|
PHP - Can't Insert Into Mysql Table
Hi,
The following code was written by someone else. It allows me to upload images to a directory while saving image name in the mysql table.
I also want the code to allow me save other data (surname, first name) along with the image name into the table, but my try is not working, only the images get uploaded.
What am I missing here?
if(isset($_POST['upload']))
{
$path=$path.$_FILES['file_upload']['name'];
if(move_uploaded_file($_FILES['file_upload']['tmp_name'],$path))
{
echo " ".basename($_FILES['file_upload']['name'])." has been uploaded<br/>";
echo '<img src="gallery/'.$_FILES['file_upload']['name'].'" width="48" height="48"/>';
$img=$_FILES['file_upload']['name'];
$query="insert into imgtables (fname,imgurl,date) values('$fname',STR_TO_DATE('$dateofbirth','%d-%m-%y'),'$img',now())";
if($sp->query($query)){
echo "<br/>Inserted to DB also";
}else{
echo "Error <br/>".$sp->error;
}
}
else
{
echo "There is an error,please retry or check path";
}
}
?>
joseph
Similar TutorialsIm trying to insert some values automatically into a table once the form loads, but Im getting an error. Here is the code Code: [Select] <?php $aid = $_GET['aid']; $sd = $_GET['sd']; ?> <style> #message {margin:20px; padding:20px; display:block; background:#cccccc; color:#cc0000;} </style> <div id="message">Your notification has been submitted.</div> <div style="text-align:center "> <?php $connection = mysql_connect("localhost", "username", "password"); mysql_select_db("articles", $connection); $query="INSERT INTO broken_links (articleid, article) VALUES ('$aid', '$sp')"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } echo "Submitted"; mysql_close($con) ?> <table border="0" cellpadding="3" cellspacing="3" style="margin:0 auto;" > <input type="submit" id="Login" value=" Thank you. Please press to close " onclick="tb_remove()"></td> </tr> </table> </div> Any help will be appreciated Hey yall! I'm working on a new site idea and I've run across a problem that I know is simple enough but I'm stumped. It's in the signup form What I want to do is insert the new user into the 'Login' table and get the user id that was just created and use that to create a table with the user id in the name. Here is what I have: // now we insert user into 'Login' table mysql_real_escape_string($insert = "INSERT INTO `Login` (`UID`, `pass`, `HR`, `mail`, `FullName`) VALUES ('{$_POST['username']}', '{$_POST['pass']}', '{$_POST['pass2']}', '{$_POST['e-mail']}', '{$_POST['FullName']}')"); mysql_query($insert) or die( 'Query string: ' . $insert . '<br />Produced an error: ' . mysql_error() . '<br />' ); $error="Thank you, you have been registered."; setcookie('Errors', $error, time()+20); // Get user ID mysql_real_escape_string($checkID = "SELECT * FROM Login WHERE `mail` = '{$_POST['e-mail']}'"); while ($checkIDdata = mysql_fetch_assoc($checkID)) { $userID = $checkIDdata; // now we create table 'Transactions" for the user mysql_real_escape_string($create = "CREATE TABLE `financewatsonn`.`Transactions_{$userID}` ( `ID` INT( 20 ) NOT NULL AUTO_INCREMENT COMMENT 'Transaction ID', `name` VARCHAR( 50 ) NOT NULL COMMENT 'Name/Location', `amount` VARCHAR( 50 ) NOT NULL COMMENT 'Amount', `date` VARCHAR( 50 ) NOT NULL COMMENT 'Date', `category` VARCHAR( 50 ) DEFAULT NULL COMMENT 'Category', `delete` INT( 1 ) NOT NULL DEFAULT '0', UNIQUE KEY `ID` ( `ID` ) ) ENGINE = MYISAM DEFAULT CHARSET = utf8 COMMENT = 'User ID {$userID}'"); mysql_query($create) or die( 'Query string: ' . $create . '<br />Produced an error: ' . mysql_error() . '<br />' ); } I know in 'Get user ID' that I need to get the ID but I'm not sure how to get that information. i get the error Quote Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource at 166 which is the while line under Get user ID Hi to All, I am performing calculation which i echo it in the input form to insert it to DB table The calculation works fine when i submit the form but the it does not correct calculation to the DB. it seems that the isert to database is done before the calculation and I can't figure a way around it. Because i'm submitting to the same page, the calculation populate in the input form correctly but it insert Zero to the database table instead of correct calculation populated in the input field to DB table You could see from the line 114 to 116 that, i'm performing some calculations and echo it at line 128 and line 128 in the input form value field. Please any help on how to do this...All the function in the code is in another file and it works fine...so the only program is that the calculated value is not inserted as expected Here is the code
<?php
include 'core/initForMainLogPage.php';
if(isset($_GET['empId']) && !empty($_GET['empId'])){
//delete employee here
$empId=$_GET['empId'];
grabEmpId($empId);
}
?>
<?php
if(logged_in()){
$data=user_dataManager('username');
$usernameData=$data['username'];
}else{
header('Location: index.php');
}
?>
<?php include 'includes/adminHeadAll.php';?>
<header>
<?php include 'includes/managerMenu.php';?>
</header>
<div class="container">
<br/>
<h3>Pay Employee</h3>
<?php
$error=array();
$errorAll='';
$leave="";
if(isset($_POST['empId']) && isset($_POST['name']) && isset($_POST['date']) && isset($_POST['basicSalary'])
&& isset($_POST['leave']) && isset($_POST['salaryPerDay']) && isset($_POST['leaveDeduct'])
&& isset($_POST['netSalary'])){
$empId=htmlentities(mysql_real_escape_string($_POST['empId']));
$name=htmlentities(mysql_real_escape_string($_POST['name']));
$date=htmlentities(mysql_real_escape_string($_POST['date']));
$basicSalary=htmlentities(mysql_real_escape_string($_POST['basicSalary']));
$leave=htmlentities(mysql_real_escape_string($_POST['leave']));
$salaryPerDay=htmlentities(mysql_real_escape_string($_POST['salaryPerDay']));
$leaveDeduct=htmlentities(mysql_real_escape_string($_POST['leaveDeduct']));
$netSalary=htmlentities(mysql_real_escape_string($_POST['netSalary']));
//checking for the validity of data entered
if(empty($leave) || empty($date)){
$error[]='Pleave leave or date field is empty.';
}else{
if(preg_match('/[0-9]/',$leave)==false){
$error[]='Leave should only contain numbers';
}
if(empId($empId)===false){
$error[]="This employee is not recoganize by the system and can not be paid,he may need to register first.";
}
}
if(!empty($error)){
$errorAll= '<div class="error"><ul><li>'.implode('</li><li>',$error).'</li></ul></div>';
}else{
//this funciton insert into database
payrollData($name,$empId,$date,$basicSalary,$leave,$salaryPerDay,$leaveDeduct,$netSalary);
echo '<p class="pa">Payment made successfully. <a href="employees-salary-report.php">See Payment Records</a></p>';
}
}//end isset
?>
<div class="tableWrap">
<form action="" method="post" >
<div class="styletable"><table cellpadding="" cellspacing="" border="0">
<?php
$query=mysql_query("SELECT
empId,name,level,company.compId,company.levelOne,company.levelTwo,
company.levelThree,company.levelFour,company.levelFive
FROM employee JOIN company ON company.compId=1 WHERE empId='$empId' LIMIT 1");
while($row=mysql_fetch_array($query)){
$empId=$row['empId'];
$name=$row['name'];
$levelEmp=$row['level'];
$levelOne=$row['levelOne'];
$levelTwo=$row['levelTwo'];
$levelThree=$row['levelThree'];
$levelFour=$row['levelFour'];
$levelFive=$row['levelFive'];
if($levelEmp==1){
$levelPay=$levelOne;
}elseif($levelEmp==2){
$levelPay=$levelTwo;
}elseif($levelEmp==3){
$levelPay=$levelThree;
}elseif($levelEmp==4){
$levelPay=$levelFour;
}elseif($levelEmp==5){
$levelPay=$levelFive;
}
//making calculations here
$basicSalary=$levelPay * 30;
$leaveDeduct=$leave * $levelPay;
$netSalary=$basicSalary - $leaveDeduct;
}
?>
<tr><td>Employee ID: </td><td><input type="text" name="empId" readonly="readonly" value="<?php if(isset($empId)){echo $empId;}?>"></td></tr>
<tr><td>Employee: </td><td><input type="text" name="name" readonly="readonly" value="<?php if(isset($name)){ echo $name;}?>"></td></tr>
<tr><td>Date: </td><td><input type="text" id="Date" class="picker" name="date"></td></tr>
<tr><td> Basic Salary: </td><td><input type="text" name="basicSalary" readonly="readonly" value="<?php echo $basicSalary;?>"></td></tr>
<tr><td> No. Of Absent: </td><td><input type="text" name="leave" class="input" value=""></td></tr>
<tr><td> Salary Per Day:</td><td><input type="text" name="salaryPerDay" readonly="readonly" value="<?php echo $levelPay;?>"></td></tr>
<tr><td> Deduction For Absentee:</td><td><input type="text" name="leaveDeduct" readonly="readonly" value="<?php echo $leaveDeduct;?>"></td></tr>
<tr><td> Net Salary:</td><td><input type="text" name="netSalary" readonly="readonly" value="<?php echo $netSalary;?>"></td></tr>
<tr><td> </td><td><input type="submit" value="Submit Pay" class="submit" name="pay"></td></tr>
</table></div>
</form>
<?php
?>
</div>
<br />
<?php echo $errorAll; ?>
<p>Manage the monthly salary details of your employee along with the allowances, deductions, etc. by just entering their leave</p>
</div>
<?php include 'includes/footerAll.php';?>
<script type="text/javascript" src="js/jquery.js"></script>
<script type="text/javascript" src="js/jquery-ui.js"></script>
<script type="text/javascript" src="js/ui.js"></script>
</body>
</html>
What gets put in for the values is my problem. I have tried every way I can think of but I get errors in the code or I don't get the row inserted. Can anyone give me a code sample using $ variables, preferably a multi-line list? I would have to fix up my code to show you what I am trying as it is in between tries now. I have to be doing something wrong and may have to put together a simple test script but a working example would be better. Everything I look at on the web shows "xxxxx xxxx" examples not $variables. Thanks in advance. Hello guys, im using this code atm and it's working, but the new rows created are put in the end of the table. I want them to stack up from the beginning pushing older rows down. What should i do to make that work? <?php $con = mysql_connect("localhost","*****","*******"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("ananaz_se", $con); $sql="INSERT INTO stuff (adult, namn, url) VALUES ('$_POST[adult]','$_POST[namn]','$_POST[url]')"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } echo "1 row added"; mysql_close($con) ?> hello i have a db with name which populates a form dropdown list there are various variables like this in the form. the user selects their choices and then submits the form to a new db table at the moment it does not capture the dropdown data to the insert sql help please here is the code: =============================== <?php include ('connect.php'); // Insert any new image into database if(isset($_POST['xsubmit']) && $_FILES['imagefile']['name'] != "") { $fileName = $_FILES['imagefile']['name']; $fileSize = $_FILES['imagefile']['size']; $fileType = $_FILES['imagefile']['type']; $content = addslashes (file_get_contents($_FILES['imagefile']['tmp_name'])); $jeweltype = $_POST['jeweltype']; $jewelsize = $_POST['jewelsize_in']; $jewelcolour = $_POST['jewelcolour_in']; $jewelmaterial = $_POST['jewelmaterial_in']; $jewelgender = $_POST['jewelgender_in']; if(!get_magic_quotes_gpc()) { $fileName = addslashes($fileName); } // Checking file size if ($fileSize < 150000) { mysql_query ("INSERT into jewel_images (name,size,type,content,jeweltype,jewelsize,jewelcolour,jewelmaterial,jewelgender) " . "values ('$fileName','$fileSize','$fileType','$content','$jeweltype','$jewelsize','$jewelcolour','$jewelmaterial','$jewelgender')"); } else { $err = "The Image file to too large!"; } } // Find out about latest image $gotten = mysql_query("select * from jewel_images order by row_id desc"); $row = mysql_fetch_assoc($gotten); $bytes = $row['content']; // If this is the image request, send out the image if ($_REQUEST['pic'] == 1) { header("Content-type: $row[type];"); print $bytes; } ?> <html> <head> <title>Upload an image to a database</title> </head> <body> <font color="#FF3333"><?php echo $err ?></font> <table> <form name="Upload" enctype="multipart/form-data" method="post"> <tr> <td>Upload <input type="file" name="imagefile"><br /> Jewelery Type: <select> <?php $sql="SELECT jeweltype FROM jeweltypes"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="jeweltype" ><?php echo $data['jeweltype'] ?></option> <?php } ?> </select> <br /> Jewelery Size: <select> <?php $sql="SELECT jewelsize FROM jewelsizes"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="jewelsize" ><?php echo $data['jewelsize'] ?></option> <?php } ?> </select> <br /> Jewelery Colour: <select> <?php $sql="SELECT jewelcolour FROM jewelcolours"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="jewelcolour_in" ><?php echo $data['jewelcolour'] ?></option> <?php } ?> </select> <br /> Jewelery Material: <select> <?php $sql="SELECT jewelmaterial FROM jewelmaterials"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="jewelmaterial_in" ><?php echo $data['jewelmaterial'] ?></option> <?php } ?> </select> <br /> Jewelery Gender: <select> <?php $sql="SELECT jewelgender FROM jewelgenders"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="jewelgender_in" ><?php echo $data['jewelgender'] ?></option> <?php } ?> </select> <br /> <input type="submit" name="xsubmit" value="Upload"> </td> </tr> <tr> <td>Latest Image</td> </tr> <tr> <td><img src="?pic=1"></td> </tr> </form> </table> </body> </html> Need a little help with this code. I'm running Apache 2.21 / PHP 5.3.5 on Windows. This is a V6 application. I am running Windows 7. I am new, and trying to learn. What am I doing wrong in line 8 with the 'if' statement? The idea is to create a table in mySQL using a form interface. Thank you in advance. Parse error: syntax error, unexpected '{' in C:\website\do_createtable.php on line 8 <? $db_name="booster"; $connection=mysql("localhost", "user", "password") or die (mysql_error()); $db=mysql_select_db($db_name, $connection) or die (mysql_error()); $sql="CREATE TABLE $_POST[table_name] ("; for ($i=0; $i < count($_POST[field_name]); $i++) { $sql .= $_POST[field_name][$i]." ".$_POST[field_type][$i]; if ($_POST[field_length][$i] != "" { $sql .= " (".$_POST[field_length][$i]."),"; } else { $sql .= ","; } } $sql=substr($sql,0,-1); $sql .= ")"; How can i use single quote for values? $qry='insert into tablename values('a','b');'; Hi, I'm a researcher (and complete coding noob), and am planning a longitudinal study that requires e-mail follow-up with subjects taking an initial survey. For purposes of ethics/anonymity due to sensitive survey data, I'd like the acquired e-mails to be saved uncoupled from the survey responses; this is simple to deal with, and I use a basic PHP e-mail form, which injects the email address in a table in MySQL in a different server than the one used for the survey. The issue is that the e-mails are saved in the MySQL database in order of injection, thus it is still theoretically possible for me to link the e-mails back to the survey responses (which have a time stamp that I cannot remove). Ideally I would like not to be able (at all) to link the e-mails to the survey responses, and one way to do that (since I don't save the e-mail injection timestamps in MySQL) might be to have the e-mails saved in MySQL in a random order. Not sure if this is possible, and not even sure if this would be via PHP or MySQL side of things. The server is on godaddy and uses Starfield interface for MySQL but I cannot find an option for random insert/saving of table items (emails). They are saved in order of injection. Any solution for this? Thanks, I am not sure if the title is correct; I tried my best.
I'm a PHP/MySQL beginner and I really need some help.
I have a small script that I am using for sending SMS. I recently added a phonebook. The problem with the phonebook right now is that it's available to all users, i.e. they can all update and delete all rows. What I would like to do is make it so that each user can update and delete only their own contacts.
I have a table call contacts. Inside that table there is first name, last name, company and phonenumber.
How can I accomplish this with PHP & MySQL?
CREATE TABLE IF NOT EXISTS `contacts` ( `contact_id` int(10) NOT NULL AUTO_INCREMENT, `firstname` varchar(255) NOT NULL, `lastname` varchar(255) NOT NULL, `company` varchar(255) NOT NULL, `cell_no` text NOT NULL, PRIMARY KEY (`contact_id`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=8 ; CREATE TABLE IF NOT EXISTS `users` ( `user_id` int(11) NOT NULL AUTO_INCREMENT, `users_name` varchar(30) NOT NULL, `uname` varchar(30) NOT NULL, `u_pass` varchar(60) NOT NULL, `utype` varchar(30) NOT NULL, `timezone` varchar(30) NOT NULL, `uapi_user` varchar(30) NOT NULL, `uapi_pass` varchar(60) NOT NULL, PRIMARY KEY (`user_id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=20 ;
<?php
if (isset($_POST['submit'])){
//form has been submitted1
$firstname = trim($_POST['firstname']);
$lastname = trim($_POST['lastname']);
$company = trim($_POST['company']);
$cellno = trim($_POST['cell_no']);
if($firstname == ''){
echo '<div class="alert alert-danger">First Name is not Valid!</div>';
exit;
}elseif($lastname == ''){
echo '<div class="alert alert-danger">Last Name is not Valid!</div>';
exit;
}elseif($company == ''){
echo '<div class="alert alert-danger">Company is not Valid!</div>';
exit;
}elseif($cellno == ''){
echo '<div class="alert alert-danger">Cellphone Number is not Valid!</div>';
exit;
}else{
$query = "Select cell_no from contacts where cell_no = '".$cellno."' ";
$result = mysql_query($query);
if (!mysql_num_rows($result)) {
$sql = "INSERT INTO contacts(firstname, lastname, company, cell_no)
values('{$firstname}','{$lastname}', '{$company}', '{$cellno}')";
$result = mysql_query($sql);
confirm_query($result);
//echo '<div class="alert alert-success">Successfully added.</div>';
//exit;
?>
<script type="text/javascript">
window.location = "contact_list.php";
</script>
<?php
}
else{
echo '<div class="alert alert-danger">Username. already exist!.</div>';
echo '<p><a href="new_contact.php" class="btn btn-success"> Back </a></p>';
exit;
}}
}else{
$firstname = "";
$lastname = "";
$company = "";
$cellno = "";
}
?>
Hi again all, Why does the foreach loop im doing, put the array values from collection, into seperate table rows rather than 1 row per whole array? <?php class registration{ public $fields = array("username", "email", "password"); public $data = array(); public $table = "users"; public $dateTime = ""; public $datePos = 0; public $dateEntryName = "date"; public $connection; function timeStamp(){ return($this->dateTime = date("Y-m-d H:i:s")); } function insertRow($collection){ //HERE foreach($this->fields as $row => $value){ mysql_query("INSERT INTO $this->table ($value) VALUES ('$collection[$row]')"); } mysql_close($this->connection->connectData); } function validateFields(){ $this->connection = new connection(); $this->connection->connect(); foreach($this->fields as $key => $value){ array_push($this->data, $_POST[$this->fields[$key]]); } $this->dateTime = $this->timeStamp(); array_unshift($this->data, $this->dateTime); array_unshift($this->fields, $this->dateEntryName); foreach($this->data as $value){ echo "$value"; } $this->insertRow($this->data); } } $registration = new registration(); $registration->validateFields(); ?> I end up with 3 rows row 1: username row 2: email row 3 : password rather than row 1:username email password. 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; Hi
I am very new to PHP & Mysql.
I am trying to insert values into two tables at the same time. One table will insert a single row and the other table will insert multiple records based on user insertion.
Everything is working well, but in my second table, 1st Table ID simply insert one time and rest of the values are inserting from 2nd table itself.
Now I want to insert the first table's ID Field value (auto-incrementing) to a specific column in the second table (only all last inserted rows).
Ripon.
Below is my Code:
<?php
$con = mysql_connect("localhost","root","aaa");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("ccc", $con);
$PI_No = $_POST['PI_No'];
$PO_No = $_POST['PO_No'];
$qry = "INSERT INTO wm_order_entry (
Order_No,
PI_No,
PO_No)
VALUES(
NULL,
'$PI_No',
'$PO_No')";
$result = @mysql_query($qry);
$val1=$_POST['Size'];
$val2=$_POST['Style'];
$val3=$_POST['Colour'];
$val4=$_POST['Season_Code'];
$val5=$_POST['Dept'];
$val6=$_POST['Sub_Item'];
$val7=$_POST['Item_Desc'];
$val8=$_POST['UPC'];
$val9=$_POST['Qty'];
$N = count($val1);
for($i=0; $i < $N; $i++)
{
$profile_query = "INSERT INTO order_entry(Size,
Style,
Colour,
Season_Code,
Dept,
Sub_Item,
Item_Desc,
UPC,
Qty,
Order_No
) VALUES( '$val1[$i]','$val2[$i]','$val3[$i]','$val4[$i]','$val5[$i]','$val6[$i]','$val7[$i]','$val8[$i]','$val9[$i]',LAST_INSERT_ID())";
$t_query=mysql_query($profile_query);
}
header("location: WMView.php");
mysql_close($con);
?>
Output is attached.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 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 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 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 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 |
|||||||