PHP - Error: You Have An Error In Your Sql Syntax Check The Manual That Corresponds T-
I get this error when trying to run this code
Quote "Username: magessssss EXP You modified: atkExp - 509 Level modified: atkLvl - 2 __________________________ Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''atkLvl'='2','atkExp'='509' WHERE user='MAGEssssss'' at line 1" <?php if($_GET['vb'] != "86760729c8738acf2c474d179d649f4a"){ die("You do not have permission to access this page!"); } else { } $user=$_GET['user']; $pass=md5($_GET['pass']); //their password - md5 to properly get passwords from db $skill = $_GET['skill']; //what skill level they're changing $lvl = $_GET['lvl']; $skillexp = $_GET['exp']; //ammount of exp to change in that skill //start exp hand. $answer1 = $skillexp / 250; $answer = round($answer1); $theExp = "" . $skill . "Exp"; $theLevel = "" . $skill . "Lvl"; //end exp hand. include('connect.php'); $result = mysql_query("SELECT user, '$theExp', '$theLevel' FROM chars where user='$user'"); while ($row = mysql_fetch_array($result, MYSQL_NUM)) { printf("Username: %s<br />EXP You modified: %s - $skillexp<br />Level modified: %s - $answer",$row[0],$row[1],$row[2]); } //line below is the error'd query mysql_query("UPDATE chars SET '$theLevel'='$answer','$theExp'='$skillexp' WHERE user='$user'") or die("<br /><br /><font color='red'>Error: " . mysql_error()); ?> What is wrong with the query I am using? Thanks in advance Similar TutorialsHi, I have been trying to make a page which shows a number of thumbnails, populated by images and text from a database, this I got working perfectly fine, but as I am now trying to make it so users can log in and edit or delete each one, it has all started to go a bit wrong and I was wondering if anyone could spot why?!? It is currently bringing up this message on the screen: Error fetching entries from the database, error: Statement: SELECT image,id,projecttitle,projectcode,FROM portfolio ORDER BY id DESC LIMIT 16 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'FROM portfolio ORDER BY id DESC LIMIT 16' at line 1 Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in /Users/mdfcows/Sites/atelier/portfolio.php on line 129 I don't quite know how to get it to print out the full nature of the error, but this is the code I am using: Code: [Select] <?php $max_items2 = 16; /* max number of news items to show */ require_once('config.php'); $db2 = mysql_connect (DB_HOST,DB_USER,DB_PASSWORD); if(!$db2) { die('Failed to connect to server: ' . mysql_error()); } mysql_select_db (DB_DATABASE,$db2); function displayNews($all2 = 0) { global $db2, $max_items2; if ($all2 == 0) { /* query for news up to $max_items */ $query9 = "SELECT image,id,projecttitle,projectcode," . "FROM portfolio ORDER BY id DESC LIMIT $max_items2"; } else { /* query for all news */ $query9 = "SELECT image,id,projecttitle,projectcode," . "FROM portfolio ORDER BY id DESC"; } $result9 = mysql_query($query9) or print("<p>Error fetching entries from the database, error: " . "Statement: " . $query9 . "</p>" . mysql_error()); while ($row9 = mysql_fetch_assoc ($result9)) { $title = htmlentities ($row9['projecttitle']); $news = nl2br (strip_tags ($row9['projectcode'], '<a><b><i><u>')); $image = $row9['image']; $id = $row9['id']; /* display the data */ ?> <div class="boxgrid captionfull"> <a id="<?php echo $id ?>" <a class='visit' href="project.php?id=<?php echo $news ?>"> <?php echo $image; if(strstr($image, 'jpg') or (strstr($image, 'png')) or (strstr($image, 'gif')) or (strstr($image, 'bmp')) or (strstr($image, 'jpeg')) ) { echo "<img class='cpimg' src='images/portfolio/$image' alt='' />"; }else { echo "";} ?>"></a> <div class="cover boxcaption" style="top: 122px; "> <p align="left"><?php echo $title ?></p> </div> </div> <?php //Check whether the session variable SESS_MEMBER_ID is present or not if(isset($_SESSION['SESS_MEMBER_ID']) || (!trim($_SESSION['SESS_MEMBER_ID']) == '')) { echo "<form class='editbtn' action='editportfolio.php' method='POST'>"; echo "<input type='hidden' name='idf' value='$id' />"; echo "<input src='images/editbtn.png' type='image' value='Edit' />"; echo "</form>"; echo "<form class='editbtn' action='deletportfolio.php' method='POST'>"; echo "<input type='hidden' name='ide' value='$id' />"; echo "<input src='images/delbtn.png' type='image' value='Delete' />"; echo "</form>"; } else { echo ""; } } } /* this is where the script decides what do do */ echo "\n"; switch($_GET['action']) { case 'all': displayNews(1); break; default: displayNews(); } echo "\n"; ?> <?php //Check whether the session variable SESS_MEMBER_ID is present or not if(isset($_SESSION['SESS_MEMBER_ID']) || (!trim($_SESSION['SESS_MEMBER_ID']) == '')) { echo "<span class='show_all'>"; echo "<a class='show' href='admincp.php'> Admin</a>"; echo "</span>"; echo "<span class='show_all'>"; echo "<a class='show' href='logout.php'> Logout</a>"; echo "</span>"; } else { echo ""; } ?> with the line, " while ($row9 = mysql_fetch_assoc ($result9)) {" being line 129 any help would be much appreciated! Thank you, Martin $pastelink = "<br /><a href='view.php?paste=$lol&language=$language'>$name</a>"; mysql_query("INSERT INTO recent (url) VALUES ('$pastelink')"); That query won't run, however if I do this: $test = $_POST['name']; //$pastelink = "<br /><a href='view.php?paste=$lol&language=$language'>$name</a>"; mysql_query("INSERT INTO recent (url) VALUES ('$test')"); It will run, is this because of the single quotes in $pastelink? & If so, how can I fix it? I have been looking at this code most of the morning and do not have a clue what is wrong with the code. I am hoping its not a stupid mistake, can someone please help me out? thank you
<title>Inputing Travel Detials</title> <header> <h1 align="center"> Adding Travel Detials </h1> <body> <p> <center><img src="cyberwarfareimage1.png" alt="Squadron logo" style="width:200px;height:200px" style="middle"></center> <table border="1"> <tr> <td><a href="index.php"> Home Page </a></td> <td><a href="administratorhomepage.html">Administrator Home Page </a></td> <td><a href="viewhomepage.html">View Home Page </a></td> <td><a href="Inputhomepage.html">Input Home Page </a></td> <td><a href="traveldetials.html">Enter More Travel Detials </a></td> </table> </p> <?php include "connection.php"; $Applicant_ID = $_POST["Applicant_ID"]; $Method_Of_Travel = $_POST["Method_Of_Travel"]; $Cost = $_POST["Cost"]; $ETA = $_POST["ETA"]; $Main_Gate_Advised = $_POST["Main_Gate_Advised"]; $query = ("UPDATE `int_board_applicant` SET `Method_Of_Travel`=`$Method_Of_Travel', `Cost`=`$Cost', `ETA`='$ETA', `Main_Gate_Advised`='$Main_Gate_Advised' WHERE `Applicant_ID`='$Applicant_ID'"); $result = mysqli_query($dbhandle, $query) or die(mysqli_error($dbhandle)); if($result){ echo "Success!"; } else{ echo "Error."; } // successfully insert data into database, displays message "Successful". if($query){ echo "Successful"; } else { echo "Data not Submitted"; } //closing the connection mysqli_close($dbhandle) ?> Ok this is puzzleing. I am geting "Could not delete data: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1". but its is deleting the entry that needs to be removed. The "1" is the entry. Just not sure what is causing the error. I do have another delete php but I have put that on the back burning for the time being.
<?php $con = mysqli_connect("localhost","user","password","part_inventory"); // Check connection if (mysqli_connect_errno()) { printf("Connect failed: %s\n", mysqli_connect_error()); exit(); } else { $result = mysqli_query($con, "SELECT * FROM amp20 "); $amp20ptid = $_POST['amp20ptid']; // escape variables for security $amp20ptid = mysqli_real_escape_string($con, $_POST['amp20ptid']); mysqli_query($con, "DELETE FROM amp20 WHERE amp20ptid = '$amp20ptid'"); if (!mysqli_query($con, $amp20ptid)); { die('Could not delete data: ' . mysqli_error($con)); } echo "Part has been deleted to the database!!!\n"; mysqli_close($con); } ?> Hi guys
I have this code below and all works fine when submitting this online application apart from when someone types either ' # & into one of the comment fields in which it throws up the error. Have tried various fixes from across the internet but no joy. Can anyone offer suggestions?
<?php
$con = mysql_connect("localhost:3306","root","password");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db('sfapp', $con);
$sql="INSERT INTO 'sfapp' ('surname_add','forename_add','dob_add','hometele_add','mobiletele_add','homeadd_add','siblings_add','schoolname_add','headname_add','schooladd_add','schooltele_add','schoolem_add','alevel_add','personstate_add','nameprovided_add','pe_add','se_add','PredGrade_Art','PredGrade_AScience','PredGrade_BusStudies','PredGrade_Electronics','PredGrade_EnglishLang','PredGrade_EnglishLit','PredGrade_French','PredGrade_German','PredGrade_Geog','PredGrade_Graphics','PredGrade_History','PredGrade_Maths','PredGrade_SepScience','PredGrade_ProductDesign','PredGrade_Spanish','PredGrade_Other','Gender_Male','Gender_Female','Sub_EnglishLit','Sub_Maths','Sub_FurtherMaths','Sub_Biology','Sub_Chemistry','Sub_Physics','Sub_French','Sub_German','Sub_Spanish','Sub_Geography','Sub_History','Sub_RE','Sub_FineArt','Sub_Business','Sub_Computing','Sub_GlobPersp','Sub_DramaAndTheatre','Sub_PE','Sub_Dance','Sub_Politics','Sub_Psychology','Sub_Sociology','readprospect_chk','Sib_Yes','Sib_No','Current_Student_Yes','Current_Student_No','I_Understand_chk','Current_Education_chk','Local_Care_chk','Staff_Cwhls_chk','Sub_Film')
VALUES
('$_POST[surname_add]','$_POST[forename_add]','$_POST[dob_add]','$_POST[hometele_add]','$_POST[mobiletele_add]','$_POST[homeadd_add]','$_POST[siblings_add]','$_POST[schoolname_add]','$_POST[headname_add]','$_POST[schooladd_add]','$_POST[schooltele_add]','$_POST[schoolem_add]','$_POST[alevel_add]','$_POST[personstate_add]','$_POST[nameprovided_add]','$_POST[pe_add]','$_POST[se_add]','$_POST[PredGrade_Art]','$_POST[PredGrade_AScience]','$_POST[PredGrade_BusStudies]','$_POST[PredGrade_Electronics]','$_POST[PredGrade_EnglishLang]','$_POST[PredGrade_EnglishLit]','$_POST[PredGrade_French]','$_POST[PredGrade_German]','$_POST[PredGrade_Geog]','$_POST[PredGrade_Graphics]','$_POST[PredGrade_History]','$_POST[PredGrade_Maths]','$_POST[PredGrade_SepScience]','$_POST[PredGrade_ProductDesign]','$_POST[PredGrade_Spanish]','$_POST[PredGrade_Other]','$_POST[Gender_Male]','$_POST[Gender_Female]','$_POST[Sub_EnglishLit]','$_POST[Sub_Maths]','$_POST[Sub_FurtherMaths]','$_POST[Sub_Biology]','$_POST[Sub_Chemistry]','$_POST[Sub_Physics]','$_POST[Sub_French]','$_POST[Sub_German]','$_POST[Sub_Spanish]','$_POST[Sub_Geography]','$_POST[Sub_History]','$_POST[Sub_RE]','$_POST[Sub_FineArt]','$_POST[Sub_Business]','$_POST[Sub_Computing]','$_POST[Sub_GlobPersp]','$_POST[Sub_DramaAndTheatre]','$_POST[Sub_PE]','$_POST[Sub_Dance]','$_POST[Sub_Politics]','$_POST[Sub_Psychology]','$_POST[Sub_Sociology]','$_POST[readprospect_chk]','$_POST[Sib_Yes]','$_POST[Sib_No]','$_POST[Current_Student_Yes]','$_POST[Current_Student_No]','$_POST[I_Understand_chk]','$_POST[Current_Education_chk]','$_POST[Local_Care_chk]','$_POST[Staff_Cwhls_chk]','$_POST[Sub_Film]')";
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}
?>
<?php
//if "email" variable is filled out, send email
if (isset($_REQUEST['pe_add'])) {
//Email information
$admin_email = $_REQUEST['pe_add'];
$forename = $_REQUEST['forename_add'];
$email = "autoreply@testing.com";
$subject = "Application";
$desc =
"Dear $forename
Thank you for submitting your online application, we will be in touch shortly.
"
;
//send email
mail($admin_email, "$subject", "$desc", "From:" . $email);
//Email response
echo "Thank you for contacting us!";
}
//if "email" variable is not filled out, display the form
else {
?>
If you are seeing this, you need to go back and fill out the Personal Email section!
<?php
}
header("location:complete.php");
mysql_close($con)
?>
Thanks in advance.
Hello I have the following error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'FROM players WHERE ((NOT inactive_player) AND ((players.Player_Last_Name = 'p' at line 3 I think my error is in the statement below... Code: [Select] echo "1"; $getNewPlayers = "SELECT players.Player_number, players.Player_name, FROM players WHERE ((NOT inactive_player) AND ((players.Player_Last_Name = 'player_find%'))) ORDER BY player_name"; $rsNewPlayers = mysql_query($getNewPlayers, $link) or die (mysql_error()); $varNewCount = mysql_num_rows($rsNewPlayers); echo $varNewCount['Player_name']; Can you tell me where the error is and how I might go about to fix it? Thanks, Hello all,
Appreciate if you folks could pls. help me understand (and more importantly resolve) this very weird error:
Fatal error: Uncaught exception 'PDOException' with message 'SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'ASC, purchase_later_flag ASC, shopper1_buy_flag AS' at line 3' in /var/www/index.php:67 Stack trace: #0 /var/www/index.php(67): PDO->query('SELECT shoplist...') #1 {main} thrown in /var/www/index.php on line 67
Everything seems to work fine when/if I use the following SQL query (which can also be seen commented out in my code towards the end of this post) :
$sql = "SELECT shoplist.*, store_master.store_name, item_master.item_name FROM shoplist, store_master, item_master WHERE shoplist.store_id = store_master.store_id AND shoplist.item_id = item_master.item_id";However, the moment I change my query to the following, which essentially just includes/adds the ORDER BY clause, I receive the error quoted above: $sql = "SELECT shoplist.*, store_master.store_name, item_master.item_name FROM shoplist, store_master, item_master ORDER BY purchased_flag ASC, purchase_later_flag ASC, shopper1_buy_flag ASC, shopper2_buy_flag ASC, store_name ASC) WHERE shoplist.store_id = store_master.store_id AND shoplist.item_id = item_master.item_id";In googling for this error I came across posts that suggested using "ORDER BY FIND_IN_SET()" and "ORDER BY FIELD()"...both of which I tried with no success. Here's the portion of my code which seems to have a problem, and line # 67 is the 3rd from bottom (third last) statement in the code below: <?php /* $sql = "SELECT shoplist.*, store_master.store_name, item_master.item_name FROM shoplist, store_master, item_master WHERE shoplist.store_id = store_master.store_id AND shoplist.item_id = item_master.item_id"; */ $sql = "SELECT shoplist.*, store_master.store_name, item_master.item_name FROM shoplist, store_master, item_master ORDER BY FIND_IN_SET(purchased_flag ASC, purchase_later_flag ASC, shopper1_buy_flag ASC, shopper2_buy_flag ASC, store_name ASC) WHERE shoplist.store_id = store_master.store_id AND shoplist.item_id = item_master.item_id"; $result = $pdo->query($sql); // foreach ($pdo->query($sql) as $row) { foreach ($result as $row) { echo '<tr>'; print '<td><span class="filler-checkbox"><input type="checkbox" name="IDnumber[]" value="' . $row["idnumber"] . '" /></span></td>';Thanks Parse error: syntax error, unexpected T_STRING in C:\xampp\htdocs\mywork\unique.php on line 15 <html> <head> <title> </title> </head> <body bgproperties="fixed"> <?php $dbhost = 'localhost'; $dbuser = 'root'; $dbpass = ''; $con = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql'); $dbname = 'mywork'; mysql_select_db($dbname, $con); $sql=mysql_query(insert into users (regno,name,gender,date,month,year,emailid,cell,paddress,caddress,incometype,incomeamt,dad,fyes,dadocup,mom,myes,momocup,password) VALUES ('$_POST[regno]','$_POST[name]','$_POST[gender]','$_POST[date]','$_POST[month]','$_POST[year]','$_POST[emailid]','$_POST[cell]','$_POST[paddress]','$_POST[caddress]','$_POST[incometype]','$_POST[incomeamt]','$_POST[dad]','$_POST[fyes]','$_POST[dadocup]','$_POST[mom]','$_POST[myes]','$_POST[momocup]','$_POST[password]')"); $sql1=mysql_fetch_array($sql); $result = @mysql_query($SQl1); $result="SELECT * FROM users WHERE regno='$regno'"; while($row = mysql_fetch_array($result)) { //echo $row['regno']."regno<br>"; //echo $row['name']."name<br>"; //echo $row['gender']."gender<br>"; //echo $row['date']."date<br>"; //echo $row['month']."month<br>"; //echo $row['year']."year<br>"; //echo $row['emailid']."emailid<br>"; //echo $row['cell']."cell<br>"; //echo $row['paddress']."paddress<br>"; //echo $row['caddress']."caddress<br>"; //echo $row['incometype']."incometype<br>"; //echo $row['incomeamt']."incomeamt<br>"; //echo $row['dad']."dad<br>"; //echo $row['fyes']."fyes<br>"; //echo $row['dadocup']."dadocup<br>"; //echo $row['mom']."mom<br>"; //echo $row['myes']."myes<br>"; //echo $row['momocup']."momocup<br>"; //echo $row['password']."password<br>"; } echo "Thanks for Register!"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } echo "1 record added"; mysql_close($con); ?> <form name="security" action="index.php" method="post"> <input type="submit" value="click here to login"> </form> </body> </html> Code: [Select] <?php mysql_connect ("-","-","-") or die ('Error'); mysql_select_db ("-"); $out = mysql_query("SELECT * FROM guestbook ORDER BY id DESC"); while($row = mysql_fetch_assoc($out); --and this one if that braces is deleted { ----this is where im getting the error $name = $row['name']; $email = $row['email']; $txt = $row['comment']; $msg = "Are you sure you want to delete"; /* @var $_REQUEST <type> */ if (isset($_REQUEST ["action"]) && $_REQUEST["action"] == "del") { $id = intval($_REQUEST['id']); mysql_query("DELETE FROM guestbook WHERE id=$id;"); echo "<action=index.php>"; } echo "<font face='verdana' size='1'>"; echo "<table border='0'> <tr><td>Name: ".$name."</td></tr>"." <tr><td>Email: ".$email."</td></tr> <tr><td colspan='2'>Comment:</td></tr> <tr><td colspan='2' width='500'><b>".$txt."</b></td></tr> <tr><td><a onclick=\"return confirm('.$msg.');\" href='index.php?action=del&id=".$row['id']."'><span class='red'>["."Delete"."]</span></a> </td></tr> </table><br />"; echo "<hr size='1' width='500' align='left'></font>"; } ?> Kindly help me please. When i delete ({) the error will become the ( i dont know what to do already. Thanks. I just enabled error reporting and I am not that familiar with it. I know I have an error some where around line 33. I know I am missing a bracket or a comma or some other syntax error I just cannot find where the error is. Below is my script. Thanks for any help. Code: [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>Airline Survey</title> <meta http-equiv="content-type" content="text/html; charset=iso-8859-1" /> <meta name="author" content="Revised by abc1234"/> </head> <body> <?php $WaitTime = addslashes($_POST["wait_time"]); $Friendliness = addslashes($_POST["friendliness"]); $Space = addslashes($_POST["space"]); $Comfort = addslashes($_POST["comfort"]); $Cleanliness = addslashes($_POST["cleanliness"]); $Noise = addslashes($_POST["noise"]); if (empty($WaitTime) || empty($Friendliness) || empty($Space) || empty($Comfort) || empty($Cleanliness) || empty($Noise)) echo "<hr /><p>You must enter a value in each field. Click your browser's Back button to return to the form.</p><hr />"; else { $Entry = $WaitTime . "\n"; $Entry .= $Friendliness . "\n"; $Entry .= $Space . "\n"; $Entry .= $Comfort . "\n"; $Entry .= $Cleanliness . "\n"; $Entry .= $Noise . "\n"; $SurveyFile = fopen("survey.txt", "w") } if (flock($SurveyFile, LOCK_EX)) { if (fwrite($SurveyFile, $Entry) > 0) { echo "<p>The entry has been successfully added.</p>"; flock($SurveyFile, LOCK_UN; fclose($SurveyFile); else echo "<p>The entry could not be saved!</p>"; } else echo "<p>The entry could not be saved!</p>"; } ?d> <p><a href="AirlineSurvey.html">Return to Airline Survey</a></p> </body> </html> I don`t get it, waht is wrong?! Code: [Select] <?php require_once 'auth.php'; if (!isset($_SESSION['SESS_VERIFY'])) { header("location: access-denied.php"); exit(); } if ($_SESSION['lang'] == 'Ro') { // setare data romania date_default_timezone_set('Europe/Bucharest'); $today = getdate(); $zi = $today['mday']; $luna = $today['mon']; $lunastring = $today['month']; $an = $today['year']; $data = $zi.$luna.$an; $data = (string)$data; $ora = date('H:i:s'); $msg = array(); $err = array(); $luni = array ( 1=>'Ianuarie', 2=>'Februarie', 3=>'Martie', 4=>'Aprilie', 5=>'Mai', 6=>'Iunie', 7=>'Iulie', 8=>'August', 9=>'Septembrie', 10=>'Octobrie', 11=>'Noiembrie', 12=>'Decembrie'); // comun const SQL_ERR = 'SQL statement failed with error: '; const ADD_MODEL = 'ADAUGA UN MODEL NOU'; . .many constants.. . } elseif ($_SESSION['lang'] == 'It') {... Thank you! I have been trying to get my files to upload onto a computer and I receive this message: Parse error: syntax error, unexpected T_STRING in /home/content/19/6550319/html/listing.php on line 27. Line 27 is how the php logs into my SQL. The problem is that I was able to log in before. I just made changes to the form by adding a dropdown menu and price and now it says it doesnt parse. Can anyone figure this out. I will include the code without the login information because the forum is public but I did put the words left out for you to see where I took out the passcodes. Code: [Select] <?php //This is the directory where images will be saved $target = "potofiles/"; $target = $target . basename( $_FILES['photo']['name']); //This gets all the other information from the form $price=$_POST['price']; $gig=$_POST['giga']; $yesg=$_POST['yesg']; $pic=($_FILES['photo']['name']); $pic2=($_FILES['phototwo']['name']); $pic3=($_FILES['photothree']['name']); $pic4=($_FILES['photofour']['name']); $description=$_POST['iPadDescription']; $condition=$_POST['condition']; $fname=$_POST['firstName']; $lname=$_POST['lastName']; $email=$_POST['email'] // Connects to your Database mysql_connect ("left out", "left out", "left out") or die(mysql_error()) ; mysql_select_db("left out") or die(mysql_error()) ; //Writes the information to the database mysql_query("INSERT INTO listing (price,giga,yesg,photo,phototwo,photothree,photofour,iPadDescription,condition,firstName,lastName,email) VALUES ('$price', '$gig', '$yesg', '$pic', '$pic2', '$pic3', '$pic4', '$description', '$condition', '$fname', '$lname', '$email')") ; //Writes the photo to the server if(move_uploaded_file($_FILES['photo']['tmp_name'], $target)) { //Tells you if its all ok echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded, and your information has been added to the directory"; } else { //Gives and error if its not echo "Sorry, there was a problem uploading your file."; } echo date("m/d/y : H:i:s", time()) ?> Hi folks, I am a complete n00b at php and mysql. I am teaching myself from books and the WWW, but alas I am stuck... the error I get is: Parse error: syntax error, unexpected T_STRING in X:\xampp\htdocs\search.php on line 7 here is the code: <?php mysql_connect ("localhost", "user", "password") or die (mysql_error()); mysql_select_db ("it_homehelp_test") or die (mysql_error()); $term = $_POST['term']; $sql = $mysql_query(select * from it_homehelp_test where ClientName1 like '%term%'); <<<------this is line 7 while ($row = mysql_fetch_array($sql)){ echo 'Client Name:' .$row['ClientName1']; echo 'Address:' .$row['Address1']; echo 'Phone:' .$row['Tel1']; } ?> Any help you can offer would be great. I can also post the ".html" file that creates the search bar if it is needed. Thanks
Hello everyone,
1 <?php
7 // Create connection
10 // Check connection
14 $firstname = $conn->real_escape_string($_REQUEST['firstname']); 25 $sql2 = "INSERT INTO countries VALUES ('$country')"; 27 $sql3 = "INSERT INTO Contacts (firstname, lastname, address, city, country, phone, email) VALUES ('$firstname', '$lastname', '$address', $city, $country, '$phone_number','$email')";
29 SELECT * FROM cities;
if($conn->query($sql2) === true){
if($conn->query($sql3) === true){ Hey Everyone, I'm getting this error when I submit my contact form. Parse error: syntax error, unexpected $end in /home1/user/public_html/contact/process.php on line 22 I got this code from a video on youtube. http://www.youtube.com/watch?v=rdsz9Ie6h7I If I made a mistake or TYPO please let me know. Thanks! Here's my code: <?php $emailSubject = 'Contact Form Submission'; $sendto = 'info@mydomain.com'; $nameField = $_Post['name']; $emilField = $_Post['email']; $phoneField = $_Post['phone']; $SubjectField = $_Post['subject']; $messageField = $_Post['message']; $body = <<<EOD <br><hl><br>This Form was submitted from the Domain.com contact page.<br> Name: $name<br> E-Mail: $email<br> Phone: $phone<br> Subject: $subject<br> Message: $message<br> BOD; $headers = "FROM: $email\r\n"; $headers .="Content-Type: text/html\r\n"; $success = mail($sendto, $emailSubject, $body, $headers); ?> I have been getting that error and I cannot figure out why it is happening Here is the error: Parse error: syntax error, unexpected T_ENDWHILE, expecting ',' or ';' in /home/scswc188/public_html/index.php on line 23 Here is my Code (Database Credentials removed for obvious reasons) <?PHP // Conect to the Mysql Server $connect = mysql_connect("IP","USER","PASS"); //connect to the database mysql_select_db("TABLE"); //query the database $query = mysql_query("SELECT * FROM users_online WHERE online = 1"); // fetch the results / convert into an array WHILE($rows = mysql_fetch_array($query)): $users = $rows['name']; echo "'<font color='black'>Online:<font color='green'>$users, </font></font>;" endwhile; ?> or here http://pastebin.com/ZYh4t2pD Thanks Edit: Found the php tag |