PHP - Add New Member Details To Mysql Database

Hi,

I have a form that allows a user to enter details and also to upload a image. The form works great and inserts into the database if i dont include the file input.

What i would like happen is:

The user enters details and selects a file

The file then gets uploaded to ../images and the path is then written to bedroom1 variable

The addresses and the bedroom1 variable that now holds the uploaded file path e.g. images/test.jpg is written to the database

Form
Code: [Select]
<form action="admin.php" method="POST">
   <table>
     <tr><td>
     Address 1:
</td><td>
     <input type="text" name="address1" id="address1" onfocus="selected(this)" onblur="notselected(this)">
</td><td>
     <div id="dynamicText1">
             </div>
</td></tr>

<tr><td>
     Address 2:
</td><td>
     <input type="text" name="address2" id="address2" onfocus="selected(this)" onblur="notselected(this)">
</td><td>
     <div id="dynamicText1">
             </div>
</td></tr>

<tr><td>
     County:
</td><td>
     <input type="text" name="county" id="county" onfocus="selected(this)" onblur="notselected(this)">
</td><td>
     <div id="dynamicText1">
             </div>
</td></tr>

<tr><td>
     Bedroom 1:
</td><td>
     <input type="file" name="bedroom1" id="bedroom1" input name="uploadedfile">
</td><td>
     <div id="dynamicText1">
             </div>
</td></tr>



<tr><td>
     <input type="submit" name="submitaddapartment" value="Add" id="registerbutton">
</td></tr>

   </table
           
        </form>

And the php code to add it to the database


Code: [Select]
<?php
  //get variables
  $submitaddapartment = $_POST['submitaddapartment'];
  $address1 = strip_tags($_POST['address1']);
  $address2 =  strip_tags($_POST['address2']);
  $county = strip_tags($_POST['county']);
  $bedroom1 = strip_tags($_POST['bedroom1']);
   
  
  

  if ($submitaddapartment) //if submit button was pressed
  {
    if ($address1&&$address2&&$county&&$bedroom1&&) //if fields arn't blank
    {
  include 'dbase.php'; //connect to database
  
  if ((($_FILES["bedroom1"]["type"] == "image/gif")
  || ($_FILES["bedroom1"]["type"] == "image/jpg")
  || ($_FILES["bedroom1"]["type"] == "image/pjpeg"))
  && ($_FILES["bedroom1"]["size"] < 20000000))
  {

  if ($_FILES["bedroom1"]["error"] > 0)
  {
  echo "Return Code: " . $_FILES["bedroom1"]["error"] . "<br />";
  }
  else
  {
  echo "Upload: " . $_FILES["bedroom1"]["name"] . "<br />";
  echo "Type: " . $_FILES["bedroom1"]["type"] . "<br />";
  echo "Size: " . ($_FILES["bedroom1"]["size"] / 1024) . " Kb<br />";
  echo "Temp file: " . $_FILES["bedroom1"]["tmp_name"] . "<br />";
  
  if (file_exists("../images/" . $_FILES["bedroom1"]["name"]))
      {
      echo $_FILES["bedroom1"]["name"] . " already exists. ";
      }
  else
      {
      move_uploaded_file($_FILES["bedroom1"]["tmp_name"],
      "/" . $_FILES["bedroom1"]["name"]);
      echo "Stored in: " . "../images/" . $_FILES["bedroom1"]["name"];
  $bedroom1=".'images/.' $_FILES["bedroom1"]["name"]";  
  }
  }
  }
  
  else
  {
  echo "Invalid file";
  }
  
  
  mysql_query ("INSERT INTO user VALUES(NULL,'$address1','$address2','$county','$bedroom1')");
          
    }
    else
    {
    }
  }    
  
?>
Any help would be greatly appreciated.

Thanks

Fred

Perhaps this is a rather lay question, but, is there a way to gather specific connection details about an open MySQL connection in php?

Example:

<?php
    $connect = mysql_connect('localhost', 'username', 'password');
?>

Using the $connect variable, could I run a command that dumps the host, and username to a log file??

Thanks in advance, I'm still searching.

E

hi,

i'm new in php/mysql. i'm stored student marks values in following format in mysql db table.


id student_code Tamil English Maths Science Social
1                1        100      75       78      88       95
2                2         85      90        88     80      100

But i want to search and display the specific student marks in following format.

id:1

student_code:1

Tamil:100

English:75

Maths:78

Science:88

Social:92

Total:?

Avg:?

please give correct code for this format.

Obviously when connecting to php Im not going to show all of my login details;

mysql_connect("details","details","password") or die(mysql_error()); 
mysql_select_db("details") or die(mysql_error());

whats the best way to hide them?
Ive seen some people using an include file with their login details on but say for eg.

<?php
  include('con.php');
?>

Whats to stop  somone looking at www.myweb/con.php and obtaining my details there instead?

i want to make a monthly report
  the user selects month from drop down and i must get the specified dates of that month from the DB
  I am using ajax to get the dates

Hello Everyone,

I recent made a simple membership website. Every page I created works exactly how I envisioned it...
All members data from my registration form goes into my database along with their md5 Encrypted passwords with a time-stamp.
Subsequent pages have a start_session included.

I am very please with it except ONE THING. Logging in is now a problem... username is recognized but NOT the password.

Now the strange thing is that when I go into the database and copy the encrypted password and paste
it into the password field in my login page, I miraculously get into my website with NO problem.

" How do I get the registered members Encrypted Passwords to be recognized by the database when
the  registered members decide to logging in with the password that they create? "

Is there a easy fix for this?

I appreciate ALL your help...


thanks
mrjap1

Hi,

I have successfully implemented a master details page with the results aligned in columns linking to a details page.
I wish to maintain the recordID passed from the master details page and make the dynamic text, which reads
Shade
A tree that is capable of.....
in the attached screen shot a link to another details page referencing the same recordID.

The detailspage2.php would look the same as the screenshot except the Shade text and description below will be highlighted, which I can do, there will be a new image and a new image description.  All other dynmaic elements on the page will remain the same.

I tried to simply save as my detailspage.php to detailspage2.php and create a link to detailspage2.php. It linked to detailspage2.php but none of the record info showed up in their respective table cells.

I have all the names desc's, images, etc setup in a table in my database.

Please let me know what code and other info you need to help me out with this procedure.

Thanks.

 

I Have a protected members section on my site for staff. I would like it so that information specific to that member of staff would appear on their homepage straight after login. Just basic information that is stored on a MySQL database so they can login and check appointments and clients etc for the week etc.

is this something that is possible to do in PHP and MySQL? or am I going mad?

At the moment I have the entire table on the members homepage and the only way I've figured out to specify a member to only show one users information is to use - (SELECT * FROM Tutors WHERE name = 'XXX'. I don't like this method as it would mean i would have to have a separate files for each member. I was hoping I could do something along the lines of - WHERE name = ' relative to the login name used'. but I am sure I am flogging a dead horse here.

So many websites do this kind of thing and I'm surprised after scouring the net for hours and hours that I haven't found any useful information or tutorials on the subject. I am fairly new to programming in PHP and MySQL and am only just getting to grips with the lingo. Maybe I've just been seacrhing the wrong thing, I don't Know.

Any help at all would be amazing and much appreciated, a nudge in the right direction would be a start as I'm getting nowhere on my own

THANKS

Hey there,   First time using MySQL database to connect to a member login.  I have a paid subscription site through ccbill which they add the username and logins to.  I have setup a database, username and password as well as a table that I have connected correctly "I believe" to my website but get this message:  warning: MySQL-fetch_array()expects parameter 1 to be resource, Boolean given in /home....   My table I have setup just has username and password to authenticate the users, which I was told by ccbill is all I need.  Maybe I need authentication 1 or 0 etc. Any help on this would be amazing.  Spent hours trying to figure this out but nothing.   Thanks for your time.   Steven

At the moment  I am creating a search function for my website.
The approach I have in mind is a pseudo-PHP database. To give an example:

A HTML form will submit the results to a PHP file.

HTML FORM - Colour: Black
PHP RESULT PAGE - if ($_POST['color'] == 'Black') {readfile("./products/black/*.html");}
HTML FORM - Price: <$50
PHP RESULT PAGE - if ($_POST['Price'] == '<$50') {readfile("./products/less50/*.html");}

The problem here is if there is an item that is black and costs less than $50, then its going to be listed twice. There is probably some code I can write to ommit the listing of duplicate entries, but it is probably going to be messy, so I am wondering if its better to use a centralized MySQL database, rather than a pseudo-PHP database? I've never used MySQL and don't know much about it and this is my first real attempt at using PHP.

How would I go about doing the following:

I have a csv like this
Quote

"Division","Section","Group","Product Code","Description","Description + Secondary Description"
"Division 1","Section 1","Group 1","BMSLPL25","Test Name","Test Description"
"Division 1","Section 1","Group 2","BMSLPL26","Test Name 2","Test Description 2"
"Division 2","Section 2","Group 2","BMSLPL27","Test Name 3","Test Description 3"



I have a database structured like this
Quote

Divisions
---
id
name
parent_id

Groups
---
id
name
division_id

Products
---
id
code
description
secondary_description



Section is a sub division. What is the best way to get the information from CSV into this database? Should I have another table and store the CSV data as is and then query that to make the other tables.
Any help much appreciated.

Hello all, first post here so i hope i'm doing this right and am putting this into the right place.
Im in the process of integrating a blog into my website, mostly it's set up however i'm currently working on the code to update the posts (theres only 2 parts a title and a comments[which is in actual fact the post content itself]). The code succsefully completes without any errors but for some reason it does not actually update the mysql database.. the code i'm using is as follows:

Code: [Select]
<?php
require_once('header.php');
include "../blog/blogconfig.php";

if(isset($_POST['submit'])){
$update="UPDATE eq_blogarticle SET title='".$_POST['title']."',comments='".$_POST['comment']."' WHERE artid='$aid'";
if(!mysql_query($update)){
echo mysql_error();
}else{
header("location:blog.php?action=listmsgs");
exit;
}
}
?>
<?php
// get value of aid that sent from address bar by blog.php?action=listmsgs
$aid=$_GET['aid'];

// Retrieve data from database
$sql="SELECT * FROM eq_blogarticle WHERE artid='$aid'";
$result=mysql_query($sql);

$row=mysql_fetch_array($result);
?>

<form name="form2" method="post" action="update.php">
<script type="text/javascript">var SITE_URL="<?php echo SITE_URL;?>";</script>
<script type="text/javascript" src="<?php echo SITE_URL;?>/includes/js/nicEdit.js"></script>
<script type="text/javascript">
//<![CDATA[
bkLib.onDomLoaded(function() { nicEditors.allTextAreas() });
//]]>
</script>
<body>
<table width="100%" border="0" cellspacing="1">
  <tr>
    <td colspan="2" class="temptitle">Equidisc Blog</td>
  </tr>
  <tr>
    <td width="74%" valign="top">
<table>
Edit Blog Post
<br>
<br>
Title:
<br>
<input name="title" type="text" class="input" id="title" value="<? echo $row['title']; ?>">
<br>
<br>
<span class="style1 style2 style3">Blog Post:</span>
<br>
<br>
<textarea name="comments" cols="55" rows="12" class="input" id="comments"><? echo stripslashes($row['comments']); ?></textarea>
<br>
<br>
<input name="aid" type="hidden" id="aid" value="<? echo $row['artid']; ?>">
<input type="submit" name="submit" value="Submit">
</form>
</table>
</td>
    <td width="26%" valign="top"><table width="100%" border="0" cellspacing="1">
      <tr>
        <td colspan="2"><img src="../blog/images/fb.gif" width="16" height="16" /> <strong>Blog Menu</strong></td>
        </tr>
       <tr>
        <td><a href="blog.php">Home</a></td>
       </tr>
      <tr>
        <td><a href="blog.php?action=newblogpost">New Post </a></td>
        </tr>
      <tr>
        <td><a href="blog.php?action=listmsgs">Manage Posts </a></td>
        </tr>
    </table></td>
  </tr>
</table>
</body>
<?php
require_once('footer.php');
?>
Explanation: header/footer.php obvious
../blog/blogconfig.php holds my mysql connection settings and connects to the sql, this is the same config as is used for creating the new posts which i have no issues with so i dont think the issue lays there. If i run the query on phpmyadmin with dummy data it works fine and updates the entry..
Any help would be very much appreciated as i'm at the end of my tether with this!.
Thanks in advance.
Jo

Ok, I got someone to help me fix this but he had no idea what the error was...

I have 2 tables, one called points and the other called members.

In members i have got:
id
name

In points i have got:
id
memberid
promo

I have the following code:
Code: [Select]
<?php
$con = mysql_connect("localhost","slay2day_User","slay2day");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("slay2day_database",$con);
$sqlquery="SELECT Sum(points.promo) AS score, members.name, members.id = points.memberid Order By members.name ASC";
$result=mysql_query($sqlquery,$con);
while ($row = mysql_fetch_array($result))
{

//get data
$id = $row['id'];
$name = $row['name'];
$score = $row['score'];

echo "<b>Name:</b> $name<br />";
echo "<b>Points: </b> $score<br />" ;
echo "<b>Rank: </b>";
if  ($name == 'Kcroto1'):
  echo 'The Awesome Leader';
else:
if ($points >= '50'):
  echo 'General';
elseif ($points >= '20'):
  echo 'Captain!';
elseif ($points >= '10'):
  echo 'lieutenant';
elseif ($points >= '5'):
  echo 'Sergeant';
elseif ($points >= '2'):
  echo 'Corporal';
else:
  echo 'Recruit';
endif;
endif;
echo '<br /><br />';

}

?>
I am getting the following error when i do the query in mysql:
Code: [Select]
#1109 - Unknown table 'points' in field list
And when i open the webpage i get the following error:
Code: [Select]
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/slay2day/public_html/points/members.php on line 18
Please Help me?