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PHP - Syntax Error:check The Manual That Corresponds To Your Mysql Server Version For The Right Syntax To Use Near ''sfapp' ('surname_add','forename_add')
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.
Similar TutorialsI 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, 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 $pastelink? & If so, how can I fix it? I get this error when trying to run this code Quote "Username: magessssss EXP You modified: atkExp - 509 Level modified: atkLvl - 2 __________________________ Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''atkLvl'='2','atkExp'='509' WHERE user='MAGEssssss'' at line 1" <?php if($_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 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, Hi guys i have the following code which is misbehaving can anyone see where its wrong? Code: [Select] <?php include 'dbc.php'; page_protect(); company(); $Referrer = mysql_query("SELECT * FROM Referrer WHERE SentOut='0' "); if (isset($_POST['submit'])) { //Assign each array to a variable $StaffMember = $_POST['StaffMember']; $referrer = $_POST['referrer']; $referred = $_POST['referred']; $SentOut = $_POST['SentOut']; $today = date("y.m.d H:i:s"); $user_id = $_SESSION['user_id']; $IssueNum = $_POST['Referrerid']; $limit = count($StaffMember); for($k=0;$k<$limit;$k++){ $msg[] = "$limit New KPI's Added"; $values[$k] = array( $StaffMember[$k],$referrer[$k],$referred[$k],$SentOut[$k],$today,$user_id ); // build the array of values for the query string } foreach( $values as $key => $value ) { $query = "UPDATE `Referrer` (StaffMember, referer, referred, SentOut, SentOutDate, SentOutBy) VALUES ('" . implode( '\', \'', $value ) . "') WHERE IssueNum= '{$IssueNum[$key]}'"; mysql_query($query) or die(mysql_error()); } } if (checkAdmin()) { ?> <html> <head> <title>Book Off Holiday</title> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1"> <script src="php_calendar/scripts.js" type="text/javascript"></script> <link href="styles.css" rel="stylesheet" type="text/css"> </head> <body> <form name="form" action="SendReferrers.php" method="post"> <table width="100%" border="0" cellspacing="0" cellpadding="5" class="main"> <tr> <td colspan="3"> </td> </tr> <td width="160" valign="top"> <?php if (isset($_SESSION['user_id'])) { } ?> <a href="admin.php">Admin CP </a> </td> <td width="732" valign="top"> <p> <h3 class="titlehdr">New KPI</h3> <table width="300px" border="0" align="Centre" cellpadding="2" cellspacing="0"> <tr bgcolor="#000050"> <td width="20px"><h3 class="Text2">Referrer ID</h3></td> <td width="20px"><h3 class="Text2">Staff Member</h3></td> <td width="20px"><h3 class="Text2">referrer</h3></td> <td width="20px"><h3 class="Text2">referred</h3></td> <td width="40px"><h3 class="Text2">Sent Out</h3></td> </tr> <?php while ($rrows = mysql_fetch_array($Referrer)) {?> <tr> <td><h3 class="Text3"><input type="" name="Referrerid[]" id="Referrerid[]" size="4" value="<?php echo $rrows['IssueNum'];?>" /></h3></td> <td><h3 class="Text3"><input type="" name="StaffMember[]" id="StaffMember[]" size="4" value="<?php echo $rrows['StaffMember'];?>" /></h3></td> <td><h3 class="Text3"><input type="" name="referrer[]" id="referrer[]" value="<?php echo $rrows['referer'];?>" /></h3></td> <td><h3 class="Text3"><input type="" name="referred[]" id="referred[]" value="<?php echo $rrows['referred'];?>" /></h3></td> <td><h3 class="Text3"><input name="SentOut[]" type="checkbox" value="1" id="SentOut[]"></h3></td> </tr> <?php } ?> </table> <input name="submit" type="submit" id="submit" value="Create"> </table> </form> </body> </html> <?php } ?> the error i get when the submit button is clicked is 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 '(StaffMember, referer, referred, SentOut, SentOutDate, SentOutBy) VALUES' at line 1 any help would be appriciative Howdy folks, I am creating a Facebook app for a bit of fun and practice and getting the following error in index.php: Code: [Select] Invalid query -- SELECT * FROM `results` WHERE `resultLow` <= AND `resultHigh`>= -- 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 'AND `resultHigh`>=' at line 1 Here is the area: Code: [Select] $res = query("SELECT * FROM `results` WHERE `resultLow` <=$user_score AND `resultHigh`>=$user_score"); Any help would be appreciated. I have been pulling my hair out for the lasy 3 hours i am trying to update a MySql table but i cant get it too work, i just keep getting MySql error #1064 - You have an error in your SQL syntax; if i just update 1 field it works fine but if i try to update more than 1 field it dosent work, Help Please! <?php $root = $_SERVER['DOCUMENT_ROOT']; require("$root/include/mysqldb.php"); require("$root/include/incpost.php"); $con = mysql_connect("$dbhost","$dbuser","$dbpass"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("$dbame", $con); mysql_query("UPDATE Reg_Profile_p SET build='$build' col='$col' size='$size' WHERE uin = '$uinco'"); ?> This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=352460.0 Guys, I'm not at all able to insert characters like " ' ` and all sorts into my database or it will always return an error. I just created a textarea field in my site of which I just want to istore all those collected datas into my database for later retrieval and all sorts. Please help! Hey all, I keep getting this 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 '' at line 2 When I use the script below. I'm finding it a bit confusing because everything about it continues to work, it's just it gives me an error. When the script runs, the outcome is "Success! Your dog now looks far more energetic! Looks like this food is all used up." Followed by the error, which cuts the remainder of the page off. Does anybody know why it's doing this? $dogyay = $_POST['dogid']; $checkenergy = "SELECT energy FROM dogs WHERE id=$dogyay"; $energylevel = mysql_query($checkenergy) or die(mysql_error()); $row = mysql_fetch_array($energylevel) or die(mysql_error()); if($row['energy'] >= 100) { echo "<b>Oops!</b> Looks like your dog is full right now...";} else{ echo "<b>Success! Your dog now looks far more energetic!</b><br><br>"; $sql11="UPDATE dogs SET energy=energy + 50 WHERE id=$dogyay"; $result11=mysql_query($sql11); $sql12="UPDATE items SET uses = uses - 1 WHERE itemid=$id"; $result12=mysql_query($sql12); $checkuses = "SELECT uses FROM items WHERE itemid=$id"; $useslevel = mysql_query($checkuses) or die(mysql_error()); $row = mysql_fetch_array($useslevel) or die(mysql_error()); if($row['uses'] == 0) { echo "Looks like this food is all used up.<bR><br>"; mysql_query("DELETE FROM items WHERE itemid='$id'") or die(mysql_error());} Thanks ! Im building a list of offers and adding them to a table in a database. Pretty much all it is is HTML. Im inserting an ahref link that has a php echo in it. So it looks like this: <div class="offerlinks"><a href="http://website.com/offer/blahblah&blah=blah&sid=<?php echo $_SESSION['uid'];?>">Offer name</a><br><b>Info:</b> Signup<br><b>Value</b> 1 pt</div> When I insert this (through my form) I get mysql error 1064 which is syntax error. I tested it without the php & it gives me 0, which worked fine. I need the php code so I can append userid to the SID var. Am I doing something wrong? Well I guess I obviously am so the real question is what am I doing wrong & how could I do it the right way? Thanks guys This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=342913.0 Hi.. I create mysql syntax for query testing before i input to my php code here is my mysql code: Code: [Select] set @t = 0; set @rqty=31968; SELECT LOT_CODE as code, DATE_ENTRY, CASE WHEN @t+OUTPUT_QTY > @rqty THEN @rqty -@t ELSE OUTPUT_QTY END as qty, @t := @t + d.OUTPUT_QTY as cumulative FROM dipping d WHERE SUBSTR(LOT_CODE, 9,4) = 'P28' AND (@t < @rqty); and i attach the sample output of the above query. Now that query test is work i will input that code to my php codes. $sql = "SELECT SKUCode, Materials, Comp, Qty FROM bom WHERE SKUCode = '$SKUCode'"; $res = mysql_query($sql, $con); ($row = mysql_fetch_assoc($res)); $Materials = $row['Materials']; $Qty = $row['Qty']; $Comp = $row['Comp']; //P28 //-----Compute Req Qty and Save to table---// $ReqQty = $Qty * $POReq; // 31968 $sql = "UPDATE bom SET ReqQty = '$ReqQty' WHERE SKUCode = '$SKUCode' AND Materials = '$Materials'"; $resReqQty = mysql_query($sql, $con); $t = 0; $sql = "SELECT LOT_CODE as code, DATE_ENTRY, CASE WHEN $t+OUTPUT_QTY > $ReqQty THEN $ReqQty -$t ELSE OUTPUT_QTY END as qty, $t := $t + d.OUTPUT_QTY as cumulative FROM dipping d WHERE SUBSTR(LOT_CODE, 9,4) = '$Comp' AND ($t < $ReqQty)"; when I echo the query: I got this: SELECT LOT_CODE as code, DATE_ENTRY, CASE WHEN 0+OUTPUT_QTY > 31968 THEN 31968 -0 ELSE OUTPUT_QTY END as qty, 0 := 0 + d.OUTPUT_QTY as cumulative FROM dipping d WHERE SUBSTR(LOT_CODE, 9,4) = 'P28' AND (0 < 31968) then I run it to the sql and I got an error: Error Code : 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 ':= 0 + d.OUTPUT_QTY as cumulative FROM dipping d WHERE SUBSTR(LOT_CODE, 9,4) = '' at line 1 (0 ms taken) Any help is highly appreciated Thank you so much The issue is there seems to be NO syntax error. 1. There is no relevant code before or after this line. 2. Yes, one would think a ! should be there as did I (I didn't write the code), however, even with the ! it still gives the same error. if (function_exists('gzcompress')) die(FUNCTION_NOT_FOUND); is giving me a syntax error, unexpected 'if', expecting 'function' or 'const' I am updating this code from PHP5.3 to PHP7.4 and I can't figure out what the syntax problem is since PHP allows this. I am using Eclipse PHP to do the conversion. Hi all, I am new to PHP and wish to use it simply to collect form data and email it to myself. From what I have gathered from PHP tutorials I have write the following script but I am getting syntax error warnings and cant figure out why. Can somebody please tell me where I am going wrong and if this script will actually work? The line I am getting errors on is Code: [Select] <?php //message contains $message = "Name:$name</br>Email:$email</br>Phone:$phone</br>Preferred Date:$date</br>Best Time To Contact:$time"; ?> Thank you in advance. Code: [Select] <body> <?php //my email $recipient = "info@hairlosslasergroup.co.uk"; //persons name $name = htmlspecialchars($_POST['name']); //persons email $email = htmlspecialchars($_POST['email']); if (!preg_match("/([\w\-]+\@[\w\-]+\.[\w\-]+)/",$email)) { die("E-mail address not valid"); } //persons phone number $phone = htmlspecialchars($_POST['phone']); //preferred date $date = htmlspecialchars($_POST['comments']); //best time $time = htmlspecialchars($_POST['time']); //check required fields $required = check_input($_POST['required'], "Please go back and complete all required fields"); //email subject $subject = ($_POST['subject'] //message contains $message = "Name:$name</br>Email:$email</br>Phone:$phone</br>Preferred Date:$date</br>Best Time To Contact:$time"; //message mail($recipient, $subject, $message); //redirect to thank you page header('Location:$redirect'); ?> </body> </html> Hi, I joined new here, and I am really having hard time figuring out what's wrong with the code below which isn't working. if (isset($_POST['privacy_submit'])) { if (($_SERVER['REQUEST_METHOD'] == "POST")) { $privacy_upd = mysql_query("UPDATE clf_privacy_terms SET 'p_bannerads' = '".$_POST['p_bannerads']."', 'p_shareinfo' = '".$_POST['p_shareinfo']."', 'p_crossmarketing' = '".$_POST['p_crossmarketing']."', 'p_tacking' = '".$_POST['p_tacking']."', 'p_sendcommunication' = '".$_POST['p_sendcommunication']."', 'p_under13' = '".$_POST['p_under13']."', 'p_internationally' = '".$_POST['p_internationally']."', 'p_discloselegal' = '".$_POST['p_discloselegal']."', 'p_server_country' = '".$_POST['p_server_country']."', 'p_forums' = '".$_POST['p_forums']."', 'p_newslettermodule' = '".$_POST['p_newslettermodule']."', 'p_membershipmodule' = '".$_POST['p_membershipmodule']."' ") or die(mysql_error()); $msg="Successfully Updated!"; } } else { $msg="<font color=\"red\">Incorrect method of submission.</font> Please try again."; } ?> <h2>Privacy Policy</h2> <?php if ($privacy_upd) { echo ($msg); } ?> <form class="box" name="privacy_upd" action="gen.php" method="post"> ......................................................................................... ........................ MySQL Error returns the following: Code: [Select] 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 ''p_bannerads' = 'do not', 'p_shareinfo' = 'do not', 'p_cross' at line 2 Let me know if any of your guys can solve it for me, I would appreciate since I am not able to solve it for last two days already. Thank you... 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> |
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