PHP - Mysql Insert Into Table Values

Hello,

I need some help. Say that I have a list in my MySQL database that contains elements "A", "S", "C", "D" etc... Now, I want to generate an html table where these elements should be distributed in a random and unique way while leaving some entries of the table empty, see the picture below. But, I have no clue how to do this... Any hints?



Thanks in advance,
Vero

Im 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="&nbsp;&nbsp;Thank you. Please press to close&nbsp;&nbsp;" onclick="tb_remove()"></td>
</tr>
</table>
</div>

Any help will be appreciated

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

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.&nbsp;&nbsp;&nbsp;<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>

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>

How can i use single quote for values?

$qry='insert into tablename values('a','b');';

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)
?>

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 .= ")";

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,

Hi,

Hello Guys,

I am trying to insert a value in my table like this
image.jpg,image2.jpg,image3.jpg
Currently it just runs the image names together

Here is my code.. I tried to use explode, but it just adds the word "array" in the table.

Please advise


$filename2 = $phstring;
}

$csv = $filename2;

$fphoto = explode(',', $csv);

// Insert ad details into mysql
$query = "INSERT INTO ads (ad_type,ad_title,ad_body,ad_category,ad_photo,ad_member,ad_status,ad_city,ad_state,ad_zip)VALUES('".$ad_type."','".$ad_title."','".$ad_body."','".$catid."','".$fphoto."','".$vname."','".$ad_status."','".$city."','".$state."','".$zip."')";

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.