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PHP - Unknown Mysql Syntax Error...
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 ! Similar TutorialsYou have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '["rid"]., SELECT refid1 FROM oto_members WHERE id='13', SELECT refid2 FR' at line 5 With this script area: $affiliate1=('.$_REQUEST["rid"].'); $affiliate2=("SELECT refid1 FROM oto_members WHERE id='".$_REQUEST["rid"]."'"); $affiliate3=("SELECT refid2 FROM oto_members WHERE id='".$_REQUEST["rid"]."'"); $affiliate4=("SELECT refid3 FROM oto_members WHERE id='".$_REQUEST["rid"]."'"); $affiliate5=("SELECT refid4 FROM oto_members WHERE id='".$_REQUEST["rid"]."'"); $affiliate6=("SELECT refid5 FROM oto_members WHERE id='".$_REQUEST["rid"]."'"); $affiliate7=("SELECT refid6 FROM oto_members WHERE id='".$_REQUEST["rid"]."'"); $affiliat8e=("SELECT refid7 FROM oto_members WHERE id='".$_REQUEST["rid"]."'"); $affiliate9=("SELECT refid8 FROM oto_members WHERE id='".$_REQUEST["rid"]."'"); $affiliate10=("SELECT refid9 FROM oto_members WHERE id='".$_REQUEST["rid"]."'"); $qry="INSERT INTO ".$prefix."members(firstname,lastname,email,address,city,state,postcode,country,telephone,username,password,refid1,refid2,refid3,refid4,refid5,refid6,refid7,refid8,refid9,refid10,geo,paypal_email,joindate,mtype,groupid,cb_id,status,signupip) VALUES('".$_REQUEST["firstname"]."','".$_REQUEST["lastname"]."','".$_REQUEST["email"]."','".$_REQUEST["address"]."','".$_REQUEST["city"]."','".$_REQUEST["state"]."','".$_REQUEST["postcode"]."','".$_REQUEST["country"]."','".$_REQUEST["telephone"]."','".$_REQUEST["username"]."','".md5($_REQUEST["password"])."', $affiliate1, $affiliate2, $affiliate3, $affiliate4, $affiliate5, $affiliate6, $affiliate7, $affiliate8, $affiliate9, $affiliate10 ,'".$_REQUEST["geo"]."','".$_REQUEST["paypal_email"]."',NOW(),".$_REQUEST["mtid"].",$eogroup,'".$_REQUEST["cb_id"]."','$memberstatus','$signupip')"; I'm a newbie, can you please help? I'm getting a mysql error "Unknown column 'E0000001' in 'where clause'" The id is in the URL: ...user-profile.php?id=E0000001 My Query $query = mysqli_query($con, "SELECT * FROM UserList WHERE UserID=".$id) or die (mysqli_error($con)); Thank you Edited September 18, 2019 by ACBMSEHey, So what im trying to do is put my database variables into a session array. So this is what im trying to accomplish... $_SESSION['Name_of_Row'] = $value This is the script I wrote: Code: [Select] Function setupSession(){ session_start(); $query = "SELECT * FROM users WHERE u_id ='{$this->u_id}'"; $result = mysql_query($query); $row = mysql_fetch_array($result); foreach($row as $key => $value){ if(!empty($value)){ $_SESSION[$key] = $value; } } } When that runs I get the following warning. Can anyone tell me what this means and how to fix it? Error: Notice: Unknown: Skipping numeric key 0 in Unknown on line 0 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.
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 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! This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=342913.0 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 Hello! I have a strange error on my PHP script and i dont how to fix it. If someone can help me, please help me then! Here is my error: Code: [Select] logout(); } else { $iq = mysql_query("SELECT * FROM users WHERE username='{$signin_username}' AND password='{$signin_password}' AND suspended='0' LIMIT 1;"); $ir = mysql_fetch_array($iq); $_SESSION['me'] = $ir; } } } } else { die("The configuration did not recieve appropriate variables to accept your request."); } if ($set['next_clearup'] < time ()) { $next_clearup = time () + 60 * 60 * 24; mysql_query ('' . 'UPDATE settings SET set_value=\'' . $next_clearup . '\' WHERE set_name=\'next_clearup\' LIMIT 1;'); mysql_query ('UPDATE users SET ads_clicked=\'\' WHERE ads_clicked!=\'\''); } } ?> Warning: include(THDIRindex.php) [function.include]: failed to open stream: No such file or directory in C:\xampp\htdocs\Upload\index.php on line 16 Warning: include() [function.include]: Failed opening 'THDIRindex.php' for inclusion (include_path='.;\xampp\php\PEAR') in C:\xampp\htdocs\Upload\index.php on line 16 And here is the PHP file the error is in: Code: [Select] <?php session_start(); include_once('lib/lib.php'); include_once('lib/configuration.php'); $ddir = THDIR.$do->get_file_url(); include($ddir); if(file_exists(HEADER)) { include_once(HEADER); } if($contents) { print $contents; } if(file_exists(FOOTER)) { include_once(FOOTER); } ?> Help ASAP if you can! I Am getting along with php better than I was previously. But this 68 year old brain still refuses to learn very fast! Here is the error I'm receiving when I'm trying to open the db: Parse error: syntax error, unexpected T_VARIABLE in /home/taft65/public_html/memProtest.php on line 197 <?php error_reporting(E_ERROR | E_PARSE | E_CORE_ERROR); $host = "localhost"; $dbname="database;" Failing ------>$username = "user"; $password="drDedf#hj"; I understand you do not need to declare varibles in PHP, Correct? I checked the db to ensure that I'm calling the correct value. NuSpherePhpEd to validate the code. I also check it with DSV PHP Editor. Both come up with the same error. I'm also using MyPhpAdmin to create the database and tables. I know also to place this calling info in another folder and include it by calling it with a php include statement. I just have it within the code to quickly test it. Thank you for any assistance. Bob... im using 000webhost as a test site and when i run this code it redirects me to there err page but with no error message. the sql query works fine in phpmyadmin and i added the rest of the code to try the php side. i "think" the problem is the echo $rows"value's" as im unsure of what the $vars should be <?php include("config.php"); // Connect to server and select database. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $sql = "SELECT make, COUNT(*) AS total, SUM(IF(comments = \'pass\', 1, 0)) AS withComments FROM dsgi_serval GROUP BY make ORDER BY COUNT(*) DESC"; $result=mysql_query($sql); echo "$sql"; echo "$result"; ?> <table><tr> <td colspan="4"><strong>List data from mysql </strong> </td> </tr> <tr> <td align="center"><strong>make</strong></td> <td align="center"><strong>Total</strong></td> <td align="center"><strong>Validated</strong></td> </tr> <?php while($rows=mysql_fetch_array($result)){ ?> <tr> <td><?php echo $rows['make']; ?></td> <td><?php echo $rows['total']; ?></td> <td><?php echo $rows['withcomments']; ?></td> </tr> <?php } ?> </table> <?php mysql_close(); ?> i am trying to upload a .sql file on godaddy database and i am getting this error . Help me.
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
Hi there I'm trying to insert audi files' records into mysql database this is how the records insertion looks like: Code: [Select] $fileName = $_FILES['uploaded']['name'];//name $tmpName = $_FILES['uploaded']['tmp_name'];//temp location $fileSize = $_FILES['uploaded']['size'];//size of the file $fileType = $_FILES['uploaded']['type'];//type of file $error = $_FILES['uploaded']['error'];//verifys errprs $ext = substr($fileName, strrpos($fileName, '.') +1); //this will get the extention out of the file name e.g. .mp3 //check that a file is passed by and no errors if(isset($fileName) && $error == 0 && $fileSize != 0){ //condition to accept only certain file types/extentions if($ext == "mp3" || $ext == "wma" || $ext == "wav"){ //get file content $fp = fopen($tmpName, 'r'); $content = fread($fp, filesize($tmpName)); $content = addslashes($content); fclose($fp); if(!get_magic_quotes_gpc()) { $fileName = addslashes($fileName); } //query to retrieve user's id $userid = mysql_query("select userID from user where username = '$username'"); //to get id by username $row = mysql_fetch_assoc($userid); $userid = $row['userID']; $query = "INSERT INTO tracks (trackName, userID, tag, price, file, fileName, fileSize, fileType) VALUES ('$tname','$userid','$tag','$price','$content','$fileName','$fileSize' , '$fileType)"; mysql_query($query) or die (mysql_error()); when i send it i get this error: Unknown column 'application' in 'field list' i did set data type for the fileType field to varchar. when i remove the fileType all together the record is inserted successfully into database though. ?!! can anyone help please I have searched this forum as well as over 200 other forums and have not found the answer that is specific to my question. I have shortened my code drastically to assist in resolving this quickly -
I have a search form that has criteria for the search criteria with "virtual" "columns" in an array but it's not working. If I search one column at a time it works just fine but when I try to search 8 columns with one select I get the following error: SELECT Error: Unknown column 'achievements' in 'where clause'.
When a user selects search in Achievements, I need it to look at all 8 columns that are associated with achievements and bring back the results that match - the same as if the user selects search in Associations, I need it to look at all 5 columns and bring back the results that match.
My shortened code is as follows:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Search</title>
</head>
<body>
<form name="search" action="" method="POST">
<p>Search:</p>
<p>
Achievements/Associations: <input type="text" name="find1" /> in
<Select NAME="field1">
<Option VALUE="achievements">Achievements</option>
<Option VALUE="associations">Associations</option>
</Select>
<br><br>
Secondary Education: <input type="text" name="find2" /> in
<Select NAME="field2">
<Option VALUE="edu1sectype">Highest Certificate Attained</option>
<Option VALUE="edu1secname">Highest Grade Passed</option>
<Option VALUE="edu1secinst">Name of High School</option>
<Option VALUE="edu1secdate">Date Completed</option>
<Option VALUE="edu1secinsttyp">Type of Institution</option>
<Option VALUE="subjects">Subjects</option>
</Select>
<br><br>
<input type="hidden" name="searching" value="yes" />
<input type="submit" name="search" value="Search" />
</p>
</form>
<?php
$searching = $_POST['searching'];
$find1 = $_POST['find1'];
$field1 = $_POST['field1'];
$find2 = $_POST['find2'];
$field2 = $_POST['field2'];
if ($searching =="yes")
{
echo "<br><b>Searched For:</b> $find1 $find2<br>";
echo "<br><h2>Results</h2><p>";
//If they did not enter a search term we give them an error
// Otherwise we connect to our Database
include_once "connect_to_mysql.php";
mysql_select_db("table_name") or die(mysql_error());
// We preform a bit of filtering
$find = strtoupper($find);
$find = strip_tags($find);
$find = trim($find);
$find = mysql_real_escape_string($find);
$field = mysql_real_escape_string($field);
$data = mysql_query("SELECT * FROM table_name
WHERE upper(".$field1.") LIKE '%$find1%'
AND upper(".$field2.") LIKE '%$find2%'
") or die("SELECT Error: ".mysql_error());
$result = mysql_query("SELECT * FROM table_name
WHERE upper($field1) LIKE '%$find1%'
AND upper($field2) LIKE '%$find2%'
") or die("SELECT Error: ".mysql_error());
$num_rows = mysql_num_rows($result);
echo "There are $num_rows records:<br>";
echo '<center>';
echo "<table border='1' cellpadding='5' width='990'>";
// set table headers
echo "<tr><th>Reference</th>
<th>First Name</th>
<th>Last Name</th>
</tr>";
//get images and names in two arrays
$name= $row["name"];
$surname= $row["surname"];
$achieve1 = $row["achieve1"];
$achieve2 = $row["achieve2"];
$achieve3 = $row["achieve3"];
$achieve4 = $row["achieve4"];
$achieve5 = $row["achieve5"];
$achieve6 = $row["achieve6"];
$achieve7 = $row["achieve7"];
$achieve8 = $row["achieve8"];
$assoc1 = $row["assoc1"];
$assoc2 = $row["assoc2"];
$assoc3 = $row["assoc3"];
$assoc4 = $row["assoc4"];
$assoc5 = $row["assoc5"];
$edu1sectype = $row["edu1sectype"];
$edu1secinst = $row["edu1secinst"];
$edu1secname = $row["edu1secname"];
$edu1secdate = $row["edu1secdate"];
$edu1secinsttyp = $row["edu1secinsttyp"];
$subject1 = $row["subject1"];
$subject2 = $row["subject2"];
$subject3 = $row["subject3"];
$subject4 = $row["subject4"];
$subject5 = $row["subject5"];
$subject6 = $row["subject6"];
$subject7 = $row["subject7"];
$subject8 = $row["subject8"];
$compsoft1name = $row["compsoft1name"];
$compsoft2name = $row["compsoft2name"];
$compsoft3name = $row["compsoft3name"];
$compsoft4name = $row["compsoft4name"];
$compsoft5name = $row["compsoft5name"];
$compsoft6name = $row["compsoft6name"];
$achievements = array('achieve1', 'achieve2', 'achieve3', 'achieve4', 'achieve5', 'achieve6', 'achieve7', 'achieve8');
$associations = array('assoc1', 'assoc2', 'assoc3', 'assoc4', 'assoc5');
$subjects = array('subject1', 'subject2', 'subject3', 'subject4', 'subject5', 'subject6', 'subject7', 'subject8' );
$compsoft = array('compsoft1name', 'compsoft2name', 'compsoft3name', 'compsoft4name', 'compsoft5name', 'compsoft6name');
while ($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td ALIGN=LEFT>" . $row['id'] . "</td>";
echo "<td ALIGN=LEFT>" . $row['name'] . "</td>";
echo "<td ALIGN=LEFT>" . $row['surname'] . "</td>";
echo "</tr>";
}
echo "</table>";
//This counts the number or results - and if there wasn't any it gives them a little message explaining that
$anymatches=mysql_num_rows($data);
if ($anymatches == 0) {
echo "Sorry, but we can not find an entry to match your query";
}
}
?>
</body>
</html>
Any assistance will be greatly appreciated as I have been working on this website for the past 4 months which has totalled over 150 pages and this is one of the last pages left to program and it's taken 6 days to get to this search page to this point.
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