PHP - Enter Details Into A Databse

Hi All,   I need to get a small function working for a site I need to build for a group of friends.   I was able to get a login page, and a way to add data to the data base.   Now I am stuck on the search and show results fuctionalities.   I have found tens of scripts on the internet, but most of them are outdated, or I cant get them to work, either it sends me back a blank page, does nothing or sends me to an error page   The GOAL User Inputs text in a text box, clicks submit, the data in the text box is looked for in the second collumn of the DB, The information of the collumns 2 to 7 of the DB for the name search are displayed either on the same page or on a new page.   Is there any chance you could hel me with that, taken into account that I dont need the html part of the code, I only need help with the php part...   Thanks in advance to all the people reading this and the ones helping me...

Ik probeer een online gehost .xml bestand in een database te plaatsen. Ik heb inmiddels alle foutmeldingen weg weten te werken (door ze op te lossen :-)) maar heb nu een blanco pagina zonder errors en geen data in mijn database   Hier is wat ik tot nu toe heb gedaan.

<?php
// Turn off all error reporting
error_reporting(0);

// Report simple running errors
error_reporting(E_ERROR | E_WARNING | E_PARSE);

// Reporting E_NOTICE can be good too (to report uninitialized
// variables or catch variable name misspellings ...)
error_reporting(E_ERROR | E_WARNING | E_PARSE | E_NOTICE);

// Report all errors except E_NOTICE
error_reporting(E_ALL & ~E_NOTICE);

// Report all PHP errors (see changelog)
error_reporting(E_ALL);

// Report all PHP errors
error_reporting(-1);

// Same as error_reporting(E_ALL);
ini_set('error_reporting', E_ALL);

// specify url of xml file   
$xmlData = file_get_contents(get_bloginfo('template_directory').'/import/external_db.xml');         
// read XML data string
$xml = simplexml_load_string($xmlData) or die("ERROR: Cannot create SimpleXML object");
// create and execute INSERT queries
foreach ($xml->item as $item) {
$id = $item->id;
$name = mysqli_real_escape_string($item->name);
$code = mysqli_real_escape_string($item->code);
$email = mysqli_real_escape_string($item->email);
$jobdescription = mysqli_real_escape_string($item->jobdescription);
$salary = mysqli_real_escape_string($item->salary);
$sql = "INSERT INTO wajv_employees (id, name, code, email, jobdescription, salary) VALUES ('$id', '$name', '$code', '$email', '$jobdescription', '$salary')";
mysqli_query($sql) or die ("ERROR:");
}
 //mysql_close($connection);

?>

Dit is de XML:

<?xml version="1.0" encoding="utf-8"?>
<data>
<Item>
	<field name="name">xxx</field>
	<field name="code">101</field>
	<field name="email">xxx@xxx.nl</field>
	<field name="jobdescription">Job Description</field>
	<field name="salary">1000</field>
</Item>
<Item>
	<field name="name">xxx</field>
	<field name="code">102</field>
	<field name="email">xxx@xxx.nl</field>
	<field name="jobdescription">Job Description</field>
	<field name="salary">1000</field>
</Item>
<Item>
	<field name="name">xxx</field>
	<field name="code">103</field>
	<field name="email">xxx@xxx.nl</field>
	<field name="jobdescription">Job Description</field>
	<field name="salary">1000</field>
</Item>
<Item>
	<field name="name">xxx</field>
	<field name="code">104</field>
	<field name="email">xxx@xxx.nl</field>
	<field name="jobdescription">Job Description</field>
	<field name="salary">1000</field>
</Item>
</data>

Hoop dat iemand mijn fout over het hoofd ziet

My db table has columns "id" (int-11), "date(varchar-50)", "author name(varchar-100)", "text (varchar-100000)"...
I am storing images + text in text field of database. Everything is going fine while i am using ckeditor to insert data into the database for that particular text field.
Now when i am retrieving data from database... i am able to extract id, date, autorname successfully (i have check it while the issue came)..but when i am trying to retrieve text field which have image, that jumps to page not found error...

what could be the issue... am i having problem with database or editor that i am using????

Code: [Select]
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head><meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Untitled Document</title>
</head>
<body>
<?php
$hostname='xxx';
$username='xxx';
$password='xxx';
$dbname='xxx;
$usertable=xxx;


$myconn=mysql_connect($hostname,$username, $password) OR DIE ('Unable to connect to database! Please try again later.');

if ((($_FILES["file"]["type"] =="image/gif")
|| ($_FILES["file"]["type"] =="image/jpeg")
|| ($_FILES["file"]["type"] == "image/png"))
&& ($_FILES["file"]["size"]< 200000))
{
if ($_FILES["file"]["error"] >0)
{
echo "Return Code: " . $_FILES["file"]["error"] . "<br/>";
}
else
{
if (file_exists("uploads/" . $_FILES["file"]["name"]))
{
echo "File already exists. Choose another name.";
}
else
{
move_uploaded_file($_FILES["file"]["tmp_name"],"uploads/" . $_FILES["file"]["name"]);
}
}
}
else
{
echo "Invalid file";
}
$path="uploads/" . $_FILES["file"]["name"];
$desc=$_POST["desc"];
if (!myconn)
{
die ('Could not connect: ' . $mysql_error());
 } 
 $db_selected=mysql_select_db('xxx',$myconn);
 if (!$db_selected) {
    die ('Can\'t use xxxx : ' . mysql_error());
} 
  mysql_query("INSERT INTO partners (desc,photopath,state) VALUES ('$desc','$path','$state')");
  mysql_close($myconn);
?>
</body>
</html>

I am building a new web app and want to make sure I do everything right from the beginning. What is the best method of storing the date and time into a database using PHP along with the table field settings?

Dear all,

Please help me.
I have two sites and both run on separate DB, both have separate signup and login process, Now i want if any user login from any site can access both sites. There are separate session for both sites now i want to use any of the session for user authentication. Please give suggestions.

Thanks,

I want to delete rows in a table of my database using check-box.
Here are my codes below:

Code: [Select]
<?php
$con = mysql_connect('localhost', 'root', '') or die ('Connection Failed');
mysql_select_db('img', $con) or die ('Connection Failed');
$display = mysql_query("SELECT * FROM photos WHERE email='$lemail'");
echo '<input type="submit" value="Delete" name="del"/>';
echo "<table>
<tr>
<th>#</th>
<th>Images</th>
<th>Image description</th>
<th>Delete</th>
</tr>";
while($row = mysql_fetch_array($display))
{
echo "<tr>";
 echo "<td>".$row['img_ID']."</td>";
  echo "<td><img src='folder/".$row['imaged']."' alt='alt text' width='100' height='100' class='thumb'/> </td>";
echo "<td>".$row['image_description']."</td>";
echo '<td><input type="checkbox" name="delete[]" value="'.$row['img_ID'].'"/></td>';
echo "</tr>";
}
echo "</table>";
echo "</form>";
if (isset($_POST['delete'])) {
$del = $row['img_ID'];
for($i=0;$i<count($_POST["delete"]);$i++)
{
if($_POST["delete"][$i] != "")
{
$str = "DELETE FROM photos WHERE img_ID='$del' ";
mysql_query($str);
echo "Record Deleted.";
}
} 
}
mysql_close($connect);
?>
Here are the problems:

1/ Not working at all, I mean no image is being deleted
2/ At then end, I display a message that the record has been deleted, but if I check multiple checkbox, it keeps writing the message "Records deleted" multiple times

My images are stored in a folder while its details in database...

Help, thank you 

Im trying to write some code for a raffle, when someone buys one ticket it works well but if someone buys ten tickets. i would like it to put each one on a new row, the last column is the ticket number which is got by another table called count and i want the new count in the last column of each row.

In the actual script there is more than two columns but this is an example just to try to let you know what im trying to do.

As you can see i want the ticket number to increment by one every time someone buys tickets. (the ticket number is in a simple table with just id and ticket number)

EXAMPLE someone buys 2 tickes

name | ticket number

John | 1

john | 2

then someone buys three tickets

jane | 3

jane | 4

jane | 5

This is what i have. (WORKING EXAMPLE of the code tha doesnt work.) as you can see the ticker number stays the same and not increment by one.

<?php
$num //is a number between 1 and 10 
$tr //is the current count got from database (this needs to count up by one every entry) 
include 'includes/connect.php'; 
$num = "3"; // number of tickets someone buys. 
$count = "5"; // count of tickets already sold (so this is start count for this transaction). 
$id = "1"; // this is the line the counter is on to keep count updated for the amount of tickets sold. 
$name = 'john'; //example name 

for($i=0;$i< $num;$i++){ 
                        $count="$count+1"; // increments count by 1 
                        $sql123 = "UPDATE count SET count=$count WHERE id='$id'"; //should update database to new count 
                        
                        $sql = "INSERT INTO test (name, number) VALUES ('$name', '$count')"; 
                        if($result = mysqli_query($con, $sql)){ 
                        echo "<br>tickets bought and entered into database,<br>Thank you<br>"; 
                        } else { 
                        echo "Error: " . $sql . "<br>" . $con->error;
  } 
  }
?>

Not sure what im doing wrong?

Thank you in advance

Nook6

I'm working on a chat using basic textarea input, post, but the problem that I'm facing is, if a person doesn't send something, then the output is not updated eg. other messages from other users updating in real time.   How is that achieved when you see new database entries pop up live, milliseconds after they are entered? Do I set a constant refresh rate of some sort?   How do you update a section of a webpage without "refreshing" the entire page like the POST method?

Hi, I have created a form (code below) to show details about members of my website. When the user enters the username of a certain member the form should retrieve these details from my database(phpmyadmin) and display them. I cant get this to work. 

Here is the code for my form:

<form id="form1" name="form1" method="post" action="getdetails.php">
   username <input type="text" name="textfield" value ='' />
  <input type="submit" name="Get Details" value="Get Details" />
   </label>
   
 </p>
</form>


Here is my getdetails.php file

<?php
mysql_connect ("localhost","root","");
mysql_select_db ("test");

$sql = "select * from memberdetails";
       $result = mysql_query ($sql);

       while ($row = mysql_fetch_array($result))
              {
              $username= $row["username"];
              $firstname= $row["firstname"];
         $surname= $row["surname"];
              $dob= $row["dob"];
              $totalwins= $row["totalwins"];   
              $totalloses= $row["totalloses"];
              $email= $row["email"];
              $country= $row["country"];
              $info= $row["info"];

             

   echo "<b><u>Username:</b></u>  $username<br>";
   echo "<b><u>Firstname:</b></u> $firstname<br>";
   echo "<b><u>Surname: </b> </u> $surname<br>";
   echo "<b><u>Date of Birth:</b></u> $dob<br>";
   echo "<b><u>Total Chess Wins:</b></u>  $totalwins<br>";   
   echo "<b><u>Total Chess loses:</b></u> $totalloses<br>";
   echo "<b><u>Email Address: </b></u> $email<br>";
   echo "<b><u>Born in: </b></u> $country<br>";
   echo "<b><u>Other Details:</b></u>  $info<br>";
           }
?>

The above code displays all the users' details from the table not just the one which was typed in.

Thanks for any help!

hi all
I have a table which i have got results from a database. My question If i want to display the details of a particular row. How do I do that?

Thanks Rgds

Hey, i want to make a system requirements check website, and i want it to get the hardware specs automatically. Are there any java-php scripts that allow this? I would do it myself, but i do not know how to code java. Any thoughts?

Hi all,

I am trying to write a script where the user can update their details.

The html form sends the details across to this php file below. All the php variables echo after the query however the values do not replace the current values in the mysql database.

Also for some reason the fname is the only one that changes no matter what the value is. It is entered in to mysql as '0'.

I think the problem will be in the php below.

As all the information below is echoed correctly, and because the value of 'fname' changes I know the connection is working fine.
I guess the error must be in the query I have written, however there is no error that comes up...

Hope you can point me in the right direction.

Thanks


<?php
include("../cxn.php");

$fname = $_POST['fname'];
$lname = $_POST['lname'];
$newemail = $_POST['newemail'];
$telephone = $_POST['telephone'];
$icao = $_POST['icao'];
$newpassword = $_POST['newpassword'];
$id = $_POST['id'];

$sql = "UPDATE Members SET fname='$fname' AND lname='$lname' AND email='$newemail' AND telephone='$telephone' AND password='$newpassword' AND icao='$icao' WHERE id='$id'";
$result = mysqli_query($cxn,$sql)
or die ("Couldn't execute query");

echo "Your new details a  <p>";
echo "$fname <br> $lname <br> $newemail <br> $telephone <br> $icao <br> $newpassword";

?>

If I were to write a code where the browser details and resolution are to be displayed, would the following code work?


Code: [Select]
//Here, I am using the "</body>" tag to end the body of the webpage and the "</html>" tag to end the HTML script
</body>
</html>


 
//By using the "<?php" code, I am creating a PHP script and when I use the  "function" keyword here, I am creating a function called Identification with no arguments in the parameters
<?php function Identification() 
//Here, I am creating a variable called "$viewer" 
//The value of this variable uses the "getenv" library function and the "HTTP_USER_AGENT" is what the "getenv" library function will be performed on and the value of this is that value of the $viewer attribute
//The code uses the ";" to end the command
$viewer = getenv ("HTTP_USER_AGENT");
$width= $_get['width'];
$height=$_get['height'];
//Here, I am using the "echo" library function and the "$viewer" attribute as well as the "<b> (bold)" tags to write a formatted string to the webpage
//I then use the ";" to end the command and I use "?>" to end the PHP script
$msg = "Browser details: $viewer";
$msg .= "Resolution is $width pixels long and $height pixels high";
echo ($msg);
?>
//Here, I am using the "<html>"<head"> and "<title>" tags as well as the "</head"> and "</title>" so that I can write a title (Browser + Platform Identification Script) for the webpage that the script will be shown on   
<html><head><title>Browser + Platform Identification Script</title></head>
//Here, I am using the "<body>" tag to create the body of the webpage
<body>
//Here, I am using the "<?php" code so that I can start a PHP script that will run the Identifcation script and show the results on a webpage
<?php Identification()
//I then use the "}?> to end the PHP script
?>
//Here, I am using the "</body>" tag to end the body of the webpage and the "</html>" tag to end the HTML script
</body>
</html>

Thanks,

Andrew

Hi,

I am writing a little code that would populate the server name based on $_SERVER['HTTP_HOST']. My code and the error I encounter are given below. Can we ever use $_SERVER in a class?


class Get_server_name {

if ($_SERVER['HTTP_HOST'] == 'localhost'):
   $hostname = 'localhost';
else:
   $hostname = 'my remote server';
endif;

}

Error I get: Parse error: parse error, expecting `T_FUNCTION' in