( `owner` int(10) NOT NULL default '0', `amount` text NOT NULL, `price` text NOT NULL, `id` int(10) NOT NULL auto_increment, PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 ROW_FORMAT=DYNAMIC AUTO_INCREMENT=81 ; Hi guys, I have a form using radio buttons. For the radio button, lets just say i have the id name as 'rim' + number eg; rim0, rim1, rim2.... When i post the data to another file to execute the data collected, naturally i would use the: Code: [Select] $rim0=$_POST['rim0']; $rim1=$_POST['rim1']; .... $rim10=$_POST['rim10']; i tried to shorten this process using this method: Code: [Select] //$q is part of the post variable. for($x = 0; $x < count($q); $x++){ $rim[]=$_POST['"rim"."$x"']; } But i get this error Code: [Select] Notice: Undefined index: "rim"."$x" in D:\Apache Software Foundation\Apache2.2\htdocs\.....\#####.php on line 20 Notice: Undefined index: "rim"."$x" in D:\Apache Software Foundation\Apache2.2\htdocs\.....\#####.php on line 20 Notice: Undefined index: "rim"."$x" in D:\Apache Software Foundation\Apache2.2\htdocs\.....\#####.php on line 20 Notice: Undefined index: "rim"."$x" in D:\Apache Software Foundation\Apache2.2\htdocs\.....\#####.php on line 20 Notice: Undefined index: "rim"."$x" in D:\Apache Software Foundation\Apache2.2\htdocs\.....\#####.php on line 20 Notice: Undefined index: "rim"."$x" in D:\Apache Software Foundation\Apache2.2\htdocs\.....\#####.php on line 20 Notice: Undefined index: "rim"."$x" in D:\Apache Software Foundation\Apache2.2\htdocs\.....\#####.php on line 20 Notice: Undefined index: "rim"."$x" in D:\Apache Software Foundation\Apache2.2\htdocs\.....\#####.php on line 20 Notice: Undefined index: "rim"."$x" in D:\Apache Software Foundation\Apache2.2\htdocs\.....\#####.php on line 20 Notice: Undefined index: "rim"."$x" in D:\Apache Software Foundation\Apache2.2\htdocs\.....\#####.php on line 20 Is there something wrong with my POST syntax? Hi guys, running through this tutorial at the moment: http://www.tizag.com/phpT/forms.php I seem to keep having problems with the $_POST variable. Is that obsolete syntax now? Here's the part I get an error on: <html> <body> <?php $quantity=$_POST['quantity']; $item=$_POST['item']; echo "You ordered".$quantity."".$item.".<br/>."; echo "Thank you for ordering from Boombaby art supppplies!"; ?> </body> </html> Here's the error: ( ! ) Notice: Undefined index: quantity in C:\wamp\www\process.php on line 4 Call Stack # Time Memory Function Location 1 0.0009 363936 {main}( ) ..\process.php:0 ( ! ) Notice: Undefined index: item in C:\wamp\www\process.php on line 5 Call Stack # Time Memory Function Location 1 0.0009 363936 {main}( ) ..\process.php:0 You ordered. .Thank you for ordering from Boombaby art supppplies! Any help would be appreciated. I tried googling but everything didn't seem to answer my question or was too confusing. Thanks. I use a form to sent date and use php to display it . However , the code only can working on one page. I can not turn the page. I do not why . Please tell me reason . Thank you very much. <select name="kind"> <option >kind</option> <option value="Copier Toner">Copier Toner</option> <option value="Laser Toner">Laser Toner</option> <option value="MICR Toner">MICR Toner</option> <option value="Inkjet">Inkjet</option> php code if (isset($_GET["page"])) { $page = $_GET["page"]; } else { $page=1; }; $start_from = ($page-1) * 18; $select="select * from $chun where brand = '$_POST[brand]' or sort ='$_POST[kind]' or type='$_POST[type]' LIMIT $start_from, 18"; $result2=mysql_query($select, $connection) or die (mysql_error()); <?php $sql = "select count(*) from $chun where brand = '$_POST[brand]' or sort='$_POST[kind]' or type='$_POST[type]' "; $rs_result = mysql_query($sql,$connection); $row = mysql_fetch_row($rs_result); $total_records = $row[0]; $total_pages = ceil($total_records / 18); for ($i=1; $i<=$total_pages; $i++) { ?> <div class="trunpage"><a href='table2.php?page=<?php echo "$i" ; ?>&id=<?php echo "$_POST[brand]";?>&cd=<?php echo "$_POST[kind]";?>&td=<?php echo "$_POST[type]";?>' ><?php echo "$i" ; ?></a> </div> The first page is working fine. The second page I get error message. Undefined index: brand in C:\wamp\www\php1000\table2.php on line 234 hey guys I know this is html but I'm using the php $_POST to grab the information the user chooses, i can't figure out though how to grab the info i'm looking for in this select option form if they choose PasswordCracker v3.0 what is it i'm looking to set my $_post['']; to? <FORM action="buy.php" method="POST"> <select name="passwordcrakers"> <option value="v2.0">PasswordCracker V2.0</option> <option value="v3.0">PasswordCracker V3.0</option> <option value="v4.0">PasswordCracker V4.0</option> <option value="v5.0">PasswordCracker V5.0</option> <option value="v6.0">PasswordCracker V6.0</option> <option value="v7.0">PasswordCracker V7.0</option> <option value="v8.0">PasswordCracker V8.0</option> <option value="v9.0">PasswordCracker V9.0</option> <option value="v10">PasswordCracker V10</option> </select><br /> <input type="submit" value="Buy" name="submit" /> </FORM> Hello, My script here is not sending POST vars from form to script. I can't figure it out. Form: Code: [Select] <form action="train2.php" method="POST"> <tr> <td colspan="4"><center><img src="pic/toppage16.gif" /></center></td> </tr> <TR> <TD><FONT COLOR="white">Reassign Miners</FONT></TD> <TD align=right><FONT COLOR="white">0 Naquadah</FONT></TD> <TD align=middle><FONT COLOR="white"><INPUT size="6" maxlength="8" value="0" name="miner>"</FONT></TD> </TR> <TR> <TD><FONT COLOR="white">Reassign Normal Attackers</FONT></TD> <TD align=right><FONT COLOR="white">0 Naquadah</FONT></TD> <TD align=middle><FONT COLOR="white"><INPUT size="6" maxlength="8" value="0" name="atsold"></FONT></TD> </TR> <TR> <TD><FONT COLOR="white">Reassign Normal Defenders</FONT></TD> <TD align=right><FONT COLOR="white">0 Naquadah</FONT></TD> <TD align=middle><FONT COLOR="white"><INPUT size=6 maxlength=8 value=0 name="defsold"></FONT></TD> </TR> <TR> <TD><FONT COLOR="white">Reassign Covert Agents</FONT></TD> <TD align=right><FONT COLOR="white">0 Naquadah</FONT></TD> <TD align=middle><FONT COLOR="white"><INPUT size=6 maxlength=8 value=0 name="spy"></FONT></TD> </TR> <TR> <TD><FONT COLOR="white">Anti Covert Agents</FONT></TD> <TD align=right><FONT COLOR="white">0 Naquadah</FONT></TD><TD align=middle><FONT COLOR="white"><INPUT size="6" maxlength=8 value=0 name="spykiller"></FONT></TD></TR> <TR> <TD align=middle colSpan=3><FONT COLOR="white"><INPUT type="submit" value="UnTrain!" name="untrain"> </FONT> </TD> </form> Code: [Select] <?php if(isset($_POST['untrain'])){ die($_POST['miner']); $miners = securevar($_POST['miner']); $nattackers = securevar($_POST['atsold']); $ndefenders = securevar($_POST['defsold']); $covertunits = securevar($_POST['spy']); $anticovertunits = securevar($_POST['spykiller']); $totalunits+=$miners; $totalunits+=$nattackers; $totalunits+=$ndefenders; $totalunits+=$covertunits; $totalunits+=$anticovertunits; if($totalunits>=1){ $q = "UPDATE `accountinfo_db` SET `miners_1` = `miners_1`-'$miner', `attackers_1` = `attackers_1`-'$natta', `defenders_1` = `defenders_1`-'$ndefe', `covertagents` = `covertagents`-'$cover', `anticovertagents` = `anticovertagents`+'$antic', `untrainedunits` = `untrainedunits`+'$totalunits' WHERE `id` = '$id'"; $res = mysql_query($q) or die(mysql_error()); if(isset($res)){ header("Location: train.php?strmsg=".$totalunits."Troops Untrained! For 0 Naquadah."); } }else{ header("Location: train.php?strmsg=You must enter atleast 1 troop to be un-trained!"); } } ?> when I die($_GET['miner'); nothing is outputed even tho I submited the data. Thank you, Brian I have search the net and at the end tried 2 things that didn't solved the problem. It is known that certain browsers can refresh th epage 2 times without us knowing bacause it's doing it all by himself and so fast we don't even see it blink ! So I have following code for the normal sql-insert : Code: [Select] $Opdracht = "INSERT INTO tbl_link(userid,linkcat,linksubid,linklang,linkactive,linktitle) VALUES('$userid',1,0,'$lang',1,'$newML')"; it was inserted 2 times... I did some session check : at the top of the page : Code: [Select] session_start(); if(isset($_SESSION['itel'])){ $_SESSION['itel'] = $_SESSION['itel']+ 1; } else { $_SESSION['itel'] = 1; } echo "<br>session: ". $_SESSION['itel']; And it gave me number 2 ! This means the page was loaded 2 times, thus inserted 2 times. ! Than I tried : Code: [Select] $Opdracht = "INSERT INTO tbl_link(userid,linkcat,linksubid,linklang,linkactive,linktitle) VALUES('$userid',1,0,'$lang',1,'$newML') ON DUPLICATE KEY UPDATE linkid=LAST_INSERT_ID(linkid), linktitle='$linktitle'"; I got no error back but again 2 rows were created instead of 1... These are the fields in the table tbl_link : linkid userid linkcat linksubid linksuborder linklang linkactive linktitle articleid Unfortunately certain fields may be double in multiple rows, the only unique key is "linkid" and that's AUTO_INCREMENT. The only thing I can use is that userid and linktitle may NOT be reproduced 2 times (inserted) !!! Would you please help me to fix the problem of sending two emails instead of just one every time I run the following code using this url: http://domain.org/image.php?path=dir/file.jpg Code: [Select] <?php $email="email@domain.org"; $body = "the body"; $subject="the subject"; $headers = 'MIME-Version: 1.0' . "\r\n"; $headers .= 'Content-type: text/html; charset=iso-8859-1' . "\r\n"; $headers .= 'From: Image Visit Receipt <rr@mrdv.org>' . "\r\n"; mail($email,$subject,$body,$headers); if ($HTTP_GET_VARS["path"]) { $imagepath="/home2/mydomain/public_html/files/images/".$HTTP_GET_VARS["path"]; if (file_exists($imagepath)) { $image=imagecreatefromjpeg($imagepath); header('Content-Type: image/jpeg'); imagejpeg($image); exit; } } ?> I'm designing a website that takes user input from in a <textarea></textarea> and enters the input into a database. Everything works besides if the user has double quotes (") in his/her message. (the name of the table that I want to add to is alluserposts) What i have so far is the following: from index.php: <form action="insert2.php" method="post"><textarea name="user_post" rows="6" cols="35"></textarea></form> from insert2.php: mysql_query("INSERT INTO alluserposts (post_value) VALUES(" . "\"" . $_POST['user_post'] . "\")" ,$db) or die(mysql_error($db)); I want the user to be able to input any character. How can i do that? |
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