Simple uploading to database need help please

civrob · #1 (**permalink**) 04-23-04, 11:28 AM

Hiya Im trying to learn PHP and would just like to upload a form onto the database.

html form called form.html

----------------------------------------------------------------------
<html>
<head>
<title>form</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>

<body>
<form action="upload.php" method="post" name="contact">
UserName<input name="UserName" type="text"></br>
Password<input name="Password" type="text"></br>
FirstName<input name="FirstName" type="text"></br>
Surname<input name="Surname" type="text"></br>
Title<input name="Title" type="text"></br>
DateOfBirth yyyyMMDD:<input name="DateOfBirth" type="text"></br>
Address<input name="Address" type="text"></br>
TelephoneHome<input name="TelephoneHome" type="text"></br>
TelephoneMobile<input name="TelephoneMobile" type="text"></br>
Email<input name="Email" type="text"></br>
Picture<input name="Picture" type="text"></br>
UserCredit<input name="UserCredit" type="text"></br>

<input type="submit" name="Submit" value="Submit"> <input type="reset">

</form>
</body>
</html>

----------------------------------------------------------------------
upload.php

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<title>Upload</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>

<body>

PHP Code:


			
<?php



$UserName=$_POST['UserName'];

$Password=$_POST['Password'];

$FirstName=$_POST['FirstName'];

$Surname=$_POST['Surname'];

$Title=$_POST['Title'];

$DateOfBirth=$_POST['DateOfBirth'];

$Address=$_POST['Address'];

$TelephoneHome=$_POST['TelephoneHome'];

$TelephoneMobile=$_POST['TelephoneMobile'];

$Email=$_POST['Email'];

$Picture=$_POST['Picture'];

$UserCredit=$_POST['UserCredit'];



$link = mysql_connect ("localhost", "twenty" , "123")

    or die("cannot connect to server");



$db = mysql_select_db("twentyfourseven")

    or die ("Cannot select the twentyfourseven database");



$query = "INSERT INTO contacttable(UserName,Password,FirstName,Surname,Title,DateOfBirth,Address,TelephoneHome,

          TelephoneMobile, Email, Picture, Usercredit 

          VALUES('$UserName','$Password','$FirstName','$Surname','$Title','$DateOfBirth','$Address','$TelephoneHome','$TelephoneMobile','$Email','$Picture',

'$UserCredit)";



                $result=mysql_query($query);

                mysql_close(); 

?>



</body>

</html>

my problem is it works but when i goto command prompt to see if it has uploaded it says empty set (0.0sec), please any help wud be appriciated!

thanks

Trevor · #2 (**permalink**) 04-23-04, 11:47 AM

Before we get too far into this, I have one question and one suggestion.

Question : Picture. Is this an actual picture. As in something you can work on in PhotoShop, or is it a picture name, or URL to a picture? Depending upon your intentions with this field, your form will need to be adjusted to handle binary uploads.

Suggestion : Check your submitted data.

I can't say this enough. Check if for funky code, empty values, or anything else that might compromise your server and site.

Let's start with a response to the question. OK?

civrob · #3 (**permalink**) 04-23-04, 11:52 AM

Quote:

Originally Posted by Trevor

Before we get too far into this, I have one question and one suggestion.

Question : Picture. Is this an actual picture. As in something you can work on in PhotoShop, or is it a picture name, or URL to a picture? Depending upon your intentions with this field, your form will need to be adjusted to handle binary uploads.

Suggestion : Check your submitted data.

I can't say this enough. Check if for funky code, empty values, or anything else that might compromise your server and site.

Let's start with a response to the question. OK?

I havent crossed the picture yet no clue how to do, it is just a char field for now.

do u mean echo my out put?

civrob · #4 (**permalink**) 04-23-04, 12:19 PM

solved it by

PHP Code:


			


$query = mysql_query("INSERT INTO contacttable(UserName,Password,FirstName,Surname,Title,DateOfBirth,Address,TelephoneHome, TelephoneMobile,Email,Picture,Usercredit)".

         "VALUES('$UserName','$Password','$FirstName','$Surname','$Title','$DateOfBirth','$Address','$TelephoneHome','$TelephoneMobile','$Email','$Picture','$UserCredit')");