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PHP - Mysql Insert Fiasco
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; } } Similar TutorialsCan 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; 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 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 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? 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 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 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 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"; } } ?> I have an old site written for PHP 5.4 and under and trying (very trying) to get it to work with PHP 7x without much luck. Due to all the changes in 7 my code is one big error message, but one thing at a time. I cannot get the follow code to work at all, even though it worked in PHP 5. Error:
QUERY 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 'viewuser.php?u=666' id='member'>THE PREDATOR [666] was added to the hit' at line 1 I have tried at least 20+ different ways of doing this but just can't get the right syntax to get it inserted into MySQL, the code below is just the latest version. If I echo the a href line out, it works perfect. I am sure it is something ridiculously simple, but I have been 4 hours and counting on this now. Thanks
gangevent_add_2($gangdata['gangID'], "<a href='viewuser.php?u=".$r['userid']."' ".$csscode[$r['userlevel']-1].">".$r['username']."</a> [".$r['userid']."] was added to your hitlist");
function gangevent_add_2($gang, $text) {
global $db; $csscode;
$db->query("UPDATE users SET gangevent = gangevent + 1 WHERE gang={$gang}");
$db->query("INSERT INTO gangevents VALUES('','$gang', UNIX_TIMESTAMP(),'$text')");
}
Hey guys for some reason this code is not working i can't see a problem myself could someone please have a look and point the issue out to me. what i mean by it not working is it won't insert into the database or show a mysql_error. thanks in advance Code: [Select] $guestip = $_SERVER['REMOTE_ADDR']; $time = date('G:i'); $date = date("y-m-d"); $query = mysql_query("SELECT * FROM IP_Address") or die(mysql_error()); while($row = mysql_fetch_assoc($query)){ if($guestip != $row['ip']){ //insert into db. mysql_query("INSERT INTO IP_Address(id, ip, date, time) VALUES(NULL,'$questip','$date','$time')") or die(mysql_error()); echo "inserted in to database"; echo mysql_error(); }else{ // add hit count and update time and date. echo "already in db"; } } I've got Code: [Select] for ($i=1; $i<=5; $i++) { if(isset($_POST['partsusedqty'.$i]) && $_POST['partsusedqty'.$i] != "" && $_POST['partsusedqty'.$i] != "0.00") { mysql_query("INSERT INTO partsused (ptnumber, partqty, partdesc, partprice) VALUES ($ticket, '$partsusedqty'.$i, '$partsuseddesc'.$i, '$partsusedprice'.$i)") or die(mysql_error()); } } I need to know the correct formatting to put these variable variables as values in the mysql query. With this particular code, I get the 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 '.1, ''.1, ''.1)' at line 2" I've tried formatting this an endless number of ways, but I used this particular example because its the one I really thought should work. Everything I've tried that doesn't throw an error put the $partsusedqty in the partqty, partdesc, and partprice fields. Thanks for any help! Hi. I think you all know me by now so I'll cut to the chase. Code: [Select] <?php $host="edited"; $username="edited"; $password="edited"; $db_name="edited"; $tbl_name="topic"; // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $name=$_POST['name']; $detail=$_POST['details']; $sql="INSERT INTO $tbl_name(topic, detail, datetime)VALUES('$name', '$detail', NOW())"; $result=mysql_query($sql); if($result){ header("location:site.html");} else{ echo("I have failed you master.");} ?> Displayed error: "I have failed you master." Anyone know a possible cause? Thanks. Bye. I have a mysql insert statement generated with php that is not populating the table. I've echoed the statement and if I copy and paste into phpmyadmin it works fine. The result of the mysql_query function is true. I've emptied the table so there are no primary key conflicts. I've put the statement in a try catch and it does not display a exception. What else can I try? Here's the statement INSERT INTO `wp_term_relationships` (object_id, term_taxonomy_id, term_order) VALUES (1597,83,0) Works absolute fine if I copy and paste into phpmyadmin. Does not populated the table if run through mysql_query I have a Form on registration.html through which i trying to get data in mysql through the below php script but there is and mysql syntax error please help me with the below code. Code: [Select] <?php $conn = mysql_connect("localhost", "onlinewe_meghraj", "password123") or die(mysql_error()); $db = mysql_select_db("onlinewe_college") or die(mysql_error()); $name1 = $_POST['name1']; $name2 = $_POST['name2']; $year = $_POST['year']; $department = $_POST['deparment']; $group = $_POST['group']; $in_name = $_POST['in_name']; $in_address = $_POST['in_address']; $phone = $_POST['phone']; $email = $_POST['email']; $mobile1 = $_POST['mobile1']; $mobile12 = $_POST['mobile2']; $comment = $_POST['comment']; $result=mysql_query("INSERT INTO register (name1, name2, year, department, group, in_name, in_address, phone, email, mobile1, mobile2, date, comment) VALUES ('$name1', '$name2', '$year', '$department', '$group', '$in_name', '$in_address', '$phone', '$email', '$mobile1', '$mobile2', '".date("Y-m-d h:i:s")."', '$comment')") or die("Insert Error: ".mysql_error()); echo "REGISTRATION DONE"; ?> Please reply. Thank you. I am trying to add a value, input into a form, to a MySQL database. However, something must be wrong with the casting, because if there is a space in the form value, then I get an error, as in: //$_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). Hey guys, this is my first post here(not going to be the last one, Im sure), im trying to insert in mysql from session array, i don't know where is my error, I leave the code below, if someone can help me please . I have problem with this code. It does absolutely nothing. When INSERT is over it should redirect to index.php but it does nothing. There is no error, when the submit is clicked the page just refresh itself and because of echo function it write all the values. What seems to be the problem (I going slightly mad ) Code: [Select] <?php require_once("public/includes/session.php"); ?> <?php require_once("public/includes/connection.php"); ?> <?php require_once("public/includes/functions.php"); ?> <?php include_once("public/includes/form_functions.php"); include_once("public/includes/header.php"); if (isset($_POST['submit'])) { // Form has been submitted. $errors = array(); $required_fields = array('nik', 'lozinka', 'ime', 'prezime', 'adresa', 'grad', 'postanskiBroj', 'fiskni', 'moblini', 'email'); $errors = array_merge($errors, check_required_fields($required_fields, $_POST)); $username = trim(mysql_prep($_POST['nik'])); $password = trim(mysql_prep($_POST['lozinka'])); $hashed_password = sha1($password); $ime = trim(mysql_prep($_POST['ime'])); $prezime = trim(mysql_prep($_POST['prezime'])); $adresa = trim(mysql_prep($_POST['adresa'])); $grad = trim(mysql_prep($_POST['grad'])); $postanskiBroj = trim(mysql_prep($_POST['postanskiBroj'])); $fiskni = trim(mysql_prep($_POST['fiksni'])); $moblini = trim(mysql_prep($_POST['mobilni'])); $email = trim(mysql_prep($_POST['email'])); echo $username . $hashed_password . $ime . $prezime . $adresa . $grad . $postanskiBroj . $fiskni . $moblini . $email; if ( empty($errors) ) { $query = " INSERT INTO `gume`.`korisnik` (`id`, `korisnicko_ime`, `lozinka`, `ime`, `prezime`, `adresa`, `grad`, `postanskiBroj`, `fiksni_telefon`, `mobilni_telefon`, `email`) VALUES (NULL, '$username', '$hashed_password', '$ime', '$prezime', '$adresa', '$grad', '$postanskiBroj', '$fiskni, $moblini', '$email' )"; $result = mysql_query($query, $connection) or die(mysql_error); if ($result) { redirect_to("index.php"); } else { $message = "The user could not be created."; $message .= "<br />" . mysql_error(); } } else { if (count($errors) == 1) { $message = "There was 1 error in the form."; } else { $message = "There were " . count($errors) . " errors in the form."; } } } else { // Form has not been submitted. $username = ""; $password = ""; } ?> <div id="telo"> <div id="kreiranjeNaloga"> <script src="SpryAssets/SpryValidationTextField.js" type="text/javascript"></script> <link href="SpryAssets/SpryValidationTextField.css" rel="stylesheet" type="text/css" /> <script src="SpryAssets/SpryValidationPassword.js" type="text/javascript"></script> <script src="SpryAssets/SpryValidationConfirm.js" type="text/javascript"></script> <link href="SpryAssets/SpryValidationPassword.css" rel="stylesheet" type="text/css" /> <link href="SpryAssets/SpryValidationConfirm.css" rel="stylesheet" type="text/css" /> <p>Polja sa * su obavezna</p> <form action="new_user.php" method="post"> <span id="sprytextfield1"> <label>Korisnicko ime: </label> <input type="text" name="nik" id="nik" size="40" value=""/> *<span class="textfieldMinCharsMsg">Korisnicko ime ne moze imati manje od 5 karaktera</span><span class="textfieldMaxCharsMsg">Korisnicko ime moze imati najvise 30 karaktera.</span></span><br /> <span id="sprypassword1"> <label>Lozinka:</label> <input type="password" name="lozinka" id="lozinka" size="40" value=""/> *<span class="passwordMinCharsMsg">Sifra mora sadrzati najmanje 5 karaktera.</span><span class="passwordMaxCharsMsg">Sifra moze imati najvise 30 karaktera.</span></span> <br /> <span id="spryconfirm1"> <label>Potvrdite lozinku:</label> <input type="password" name="password1" id="password1" size="40" value=""/> <span class="confirmRequiredMsg">*</span>Obe lozinke moraju da budu iste.</span> <br /> <span id="sprytextfield2"> <label>Ime:</label> <input type="text" name="ime" id="ime" size="40" value=""/> * </span> <br /> <span id="sprytextfield3"> <label>Prezima</label> <input type="text" name="prezime" id="prezime"size="40" value="" /> *</span> <br /> <span id="sprytextfield4"> <label>Adresa:</label> <input type="text" name="adresa" id="adresa" size="40" value=""/> * </span> <br /> <span id="sprytextfield7"> <label>Grad:</label> <input type="text" name="grad" id="grad" size="40" value="" /> * </span> <br /> </span><span id="sprytextfield9"> <label>Postanski Broj: </label> <input type="text" name="postanskiBroj" id="postanskiBroj" size="10" value=""/> * <span class="textfieldInvalidFormatMsg">Postanski broj nije pravilno upisan.</span></span><br /> <span id="sprytextfield5"> <label>Broj fiksnog telefona: </label> <input type="text" name="fiksni" id="Broj fiksnog telefona" size="40" value="" /> *<span class="textfieldInvalidFormatMsg">Broj telefona nije pravilno upisan</span></span> <br /> <span id="sprytextfield6"> <label>Broj mobilnog telefona: </label> <input type="text" name="mobilni" id="mobilni" size="40" value="" /> *<span class="textfieldInvalidFormatMsg">Broj telefona nije pravilno upisan</span></span><br /> <span id="sprytextfield10"> <label>Email:</label> <input type="text" name="email" id="email" size="40" value="" /> *<span class="textfieldInvalidFormatMsg">Email adresa nija pravilno upisana.</span></span><br /> <input name="submit" type="submit" id="submit" value="Kreiraj korisnika" /> </form> </div> </div> <?php include("public/includes/footer.php"); ?> I'm a newbie on php. I'm really a system administrator and I was just task to do this simple task. For me its hard but I'm sure for a programmer this is very simple. My agenda is to pull out data on one of my column in mysql, select it and dump it on mysql. Here is the php for retrieving mysql data Code: [Select] <?php function database_connect($users) { $resource_link = mysql_connect("localhost", "root", "root"); if (mysql_select_db($users, $resource_link)) { return $resource_link; } else { echo "Cannot connect to DB"; return false; } } function print_dropdown($query, $link){ $queried = mysql_query($query, $link); $menu = '<select username="username">'; while ($result = mysql_fetch_array($queried)) { $menu .= ' <option value="' . $result['id'] . '">' . $result['username'] . '</option>'; } $menu .= '</select>'; return $menu; } //Some other form elements, or just start a form. echo '<form method="post" action="create2.php">'; //The important bit echo print_dropdown("SELECT username FROM mailbox;", database_connect("users")); //Some other form elements, or just end the form. echo '<input type="submit" name="submit" value="submit"/></form>'; Here is the content of my create2.php. This is the php page who do the insert on my mysql. Code: [Select] <?php // open the connection $conn = mysql_connect("localhost", "root", "root"); // pick the database to use mysql_select_db("users",$conn); // create the SQL statement $sql2 = "INSERT INTO mailbox values ('','locked','','$_POST[username]','',NOW(),'','locked','')"; // for troubleshooting $result = mysql_query($sql2, $conn) or die(mysql_error()); // execute the SQL statement //if (mysql_query($sql2, $conn)) { // echo "Success"; //} else { // echo "Fail"; //} } ?> When I click the submit button, I don't see any record being inserted on my table. I'm using the create2.php on my other page though it is only an insert/fill up form not like this one that I need to pull up the date, select and insert to mysql. |
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