PHP - Mysql Errors In Insert.php
I have a problem where I am getting these errors after a form has been submitted from join.html
Here are the errors: Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: Access denied for user 'scswccla'@'localhost' (using password: NO) in /home/scswccla/public_html/insert.php on line 9 Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: A link to the server could not be established in /home/scswccla/public_html/insert.php on line 9 Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: Access denied for user 'scswccla'@'localhost' (using password: NO) in /home/scswccla/public_html/insert.php on line 9 Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: A link to the server could not be established in /home/scswccla/public_html/insert.php on line 9 Here is the insert.php file http://pastebin.com/EVtgidQC And the join.html file - http://pastebin.com/hqDtGLe5 How would I fix those errors - I got the code from http://forum.hey0.net/showthread.php?tid=2509 The form is at scswc.com/join.html if you want to look at it. I really want to get this working so if you need any more info to help fix it just ask thanks! Similar TutorialsWhen submitting the form, the records are sometimes inserted and sometimes not. I receive no errors stating what the problem is. Code: [Select] <?php if (!function_exists("GetSQLValueString")) { function GetSQLValueString($theValue, $theType, $theDefinedValue = "", $theNotDefinedValue = "") { if (PHP_VERSION < 6) { $theValue = get_magic_quotes_gpc() ? stripslashes($theValue) : $theValue; } $theValue = function_exists("mysql_real_escape_string") ? mysql_real_escape_string($theValue) : mysql_escape_string($theValue); switch ($theType) { case "text": $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL"; break; case "long": case "int": $theValue = ($theValue != "") ? intval($theValue) : "NULL"; break; case "double": $theValue = ($theValue != "") ? doubleval($theValue) : "NULL"; break; case "date": $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL"; break; case "defined": $theValue = ($theValue != "") ? $theDefinedValue : $theNotDefinedValue; break; } return $theValue; } } $editFormAction = $_SERVER['PHP_SELF']; if (isset($_SERVER['QUERY_STRING'])) { $editFormAction .= "?" . htmlentities($_SERVER['QUERY_STRING']); } if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "form1")) { $insertSQL = sprintf("INSERT INTO customer_tbl (customer_first_name, customer_last_name, customer_company, customer_ac, customer_phone, customer_fax_ac, customer_fax_phone, customer_email) VALUES (%s, %s, %s, %s, %s, %s, %s, %s)", GetSQLValueString($_POST['customer_first_name'], "text"), GetSQLValueString($_POST['customer_last_name'], "text"), GetSQLValueString($_POST['customer_company'], "text"), GetSQLValueString($_POST['customer_ac'], "text"), GetSQLValueString($_POST['customer_phone'], "text"), GetSQLValueString($_POST['customer_fax_ac'], "text"), GetSQLValueString($_POST['customer_fax_phone'], "text"), GetSQLValueString($_POST['customer_email'], "text")); $customer_id = false; if(mysql_query($insertSQL, $connCid)) $customer_id = mysql_insert_id(); else echo "There was an error."; } if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "form1")) { $insertSQL = sprintf("INSERT INTO quote_request_tbl (customer_id, quote_trans_date, quote_from_company, quote_from_address, quote_from_city, quote_from_state, quote_from_zip, quote_to_company, quote_to_address, quote_to_city, quote_to_state, quote_to_zip, quote_comments) VALUES ('$customer_id', %s, %s, %s, %s, %s, %s, %s, %s, %s, %s, %s, %s)", GetSQLValueString($_POST['quote_trans_date'], "text"), GetSQLValueString($_POST['quote_from_company'], "text"), GetSQLValueString($_POST['quote_from_address'], "text"), GetSQLValueString($_POST['quote_from_city'], "text"), GetSQLValueString($_POST['quote_from_state'], "text"), GetSQLValueString($_POST['quote_from_zip'], "text"), GetSQLValueString($_POST['quote_to_company'], "text"), GetSQLValueString($_POST['quote_to_address'], "text"), GetSQLValueString($_POST['quote_to_city'], "text"), GetSQLValueString($_POST['quote_to_state'], "text"), GetSQLValueString($_POST['quote_to_zip'], "text"), GetSQLValueString($_POST['quote_comments'], "text")); $quote_id = false; if(mysql_query($insertSQL, $connCid)) $quote_id = mysql_insert_id(); else echo "There was an error."; } if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "form1") && (isset($quote_id))) { $insertSQL = sprintf("INSERT INTO building_tbl (quote_id, building_quantity, building_width, building_length, building_height, building_overhang, slope_id, building_type_id, manufacture_id, foundation_id) VALUES ('$quote_id', %s, %s, %s, %s, %s, %s, %s, %s, %s)", GetSQLValueString($_POST['building_quantity1'], "int"), GetSQLValueString($_POST['building_width1'], "int"), GetSQLValueString($_POST['building_length1'], "int"), GetSQLValueString($_POST['building_height1'], "int"), GetSQLValueString($_POST['building_overhang1'], "int"), GetSQLValueString($_POST['slope_id1'], "int"), GetSQLValueString($_POST['building_type_id1'], "int"), GetSQLValueString($_POST['manufacture_id1'], "int"), GetSQLValueString($_POST['foundation_id1'], "int"); mysql_select_db($database_connCid, $connCid); $Result1 = mysql_query($insertSQL, $connCid) or die(mysql_error()); header('Location: index.php?view=article&id=7'); } mysql_select_db($database_connCid, $connCid); $query_rsSlope = "SELECT * FROM slope_tbl ORDER BY slope_id ASC"; $rsSlope = mysql_query($query_rsSlope, $connCid) or die(mysql_error()); $row_rsSlope = mysql_fetch_assoc($rsSlope); $totalRows_rsSlope = mysql_num_rows($rsSlope); $query_rsSlope = "SELECT * FROM slope_tbl ORDER BY slope_id ASC"; $rsSlope = mysql_query($query_rsSlope, $connCid) or die(mysql_error()); $row_rsSlope = mysql_fetch_assoc($rsSlope); $totalRows_rsSlope = mysql_num_rows($rsSlope); mysql_select_db($database_connCid, $connCid); $query_rsBuildingType = "SELECT * FROM building_type_tbl ORDER BY building_type_id ASC"; $rsBuildingType = mysql_query($query_rsBuildingType, $connCid) or die(mysql_error()); $row_rsBuildingType = mysql_fetch_assoc($rsBuildingType); $totalRows_rsBuildingType = mysql_num_rows($rsBuildingType); mysql_select_db($database_connCid, $connCid); $query_rsManufacture = "SELECT * FROM manufacture_tbl ORDER BY manufacture_name ASC"; $rsManufacture = mysql_query($query_rsManufacture, $connCid) or die(mysql_error()); $row_rsManufacture = mysql_fetch_assoc($rsManufacture); $totalRows_rsManufacture = mysql_num_rows($rsManufacture); mysql_select_db($database_connCid, $connCid); $query_rsFoundation = "SELECT * FROM foundation_tbl ORDER BY foundation_name ASC"; $rsFoundation = mysql_query($query_rsFoundation, $connCid) or die(mysql_error()); $row_rsFoundation = mysql_fetch_assoc($rsFoundation); $totalRows_rsFoundation = mysql_num_rows($rsFoundation); ?> I belve the problem is in this part of the code: if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "form1") && (isset($quote_id))) Many thanks in advance. Hi, I"m trying to create insert images files name into the database and save images in folder, however, after i created databse connection, and excuted it... I'm getting error message said Parse error: syntax error, unexpected $end in C:\xampp\htdocs\ourdeaf\uploadimages\upload.php on line 76 That line 76 is at end of code after </html>, what went wrong! Code: [Select] <form action="<? echo $_SERVER['PHP_SELF']; ?>" method="post" ENCTYPE="multipart/form-data"> Upload:<br><br> <input type="file" name="image"><br><br> <input type="hidden" name="uploaded" value="1"> <input type="submit" value="Upload"> </form> <? }else{ //if the form hasn't been submitted then: //from here onwards, we are copying the file to the directory you made earlier, so it can then be moved //into the database. The image is named after the persons IP address until it gets moved into the database //get users IP $ip=$REMOTE_ADDR; //don't continue if an image hasn't been uploaded if (!empty($image)){ //copy the image to directory copy($image, "./temporary/".$ip.""); //open the copied image, ready to encode into text to go into the database $filename1 = "./temporary/".$REMOTE_ADDR; $fp1 = fopen($filename1, "r"); //record the image contents into a variable $contents1 = fread($fp1, filesize($filename1)); //close the file fclose($fp1); //encode the image into text $encoded = chunk_split(base64_encode($contents1)); //insert information into the database mysql_query("INSERT INTO images (img,data)"."VALUES ('NULL', '$encoded')"); //delete the temporary file we made unlink($filename1); } //end } ?> I have a form with a series of text boxes where the User enters specific data, be it text, numeric, or date/time. It is not mandatory to fill in every box, but when I submit the form to be put into a mysql table, I get errors such as "Incorrect time value: '' for column 'starttime' at row 1" "Incorrect decimal value: '' for column 'workkj' at row 1" How can I setup the mysql table to accept blank answers? Or is there something I need to do in php to fix this? Sorry if this is simple... I'm still learning. Thanks for the help! I cannot get this INSERT to work. Not records are being added to the sys_city_dev table. No query errors are being thrown. I am simply trying to add ALL records from all_illinois to sys_city_dev and Mid should have be add as 11 in all inserts. city_name of course will be field city_name from all_illinois. Here is my code: $query = "SELECT * FROM all_illinois"; if ($results = mysqli_query($cxn, $query)) { $row_cnt = mysqli_num_rows($results); echo $row_cnt . " Total Records in Query.<br /><br />"; if (mysqli_num_rows($results)) { while ($rows = mysqli_fetch_array($results)) { echo $rows['city_name'] . "<br />"; $mid_id = '11'; $insert_city_query = "INSERT INTO sys_city_dev (ID, Mid, cityName, forder, disdplay, cid) VALUES (' ','" . $mid_id . "','" . $rows['city_name'] . "', '', '','')"; if (!$insert_city_query) exit(mysql_error()); } } } Here is my insert table structure and 5 records dump: -- -- Table structure for table `sys_city_dev` -- CREATE TABLE IF NOT EXISTS `sys_city_dev` ( `ID` int(11) NOT NULL AUTO_INCREMENT, `Mid` int(11) NOT NULL DEFAULT '0', `cityName` varchar(30) NOT NULL DEFAULT '', `forder` int(4) NOT NULL DEFAULT '0', `disdplay` int(4) NOT NULL DEFAULT '0', `cid` int(11) NOT NULL DEFAULT '0', PRIMARY KEY (`ID`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=113970 ; -- -- Dumping data for table `sys_city_dev` -- INSERT INTO `sys_city_dev` (`ID`, `Mid`, `cityName`, `forder`, `disdplay`, `cid`) VALUES (84010, 1, 'Dothan', 0, 0, 0), (84011, 1, 'Alabaster', 0, 0, 0), (84012, 1, 'Birmingham', 0, 0, 0), (84013, 2, 'Flagstaff', 0, 0, 0), (84014, 1, 'Auburn', 0, 0, 0); And the all_illinois dump w/ 5 records: -- -- Table structure for table `all_illinois` -- CREATE TABLE IF NOT EXISTS `all_illinois` ( `state_id` varchar(255) NOT NULL, `city_name` varchar(255) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; -- -- Dumping data for table `all_illinois` -- INSERT INTO `all_illinois` (`state_id`, `city_name`) VALUES ('135', 'Abingdon'), ('135', 'Adair'), ('135', 'Addieville'), ('135', 'Addison'), ('135', 'Adrian'); Can anyone tell me why this is not INSERTing? My array data is coming out just fine.. I've tried everything I can think of and cannot get anything to insert.. Ahhhh! <?php $query = "SELECT RegionID, City FROM geo_cities WHERE RegionID='135'"; $results = mysqli_query($cxn, $query); $row_cnt = mysqli_num_rows($results); echo $row_cnt . " Total Records in Query.<br /><br />"; if (mysqli_num_rows($results)) { while ($row = mysqli_fetch_array($results)) { $insert_city_query = "INSERT INTO all_illinois SET state_id=$row[RegionID], city_name=$row[City] WHERE id = null" or mysqli_error(); $insert = mysqli_query($cxn, $insert_city_query); if (!$insert) { echo "INSERT is NOT working!"; exit(); } echo $row['City'] . "<br />"; echo "<pre>"; echo print_r($row); echo "</pre>"; } //while ($rows = mysqli_fetch_array($results)) } //if (mysqli_num_rows($results)) else { echo "No results to get!"; } ?> Here is my all_illinois INSERT table structu CREATE TABLE IF NOT EXISTS `all_illinois` ( `state_id` varchar(255) NOT NULL, `city_name` varchar(255) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; Here is my source table geo_cities structu CREATE TABLE IF NOT EXISTS `1` ( `CityId` varchar(255) NOT NULL, `CountryID` varchar(255) NOT NULL, `RegionID` varchar(255) NOT NULL, `City` varchar(255) NOT NULL, `Latitude` varchar(255) NOT NULL, `Longitude` varchar(255) NOT NULL, `TimeZone` varchar(255) NOT NULL, `DmaId` varchar(255) NOT NULL, `Code` varchar(255) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; Im trying to make a login box for my website and i have a problem here it is: I put all my info in the file. Code: [Select] <?php $host = "localhost"; $user = "scape4le"; <--- thats my username $pass = "********";<--- my pass covered it out $db = "scape4le_members";<------ My DATABASE. ?> Is there any problem with the code? When i put it on my website i get this error. Code: [Select] Warning: mysql_connect() [function.mysql-connect]: Unknown MySQL server host 'DB_HOST' (1) in /home/scape4le/public_html/Login/login-exec.php on line 15 Failed to connect to server: Unknown MySQL server host 'DB_HOST' (1) Theres more codes for the whole login box but too much. But thats the file were the problems at. Here is the code that it says it has a error in Login-exec.php Code: [Select] <?php //Start session session_start(); //Include database connection details require_once('config.php'); //Array to store validation errors $errmsg_arr = array(); //Validation error flag $errflag = false; //Connect to mysql server $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } //Select database $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } //Function to sanitize values received from the form. Prevents SQL injection function clean($str) { $str = @trim($str); if(get_magic_quotes_gpc()) { $str = stripslashes($str); } return mysql_real_escape_string($str); } //Sanitize the POST values $login = clean($_POST['login']); $password = clean($_POST['password']); //Input Validations if($login == '') { $errmsg_arr[] = 'Login ID missing'; $errflag = true; } if($password == '') { $errmsg_arr[] = 'Password missing'; $errflag = true; } //If there are input validations, redirect back to the login form if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; session_write_close(); header("location: login-form.php"); exit(); } //Create query $qry="SELECT * FROM members WHERE login='$login' AND passwd='".md5($_POST['password'])."'"; $result=mysql_query($qry); //Check whether the query was successful or not if($result) { if(mysql_num_rows($result) == 1) { //Login Successful session_regenerate_id(); $member = mysql_fetch_assoc($result); $_SESSION['SESS_MEMBER_ID'] = $member['member_id']; $_SESSION['SESS_FIRST_NAME'] = $member['firstname']; $_SESSION['SESS_LAST_NAME'] = $member['lastname']; session_write_close(); header("location: member-index.php"); exit(); }else { //Login failed header("location: login-failed.php"); exit(); } }else { die("Query failed"); } ?> If anyone could help i would appreciate that, Thanks This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=321772.0 Hello, I'm having a bit of a problem here, all help to this issues would be much appreciated I am trying to use text boxes to insert numbers into the database based on what is inputed. If I have a string, like this for example: $variable = 09385493; And I want to insert it into the database like this: mysql_query("INSERT INTO integers(number) VALUES ('$variable')"); When checking the integers table in my database, looking at the number field, the $variable that was inserted is outputted as 9385493 Notice the number zero was taken out of the front of the number. If the number is double 0's (009385493), both of those zero's would disappear, too. Thanks I don't understand where the empty value is. I've substituted the variables for text and still have the same problem. Code: Code: [Select] $sql = "INSERT INTO courses (course#, name, subject, semester, ap)VALUES('$courseNum', '$courseName', '$subject', '$semester', '$ap')"; 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 1 I have this code: <?php $con = mysql_connect("localhost","hhh","hhh"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("hhh", $con); // -------------------- // Avatar insert check // -------------------- session_start(); $name = $_POST[name]; $group = $_POST[group]; $age = $_POST[age]; $usernameid = $_SESSION[id]; $result = mysql_query("SELECT * FROM avatars WHERE name='$_POST[name]'"); $num = mysql_numrows($result); if ($num == 0) { mysql_query("INSERT INTO avatars (id, usernameid, name, group, age, xp) VALUES ('', '$usernameid', '$name', '$group', '$age', '0')"); header( 'Location: me/' ) ; } else echo 'Sorry, please pick a new name'; ?> And it does everything but put the data into the datebase. If I add a session befor and after '$request' they both run, but the sql doesn't. No error returns, if just redirects to the other page. Any help? Hi guys I have a registration form working fine, my database is as below: userid username password repeatpassword I have added another column which is "name", users can update their profile once they have logged in so I have created updateprofile.php and when I login-->go to update profile and insert my name nothing adds to mysql name column this is my code below: <?php include ("global.php"); //username session $_SESSION['username']=='$username'; $username=$_SESSION['username']; //welcome messaage echo "Welcome, " .$_SESSION['username']."!<p>"; if ($_POST['register']) { //get form data $name = addslashes(strip_tags($_POST['name'])); $update = mysql_query("INSERT INTO users (name) VALUES ('$_POST[name]') WHERE username='$username'"); } ?> <form action='updateprofile.php' method='POST'> Company Name:<br /> <input type='text' name='name'><p /> <input type='submit' name='register' value='Register'> </form> can you please tell me where in this code is wrong? Im new in php so please excuse me if I have silly mistakes. thanks in advance I need help badly! What I want to do is insert into database the value from the selected radio group buttons.. All of them. There are 10 radio groups total (they can be less, but not more). Thanks! Code: [Select] <?php require_once('Connections/strana.php'); mysql_select_db($database_strana, $strana); ?> <link href="css/styles.css" rel="stylesheet" type="text/css" /> <table width="100%" height="100%" style="margin-left:auto;margin-right:auto;" border="0"> <tr> <td align="center"> <form action="" method="post" enctype="multipart/form-data" name="form1"> <table> <?php $tema = mysql_query("SELECT * from prasanja where tip=2")or die(mysql_error()); function odgovor1($string) { $string1 = explode("/", $string); echo $string1[0]; } function odgovor2($string) { $string1 = explode("/", $string); echo $string1[1]; } while ($row=mysql_fetch_array($tema)) { $id=$row['prasanje_id']; $prasanje=$row['prasanje_tekst']; $tekst=$row['odgovor']; ?> <tr> <td> </td> </tr> <tr> <td class="formaP"> <?php echo $prasanje?> </td> </tr> <tr> <td class="formaO"> <p> <label> <input type="radio" name="Group<?php echo $id?>" value="<?php odgovor1($tekst) ?>" /> <?php odgovor1($tekst) ?></label> <br /> <label> <input type="radio" name="Group<?php echo $id?>" value="<?php odgovor2($tekst) ?>" /> <?php odgovor2($tekst) ?></label> <br /> </p></td> </tr> <tr> <td> <br /> </td> </tr> <?php } ?> </table> <input align="left"type="submit" name="submit" value="Внеси" > </form> </td> </tr> </table> prasanje = question tekst/odgovor = answer The answer table: id - primary question_id - the questions ID whose answer is selected in the radio group user_id - cookie takes care of this answer - the value from radio group date - automatic well this is truely embarrising...i have a insert statement which works within phpmyadmin but when using mysqli_query it returns a error.
INSERT INTO users (username, timestamp) VALUES ('test', UTC_TIMESTAMP())Unknown column 'timestamp' in 'field list' i've been playing about with this for a few hours now ...tried changing the column name (timestamp), adding ` around column names as well as table name. the column exists which is the strangest part, and ive even checked there is no space after the column name in the db. whats going on please? Hi I have a XML file as follows: <?xml version="1.0"?> <inspection_form> <inspection_type> <inspection_area_tlb>yard_and_lot</inspection_area_tlb> <inspection_area>Yard and Lot</inspection_area> <items> <item> <item_name>PID Signage/unauthorized sign on pole</item_name> <item_value>0</item_value> </item> <item> <item_name>Landscape well maintained</item_name> <item_value>0</item_value> </item> </items> </inspection_type> <inspection_type> <inspection_area_tlb>pump_island</inspection_area_tlb> <inspection_area>Pump Island and Canopies</inspection_area> <items> <item> <item_name>pumps clean and free of dirt</item_name> <item_value>0</item_value> </item> <item> <item_name>Approved trash cans/clean</item_name> <item_value>0</item_value> </item> </items> </inspection_type> </inspection_form> I want to insert into DB as follows: inspection_area_tlb inspection_area item_name item_value yard_and_lot yard and Lot PID Signage/unauthorized sign on pole 0 yard_and_lot yard and Lot Landscape well maintained 0 pump_island Pump Island and Canopies pumps clean and free of dirt 0 pump_island Pump Island and Canopies Approved trash cans/clean 0 I have written some php code. But every item node as insert for every 'inspection_type'. This is my code $filename="sample.xml"; if(filesize($filename)>0) { $oDOM = new DOMDocument(); $oDOM->loadXML(file_get_contents($filename)); foreach ($oDOM->getElementsByTagName('inspection_type') as $oBookNode) { foreach ($oDOM->getElementsByTagName('item') as $itmNode) { $sSQL = sprintf( "INSERT INTO inspections_master_tablename_import (INSPECTION_TYPE_DB_C_NAME, INSPECTION_TYPE_C_NAME, INSPECTION_TYPE_ITEM_C_NAME,INSPECTION_TYPE_ITEM_VALUE_C_NAME) VALUES ('%s', '%s', '%s', '%s')", mysql_real_escape_string($oBookNode->getElementsByTagName('inspection_area_tlb')->item(0)->nodeValue), mysql_real_escape_string($oBookNode->getElementsByTagName('inspection_area')->item(0)->nodeValue), mysql_real_escape_string($itmNode->getElementsByTagName('item_name')->item(0)->nodeValue), mysql_real_escape_string($itmNode->getElementsByTagName('item_value')->item(0)->nodeValue) ); $rResult = mysql_query($sSQL); if(mysql_errno() > 0) { printf( '<h4 style="color: red;">Query Error:</h4> <p>(%s) - %s</p> <p>Query: %s</p> <hr />', mysql_errno(), mysql_error(), $sSQL ); } } } } Can anyone help me pls. Can you guys tell me where is the error please...
$sql2= ('CREATE TABLE `'.$pub_unik.'` (ID SERIAL,ID_Use CHAR(30),Comment TEXT,Like INT,Score CHAR(30))');I can't find... Thank you guys I am trying to add a value, input into a form, to a MySQL database. However, something must be wrong with the casting, because if there is a space in the form value, then I get an error, as in: //$_POST['string'] == '1blah 2blah'; sql = "INSERT INTO table (some_string) VALUES ($_POST[string])"; $sql_result = mysql_query($sql) or die ('The error is as follows: ' . mysql_error() . '<br /><br />Value could not be added.'); Then I get the following error: The error is as follows: 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 '2blah' at line 1 I've entered paragraphs into a database before, so this error is now to me. The column 'some_value' is a type: varchar(50). Hi guys, new to the forum. Im in need of some advice. I am setting up a shopping cart using opencart. I have an xml product feed, and am writing a parser script to import/update products. The xml parsing via SimpleXML, image downloads, and category inserts work without a hitch. The product Inserts, not so much. there's 4 tables involved: product, product_description, product_to_category, product_to_store all INSERTs are successful for 3 out of 4 tables. 2,540 products get inserted. However, the product_description table shows only 2,143 rows. For some reason, not all INSERTs to this table produce a valid row. I cleared the tables and run the script many times, and each time produces the same results which leads me to believe the problem has something to do with the data instead of the script, but i have no idea where to start troubleshooting. ive included the applicable code class SpicyDB extends mysqli{ function doesExist ($name, $type) { switch ($type) { case "product": $query = "SELECT product_id from spicyvib_product WHERE model='$name'"; break; case "category": $query = "SELECT category_id from spicyvib_category_description WHERE name='$name'"; break; } $result = $this->query($query); if ($result->num_rows >= 1) $status = $result->fetch_row(); else $status = false; return $status; } function addCategory ($name) { $query = "INSERT INTO spicyvib_category (parent_id) VALUES (0)"; $this->query($query); $last_id = $this->insert_id; $query2 = "INSERT INTO spicyvib_category_description (category_id, name) VALUES ($last_id, '$name')"; $this->query($query2); $query3 = "INSERT INTO spicyvib_category_to_store (category_id, store_id) VALUES ($last_id, 0)"; $this->query($query3); } function getCatIDS () { $catIDS = array(); $result = $this->query("SELECT category_id, name FROM spicyvib_category_description"); while ($row = $result->fetch_assoc()) { $name = $row['name']; $id = $row['category_id']; $catIDS[$name] = $id; } return $catIDS; } function addProduct ($data, $catIDS) { $this->query("INSERT INTO spicyvib_product SET model = '" . $data['model'] . "', quantity = '1', minimum = '1', subtract = '0', stock_status_id = '7', image = '" . $data['image'] . "', date_available = NOW(), manufacturer_id = '0', price = '" . (float)$data['price'] . "', cost = '" . (float)$data['cost'] . "', weight_class_id = '5', length_class_id = '3', status = '1', tax_class_id = '0', date_added = NOW()"); $last_id = $this->insert_id; $this->query("INSERT INTO spicyvib_product_to_store SET product_id = '" . (int)$last_id . "', store_id = '0'"); $this->query("INSERT INTO spicyvib_product_description SET product_id = '" . (int)$last_id . "', language_id = '1', name = '" . $data['name'] . "', description = '" . $data['description'] . "'"); $categoryList = $data['category']; $categoryArr = explode(';', $categoryList); foreach ($categoryArr as $currCategory) { if (!empty($currCategory)) { if (!isset($catIDS[$currCategory])) { $currCategory = 'Miscellaneous'; } $this->query("INSERT INTO spicyvib_product_to_category SET product_id = '" . (int)$last_id . "', category_id = '" . (int)$catIDS[$currCategory] . "'"); } } } } $db = new SpicyDB(DB_HOST, DB_USER, DB_PW, DB_NAME); $categoryXml = new SimpleXMLElement(CAT_FNAME, NULL, TRUE); foreach ($categoryXml->category as $cat) { $name = (string)$cat->name; $exists = $db->doesExist($name, 'category'); if (!$exists) { $db->addCategory($name); } } $category_ids = $db->getCatIDS(); $finalXML = new SimpleXMLElement(OLD_FNAME, NULL, TRUE); foreach ($finalXML->items->item as $item) { $data = array(); $updated = (string)$item->lastupdated; $data['model'] = (string)$item->model; $data['name'] = (string)$item->title; $image_str = (string)$item->image; $image = basename($image_str); $data['image'] = 'data/' . $image; $data['price'] = (float)$item->suggested_retail; $data['cost'] = (float)$item->price; if (empty($data['price'])) { $data['price'] = ($data['cost'] * 2); } $data['description'] = (string)$item->description; $data['category'] = (string)$item->category; switch ($updated) { case "UPDATE": $model = $data['model']; if (!$db->doesExist($model, 'product')) { $db->addProduct($data, $category_ids); } break; } } Hi anyone here know how to insert $GET variable into mysql, i don't know how to put this variable between curly bracket, when i put on top insert query, i got error 'Could not insert admin'...please help Code: [Select] <?php mysql_connect("localhost","root","") or die(mysql_error()); mysql_select_db("healthsystem") or die(msql_error()); [color=red]//GET varibable $id = $_GET['id'];[/color] // file properties $file = $_FILES['image']['tmp_name']; if (!isset($file)) echo "Please select an image"; else { $image = addslashes(file_get_contents($_FILES['image']['tmp_name'])); $image_name = addslashes($_FILES['image']['name']); $image_size = getimagesize($_FILES['image']['tmp_name']); if ($image_size==FALSE) echo "That's not an image."; else { $insert = "INSERT INTO image_tbl(m_id,name,image) VALUES ('$id','$image_name','$image')"; $insert2=mysql_query($insert) or die("Could not insert admin"); print "Personal Wellness Successfully Submitted"; } } ?> |