PHP - Error Displayed In My Query
I have following error when i try to enter my student info into the database.
! ) Warning: Header may not contain more than a single header, new line detected. in C:\wamp\www\Student registration\new student registration\newStudentRegistrationFormvalidation.php on line 43 newStudentRegistrationFormvalidation.php <?php $admission_no=$_POST['admission_no']; $admission_date=$_POST['admission_date']; $full_name=$_POST['full_name']; $name_with_initial=$_POST['name_with_initial']; $date_of_birth=$_POST['date_of_birth']; $religion=$_POST['religion']; $address=$_POST['address']; $telephone=$_POST['telephone']; $grade_on_admission=$_POST['grade_on_admission']; $grade_ID=$_POST['grade_ID']; $stream_ID=isset($_POST['stream_ID']) ? $_POST['stream_ID'] :''; $class_ID=$_POST['class_ID']; $student_house=$_POST['student_house']; $password=$_POST['password']; $description_about_st=$_POST['description_about_st']; $payment=$_POST["payment"]; $currentdate=getdate(time()); $year=$currentdate["year"]; //admission number validation $answer=''; $con=mysql_connect("localhost","root",""); mysql_select_db("student_management",$con); $query="SELECT admission_no FROM student_info WHERE student_info.admission_no='$admission_no'"; $result=mysql_query($query); while($row=mysql_fetch_array($result)){ $answer=$row['admission_no'];} if($answer==0) { //line 43 header("location:student registrationDatabase.php?admission_no=".$admission_no."&year=".$year."&admission_date=".$admission_date."&full_name=".$full_name."&name_with_initial=".$name_with_initial."&date_of_birth=".$date_of_birth."&religion=".$religion."&address=".$address."&telephone=".$telephone."&grade_on_admission=".$grade_on_admission."&grade_ID=".$grade_ID."&stream_ID=".$stream_ID."&class_ID=".$class_ID."&student_house=".$student_house."&password=".$password."&description_about_st=".$description_about_st."&payment=".$payment); exit(); }else{ ?> <body> <?php echo "Admission number".$admission_no."This student has been alredy entered to the system ."."<BR>"."<BR>"."<BR>"; echo "<a href='newStudentRegistrationForm.php'>GO to manage student details page</a> "; exit(); }?> studentregistrationDatabase.php <?PHP $admission_no=$_POST['admission_no']; $admission_date=$_POST['admission_date']; $full_name=$_POST['full_name']; $name_with_initials=$_POST['name_with_initial']; $dob=$_POST['date_of_birth']; $religion=$_POST['religion']; //$gender=$_GET['gender']; $address=$_POST['address']; $telephone=$_POST['telephone']; $grade_on_admission=$_POST['grade_on_admission']; $present_grade=$_POST['grade_ID']; $stream=$_POST['stream_ID']; $present_class=$_POST['class_ID']; $student_house=$_POST['student_house']; $password=$_POST['password']; $description_about_st=$_POST['description_about_st']; $payment=$_POST["payment"]; $current=getdate(time()); $year=$current["year"]; $today = date("Y-m-d"); $con=mysql_connect("localhost","root",""); mysql_select_db("student_management",$con); //insert to database $query="select count(*) from student_info where admission_no='$admission_no'"; $result=mysql_query($query); $row=mysql_fetch_array($result); if($row[0]==0) { $query="insert into student_info values(null,'$admission_no','$admission_date','$full_name','$name_with_initials','$dob','$religion','$address','$telephone','$grade_on_admission','$password','$student_house','$description_about_st')"; $result=mysql_query($query); if($stream!=null){ $query="select class_id from class where class.grade_id='$present_grade' and class.class_name='$present_class' and class.stream='$stream'"; }else{ $query="select class_id from class where class.grade_id='$present_grade' and class.class_name='$present_class'"; } $result=mysql_query($query); while($row=mysql_fetch_array($result)){ $class_id=$row['class_id']; } $query="insert into student_class values('$admission_no','$class_id','$year')"; $result=mysql_query($query); if($payment=="2000") { $query="insert into payment_amount values(null,'2000','$today')"; $result=mysql_query($query); $query="SELECT amount_id FROM payment_amount order by amount_id DESC limit 1"; $result=mysql_query($query); while($row = mysql_fetch_array($result)) { $amount_id=$row['amount_id']; } $query="insert into payment values('$admission_no','$amount_id')"; $result=mysql_query($query); }else{ $query="insert into payment_amount values(null,'0','$today')"; $result=mysql_query($query); $query="SELECT amount_id FROM payment_amount order by amount_id DESC limit 1"; $result=mysql_query($query); while($row = mysql_fetch_array($result)) { $amount_id=$row['amount_id']; } $query="insert into payment values('$admission_no','$amount_id')"; $result=mysql_query($query); } header("location:../new student registration-parent details/studentRegistrationParentDetailsForm.php?admission_no=".$admission_no); exit(); } else{ ?> <?php echo "Admission number".$admission_no."This student has been already entered to the system"."<BR>"."<BR>"."<BR>"; echo "<a href='student registration.php'>GO to registration page</a>"; exit(); } ?> Similar TutorialsHi, when I browse my php website I recieve sometimes error 504. when it happens I try to refresh and get this error: Gateway Time-out The gateway did not receive a timely response from the upstream server or application. Additionally, a 404 Not Found error was encountered while trying to use an ErrorDocument to handle the request. This website is in a shared hosting. Where can I find logs that will help me correct the issue? Thank you. I'm trying to display error messages on the same page that submits a form (register_page.php). So in the script that executes the form submission, I create an array to hold all the error messages, and towards the end of the script I include a conditional at the end, which does the following if there are errors: First I convert the $errors array into a string, url encode it, then pass it into the url of the target page (register_page.php). Then on the target page, I retrieve the message through GET, url decode it, reconvert it to an array and try to print it out the contents of the array. So here is the excerpt of the conditional on the execution page: Code: [Select] //Display error messages. } else {// if errors array is not empty //Conver the errors array into a string $errors_string = implode(" ", $errors); //Encode the imploded string. $message = urlencode($errors_string); header("Location: register_page.php?message=$message"); } And here is the code for the target page: Code: [Select] </div> <!--closes register box--> <div id="error"> <?php if(isset($_GET['message'])) { //Decode the url $errors_string = urldecode($_GET['message']); //Explode the decoded errors string back to an array. $errors = explode(" ", $errors_string); echo '<h3>Error!</h3> The following error(s) occured:<br />'; foreach ($errors as $msg) { echo "$msg<br/>\n"; } } ?> </div> <!--closes error--> Now everything works just fine, except that the error messages are printed one word per line for example Your is not a valid address. I can't seem to find the fluke in the code. Any ideas? Also is there any potential security risk for passing error messages in a url? No sensitive information is being transmitted. Just error messages. Here is my code: // Start MySQL Query for Records $query = "SELECT codes_update_no_join_1b" . "SET orig_code_1 = new_code_1, orig_code_2 = new_code_2" . "WHERE concat(orig_code_1, orig_code_2) = concat(old_code_1, old_code_2)"; $results = mysql_query($query) or die(mysql_error()); // End MySQL Query for Records This query runs perfectly fine when run direct as SQL in phpMyAdmin, but throws this error when running in my script??? Why is this??? 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 '= new_code_1, orig_code_2 = new_code_2WHERE concat(orig_code_1, orig_c' at line 1 A friend of mine must of changed something on the site while I was asleep last night and now all the site says when you go to it is: Error Database query error Warning: mail() [function.mail]: SMTP server response: 530 SMTP authentication is required. in C:\xampp\htdocs\inc\utils.inc.php on line 449 I'm not exactly sure what he did since I can't contact him. Can anyone help me fix this? Hello, i have one little problem and can't pass it. Problem is i need to call new sql query inside another query. Am trying to make accordion which will put result of first query($sql) like title and result of second query($sql2) like list of current item. All time am getting error : Quote Warning: odbc_exec() [function.odbc-exec]: SQL error: [Microsoft][ODBC SQL Server Driver]Connection is busy with results for another hstmt, SQL state S1000 in SQLExecDirect in ..... I know reasone is because i use query inside query so am trying to figure is there any way to bypass it or make it work. Example: Code: [Select] $sql="EXECUTE _PROCEDURE1 '".$date."',''.$code."; $rs=odbc_exec($conn,$sql); if (!$rs){exit("Error in SQL");} while (odbc_fetch_row($rs)){ $id_number=odbc_result($rs,"ID"); $name=odbc_result($rs,"NAME"); echo $id_number.' - '.$name; $sql2="EXECUTE _PROCEDURE2 '".$id_number."',''.$name."; $name=odbc_exec($conn,$sql2); while(odbc_fetch_row($popust)){ $detail = odbc_result($sql2,"DETAILS"); $detail2 = odbc_result($sql2,"DETAILS2"); $detail3 = odbc_result($sql2,"DETAILS3"); echo $detail.' - '.$detail2.' - '.$detail3; } } I hope i explained it well. Thanks. Hi all, having a strange problem with my query Its only returning some of my data, and in the format User 1 <br /> <br /> User 2 <br /> <br /> <br /> <br /> Code: [Select] $newmembers = "SELECT * FROM users WHERE linked_user IS NOT NULL ORDER BY datejoined LIMIT 6"; $nmresult = mysql_query($newmembers); while($row = mysql_fetch_array($nmresult)){ echo $row['linked_user']; echo "<br />";} ?>  Warning: mysqli_query() expects parameter 1 to be mysqli, null given in /home/omerbsh/hallofblogs.com/autoTwit/Library/database.php on line 13 Warning: mysqli_error() expects exactly 1 parameter, 0 given in /home/omerbsh/hallofblogs.com/autoTwit/Library/database.php on line 13 Error: The Query:INSERT INTO twitter_profiles VALUES('dfggffg','dgfgfdg','1') Code: [Select] mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, boolean i get that error with this code: $online_query = $link->simple_query('u_username', 'users', 'u_online=1 AND u_hidden != 1', '0u_username'); while($online_info = $link->fetch_array($online_query)) //THIS LINE THROWS THE ERROR the simple_query function is: function simple_query($fields, $table, $clause, $order) { global $link, $config; if(!empty($clause)) { $clause = "WHERE $clause"; } else { $clause = ''; } if(!empty($order)) { $direction = $order[0]; switch($direction) { case '0': $direction = 'ASC"'; break; case '1': $direction = 'DESC'; break; } $order = substr($order, 1, strlen($order)); $order = "ORDER BY $order $direction"; } $query = mysqli_query($this->link, "SELECT $fields FROM ".TBL_PREFIX."$table $clause $order"); return $query; } when i use a normal query instead of my simple_query function it works fine. it also worked fine before i made the database class. Where am i going wrong? Hi, i'm currently coding a new inbox for my website, but ive got an error which says: 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 'to='NoName' ORDER BY id DESC' at line 1 Im not sure why ive got that as everything seems to be fine :S My Code: $user=$_SESSION['username']; $get_messages = mysql_query("SELECT `id` FROM `inbox` WHERE to='$user' ORDER BY `id` DESC") or die("Error on line 9 - " . mysql_error()); Thanks for any help/advise given. Quote
Unknown column 'Carmel' in 'where clause'
Passing variable via the URL: https://www.courtsideindiana.com/season-preview/19-20/sectional1920/?sectional=8&school=Carmel Sectional = 8 School = Carmel Before I added the &school=Carmel, it was working, just echoing the total list of schools in the table. $sectional = $_GET['sectional']; $school = $_GET['school']; echo $school; // Query $query = "SELECT * FROM a_schools WHERE sectional=".$sectional." AND school=" . $school .""; $results = mysqli_query($con,$query); echo mysqli_error($con); while($row = mysqli_fetch_assoc($results)) { echo $row['school'] . '<br>'; }
I am getting an empty query error on this line Code: [Select] $mysql->query($insert); here is my code Code: [Select] if(isset($_POST['submit'])){ $firstName = $_POST['firstName']; $lastName = $_POST['lastName']; $companyName = $_POST['companyName']; $homePhone = $_POST['homePhone']; $cellPhone = $_POST['cellPhone']; $companyPhone = $_POST['companyPhone']; //checking the values are filled //echo $firstName. " " . $lastName . " " . $companyName . " " . $homePhone . " " . $cellPhone . " " . $companyPhone; $insert = $mysql->query("INSERT INTO names('firstName','lastName','companyName'), phone('home','cell','company') values('$firstName','$lastName','$companyName','$homePhone','$cellPhone','$companyPhone'"); $mysql->query($insert); if($insert){ echo "success"; } else { mysql_error(); } } I got a connection to the database, I tested for it. I can't find the problem. this is the error I get Warning: mysqli::query() [mysqli.query]: Empty query in C:\wamp\www\test\formData.php on line 23 which is this line Code: [Select] $mysql->query($insert); Hello all, I am having this very frustrating issue, I am trying to print this query i got from my SQL table. However, If the query returns 10 [1,2,3,4,5,6,7,8,9] results it just prints 5 [2,4,6,8,10] . I have no idea what to do. Code: [Select] <!--Error Reporting Production Only[start]--> <?php error_reporting(E_ALL & ~E_NOTICE); ini_set('display_errors','1'); //Error Reporting Production Only[end]--> //Connect to Database [start]--> require_once 'sqllogin.php'; $db_server = mysql_connect($db_hostname, $db_username, $db_password); if (!$db_server) die("Unable to connect to MySQL: " . mysql_error()); mysql_select_db($db_database) or die("Unable to select database: " . mysql_error()); //Connect to Database [end]--> //Process the search [start]--> //Initialize the search output variable $search_output = ""; $sql_command = "SELECT * FROM `Alpha` WHERE 1"; //See if the posted search field is set and has a value if (isset($_POST['searchquery']) && $_POST['searchquery'] !="") {//IF START 1 //run code if condition is met //Filter the search query user input $searchquery = preg_replace('#[a-z 0-9 A-Z]#i', '', $_POST['searchquery']); // Search Query********************* if($_POST['filter1'] == "All") {//all filter code goes her $sql_command = "SELECT id, lastname, firstname, building, room FROM `Alpha` WHERE firstname LIKE '%$_POST[searchquery]%' OR lastname LIKE '%$_POST[searchquery]%' OR id LIKE '%$_POST[searchquery]%'"; } else if($_POST['filter1'] == "Last Name") {//Last Name filter code goes her $sql_command = "SELECT id, lastname, firstname, building, room FROM `Alpha` WHERE lastname LIKE '%$_POST[searchquery]%'"; } else if($_POST['filter1'] == "First Name") {//First Name filter code goes her $sql_command = "SELECT id, lastname, firstname, building, room FROM `Alpha` WHERE firstname LIKE '%$_POST[searchquery]%'"; } else if($_POST['filter1'] == "First Name") {//ID filter code goes her $sql_command = "SELECT id, lastname, firstname, building, room FROM `Alpha` WHERE id LIKE '%$_POST[searchquery]%'"; } }//IF END 1 $query = mysql_query($sql_command) or die(mysql_error()); $count = mysql_num_rows($query); if ($count>0) { $search_output.= "<hr /> $count results for <strong>$_POST[searchquery]</strong><hr /> $sql_command <hr />**IF**POST = $_POST[searchquery] Count=$count, Query= $query<hr />"; /*for ($j = 0 ; $j < $rows ; ++$j) { $id = $row["id"]; $lastname = $row["lastname"]; $firstname = $row["firstname"]; $building = $row["building"]; $room = $row["room"]; $search_output .= "$id \t $firstname \t\t $lastname \t\t $building \t$room<br />"; } *************** Can be used instead but not here try putting it where $search_output is echo 'ID: ' . $row[0] . '<br />'; echo 'Last Name: ' . $row[1] . '<br />'; echo 'First Name: ' . $row[2] . '<br />'; echo 'Building: ' . $row[3] . '<br />'; echo 'Room: ' . $row[4] . '<br /><br />'; *********************** */ while($row = mysql_fetch_array($query)) { $row = mysql_fetch_array($query); $id = $row[0]; $lastname=$row[1]; $firstname=$row[2]; $building=$row[3]; $room=$row[4]; $search_output .= "$id $firstname $lastname $building $room<br />"; } } else { $search_output ="<hr /> 0 results for <strong>$searchquery</strong><hr /> $sql_command <hr />****ELSE**POST = $_POST[searchquery] Count=$count, Query= $query<hr /" ; } ?> <!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>MC Reslife WEB APP Developed by; JB</title> <link href="styles.css" rel="stylesheet" type="text/css" /> </head> <body> <div id="Header">Monroe College Resident's Life Web App</div> <div id="Page"> <div id="Menu"> <h1>Search</h1> <form action="<?php echo $_SERVER['PHP_SELF'];?>" method="POST"> <input name="searchquery" type="text" maxlength="19" size="19"/> <input name="myBTN" type="submit" /> <br /> Search By: <select name="filter1"> <option value="All">All</option> <option value="Last Name">Last Name</option> <option value="First Name">First Name</option> <option value="ID Number">ID Number</option> </select> </form> <hr /> <h1>Menu</h1> <p><a href="#">Alpha List</a></p> <p><a href="#">Lock Out Log</a></p> <p><a href="#">Current Probations</a></p> </div> <div id="Content"> <h1>Alpha List</h1> <?php echo $search_output; ?> <p>*************************</p> <?php echo "this is what in post $_POST[searchquery]"?> </div> </div> <div id="Footer">Content for id "Footer" Goes Here</div> </body> </html> please help me Here's my build a querty code... Code: [Select] $sql = "UPDATE $tablename SET DIST_PART_NUM = '$dist_part_num', DIST_PUB = '$dist_pub', MFR_PART_NUM = '$mfr_part_num', MFR_PUB = '$mfr_pub', ITEM_DESC = '$item_desc', ITEM_STD_DESC = '$item_std_desc', ITEM_COPY = '$item_copy', COST = '$cost', GP_MULT = '$gp_mult', ITEM_IMAGE = '$item_image', ITEM_URL = '$item_url', COUNTRY_OF_ORIG = '$country_of_orig', CATEGORY_ID = '$category_id', PICGROUP_KEY = '$picgroup_key', UPC_CODE = '$upc_code', LEADTIME = '$leadtime', UNITS = '$units', LENGTH = '$length', WIDTH = '$width', HEIGHT = '$height', WEIGHT = '$weight', FLAG_HAZARDOUS = '$flag_haz', FLAG_LTL = '$flag_ltl', FLAG_NON_RETURNABLE = '$flag_non', CAMPAIGN_KEY = '$campaign_key', "; foreach($atr as $key => $value) { $sql .= $value . "='" . addslashes($_POST[$value]) . "',"; } $sql2 = rtrim($sql,","); $sql2 .= " WHERE DIST_PART_NUM = '$partnumber' OR MFR_PART_NUM = '$partnumber'"; Now the query get's built 100% fine that's not the issue... my issue is that sine if the table fields are named something like #_of_shelves... and the # sign causes a MySQL 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 26 How can i allow for the # sign, or any other character that can cause issues like () . , * ~ etc.... i tried encompassing the field name with single quotes 'fieldname' and with those slanted ones `tablename` but that did not help. Any ideas? Thanks! Hi all !,
I am stuck on the following piece of code which does not give an error nor does it give a result. ( i.e. it gives 0 num_rows which should be > 1).
If, however, I execute the query in phpmyadmin by simply substituting the values of $pp,$ll and $room_no in the query it gives the correct result.
Please can someone tell me what I may be doing wrong here. Thanks !
$fcon = mysqli_connect($db_host,$db_user,$db_pass,$db_database) or die('Unable to establish a DB connection'); $pp = "(ms.level = 'Beginner' || ms.level = 'Intermediate')"; $ll = 'ms.diff <= 7'; $room_no = 4; $query = "SELECT md.Member_reg_id, md.fname, md.lname, md.email, md.cell, ms.level, ms.diff, ms.score, r.ID_Status FROM register as r JOIN member_detail as md ON r.ID = md.Member_reg_id JOIN memstatus as ms On r.ID = ms.ID WHERE r.CENTERCODE = ? AND r.ID_Status ='A' AND ? AND ? ORDER by level, diff, score DESC"; $stmt=$fcon->prepare($query); $stmt->bind_param('iss',$room_no,$pp,$ll); if(!$stmt->execute()) die('Failed to execute the query'.$fcon->error); else { echo "Executed"; $stmt->bind_result($Member_reg_id,$fname,$lname,$email,$cell,$level,$diff ,$score,$ID_Status); $numrows = $stmt->num_rows; $stmt->store_result(); // echo $numrows; while($stmt->fetch()) { echo "<br>".$fname.' '.$lname; echo "<br>".$level; echo "<br>".$diff; echo "<br>".$score; echo "<br>".$cell; echo "<br>".$email; } } Edited by ajoo, 03 January 2015 - 08:00 AM. So I wrote a question/answer script, and I am having issues with inputting symbols. It works great it you only have letters, number, dots, commas, and likely other symbols. However query fails every time you insert certain symbols. Here is an error: Quote 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 3 My test input was following: Quote ' " ' ''' \ | / ) // I am not sure which symbol causes this, and I need help fixing this. How do I go about this without stripping any symbols? Maybe change table structure somehow?
<?php
<!DOCTYPE html>
I am having an error with this code You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near ''name','email','password','profile') SET ('Sasural','kill@1234.com','kill','ANDK' at line 1 I am stuck with this for last 5hrs Please deal with me for a moment as I try to explain what is occurring. I have created drop down menu that you choose a name and the php/MySQL will run multiple queries to display information regarding the chosen name. The query works great. There is no problem with the query. However when you leave the page, either by navigation or by X'ing out. When you return there are two errors that pop up: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in You can then use the drop down menu and choose a name and it all works great. It is my belief that these errors occur due to how the query was written, as the two queries that have these errors both have math in them. My question is, how do I block these errors from occurring when you return to the page? Here is the query: <?php //Worst Regular Season Record include_once('../other/functions.php'); $con = mysql_connect($hostname, $username, $password) OR DIE ('Unable to connect to database! Please try again later.'); $db = mysql_select_db($dbname, $con); $query = "SELECT win, loss, year, teamname FROM standings, owners WHERE owners.owner_id = standings.owner_id AND win = (SELECT MIN(win) FROM standings) AND standings.owner_id = $thing ORDER BY year"; $result = mysql_query($query); $row = mysql_fetch_array($result); @ mysql_data_seek($result, 0); if(mysql_num_rows($result)>0) { echo "<table CELLPADDING=5 border =1>"; echo "<tr>"; echo "<th align=center colspan=4 > Worst Regular Season Record </th>"; echo "</tr>"; while ($row = mysql_fetch_array($result)) { echo "<tr>"; echo '<td align=center> In '.$row['year'].', the '.$row['teamname'].' had '.$row['win'].' wins, and '.$row['loss'].' losses.</td>'; echo "</tr>"; } } else { } echo "</table>"; mysql_close($con); ?> Here is the error I get ERROR: Could not execute query: INSERT INTO items (quickti, longti, desc, maincat, subcat1, subcat2, colour, size, style, price, stock, pic1, pic2, pic3, pic4, pic5, pic6, sold, saledate) VALUES ('quick', 'main title', 'description', 'Jewelry', 'Gemstones', 'Cubic Zirconia', 'Black', '50m', 'red', '1.50', '100', 'quick1', 'quick2', 'quick3', 'quick4', 'quick5', 'quick6', '0', '0'). 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 'desc, maincat, subcat1, subcat2, colour, size, style, price, stock, pic1, pic2, ' at line 1 Here is the relevant code $sql = "INSERT INTO items (quickti, longti, desc, maincat, subcat1, subcat2, colour, size, style, price, stock, pic1, pic2, pic3, pic4, pic5, pic6, sold, saledate) VALUES ('$quickheading', '$heading', '$content', '$cat1', '$cat2', '$cat3', '$colour', '$size', '$style', '$total', '$stock', '$quick1', '$quick2', '$quick3', '$quick4', '$quick5', '$quick6', '0', '0')"; if I remove desc and $content from the query, then it works. desc is a text field. This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=325938.0 Can anyone tell me whats wrong with this code? The mysql error I get is generic and says
"Could not enter 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 'Tickets (Name,Tech,Node,Address,Tap,Ped,Lash,Hardline,Other,Comments) VALUES ('C' at line 1"
<?php $name=$_POST['name']; $id=$_POST['id']; $node=$_POST['node']; $ped=$_POST['ped']; $tap=$_POST['tap']; $lash=$_POST['lash']; $hardline=$_POST['hardline']; $other=$_POST['other']; $address=$_POST['address']; $city=$_POST['city']; $comments=$_POST['comments']; $dbhost = 'xxxx'; $dbuser = 'xxxx'; $dbpass = 'xxxx'; $conn = mysql_connect($dbhost, $dbuser, $dbpass); if(! $conn ) { die('Could not connect: ' . mysql_error()); } $sql = "INSERT INTO Damage Tickets (Name,Tech,Node,Address,Tap,Ped,Lash,Hardline,Other,Comments) VALUES ('$name','$id','$node','$address','$city','$tap','$ped','$lash','$hardline','$other','$comments')"; mysql_select_db('xxxxx'); $retval = mysql_query( $sql, $conn ); if(! $retval ) { die('Could not enter data: ' . mysql_error()); } echo "Your Damage has been Submitted! <a href=\"damage.php\">Go Back</a>"; mysql_close($conn); ?>No idea whats wrong with it? Thanks for any help |