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PHP - Please Help - Mysql Syntax Error In The Insertion
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! 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 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.
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'"); ?> 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. 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 This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=352460.0 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 This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=320009.0 Hi guys.I am making a Query Analyzer in php which takes the query string inside a form and out puts the result table in that particular page.How do i output the notification by MySQL when you do a successful insertion of data into mysql.i want to know if there are any methods in php which gives you that such as "1 row affected","Query took 0.00sec".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> 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.. 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 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 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()) ?> 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> |
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