Most likely this is simple

Anonymous · Oct 20, 2002

I think that this is easy to do just I am not to quick on PHP and MYSQL stuff yet, newbie. I need a page on my website where it will list every member thats in my databse in on col of a table and in the other col I need it to display how much money they owe to my orginization, but I need to be able to edit this 2nd col and then submit the page so it updates the database. If it matters my Database consists of the following field names : id , login, password, name, email, level, dat, icq, occupation, interests, likes dislikes, address, newsletter, picurl, epassword, active, ip, lastmemo , lastnews, lastboard, lastviewed, lastmemoread, lastboardread. Now if you havn't guessed it yet but I am using a prebuilt MySql & PHP community engine. I use the likes field for their account balance (just b/c why back when I made the site I didn't know how to use MySql at all but edit current fields). If you could make this for me I would greatly appreciate it.
~Carl
P.S. if I was hard to understand up there I will attempt to make the webpage I need in ASCII art !!!!!!

_________________________________________________
The Super Edit Page
__________________________________________________

********************************************
*Users........................*Account Balance......................*
********************************************
* User 1......................*.............................................34*
********************************************
*User 2.......................*.......................................23432*
********************************************
*Ect............................*.......................................34323*
********************************************
----------------------
Submit Changes -
----------------------

Anonymous · Oct 20, 2002

What you need to do is pull all the info out of the database, and it is goes through each row, you want to output it to a table. Except as you're outputting it, you're doing it in a form fashion (with all the fields filled in with the database values). Then, at the bottom you can have the option to submit the form which will update all the values in the database. It's a bit complicated but this is just to give you the idea of what you need to do!

Anonymous · Oct 21, 2002

I kind of guessed that much, just I dont know enough PHP yet to do it, if someone has the spare time to write this script

that would be great.

Anonymous · Oct 21, 2002

You'd be hard pushed to find someone who'll write it out for you. We're here to help, not provide work for free! But you might be lucky

Anonymous · Oct 22, 2002

Ya I know, but I can always hope

Anonymous · Oct 23, 2002

Well dont laugh but I attempted to write a script on my own, its not very eligent (niether is my spelling) but I thouhgt it would work.... it doesn't so here's what I got.

Code:

<?PHP
include ('edge.header.php');
mysql_select_db("edge",$db);

  // query the DB

  $sql = "SELECT * FROM people WHERE id=$id";

  $result = mysql_query($sql);	

  $myrow = mysql_fetch_array($result);
  while ($myrow = mysql_fetcharray($result)) {
	  echo "<form method='post' action=$PHP_SELF?>" ;
	  echo "<input type=hidden name='userid' value=$myrow['id'] > ";
	  echo "<table><tr><td width=100>User Name</td><td>Account Balance</td>";
	  echo $myrow["id"];
	  echo "<input type='Text' name='account' value=$myrow['likes']><br>";

	  }
	  echo "<input type='Submit' name='submit' value='Enter information'>";
	  echo "</form>";
?>

now I know this wont update the information yet, but I hoped this would at least show something on the page, instead it gives me an error.
~Carl

Anonymous · Oct 24, 2002

Not bad for a first attempt, a couple of points though

Firstly, don't make a separate form for each record. Just use one form (ie put your form tags outside the loop) and separate each record by a userID or something in front of the variable name.

mysql_fetcharray should be mysql_fetch_array (this is where your error lies)

Skip the $myrow = mysql_fetch_array($result); in front of the loop!

Anonymous · Oct 25, 2002

ALright I have changed it to this yet it does'nt work at all .

Code:

<?PHP
include ('edge.header.php');
mysql_select_db("edge",$db);

  // query the DB

  $sql = "SELECT * FROM people WHERE id=$id";

  $result = mysql_query($sql);	
echo "<form method='post' action=$PHP_SELF?>" ;
  
  while ($myrow = mysql_fetch_array($result)) {
	  
	  echo "<input type=hidden name='userid' value=$myrow['id'] > ";
	  echo "<table><tr><td width=100>User Name</td><td>Account Balance</td>";
	  echo $myrow["id"];
	  echo "<input type='Text' name='account' value=$myrow['likes']><br>";

	  }
	  echo "<input type='Submit' name='submit' value='Enter information'>";
	  echo "</form>";
?>

.. I get a Parse error: parse error, expecting `T_STRING' or `T_VARIABLE' or `T_NUM_STRING' in editaccount.php on line 14

and of course I have no idea what this means
~Carl

Anonymous · Oct 25, 2002

Code:

echo "<input type=hidden name='userid' value=$myrow['id'] > ";

You can't add elements from an array in a string like this, they have to be outside the string, like this:

Code:

echo "<input type=hidden name='userid' value=" . $myrow['id'] . " > ";

Greetz Daan

Anonymous · Oct 29, 2002

why wont this work

Code:

<?PHP
    include ('edge.header.php');
   # mysql_select_db("edge",$db);
    
      // query the DB
    
      $sql = "SELECT * FROM people WHERE id=$id";
    
      $result = mysql_query($sql);	
    
    $result = mysql_db_query($edge_database,$q,$mysql);
    echo "<form method='post' action=$PHP_SELF?>" ;
      
      while ($row = mysql_fetch_array($result)) {
    	  
    #	  echo "<input type=hidden name='userid' value=" . $row['id'] . " > ";
    	  echo "<input type=hidden name='userid' value=" . $row->id . " > ";
    	  echo "<table><tr><td width=100>User Name</td><td>Account Balance</td>";
    	  echo $row["userid"];
    	  echo "<input type='Text' name='account' value=" . $row['likes'] . "><br>";
    
    	  }
    	  echo "<input type='Submit' name='submit' value='Enter information'>";
    	  echo "</form>";
    ?>

Anonymous · Oct 29, 2002

Carl689 said:
why wont this work

Code:

$result = mysql_query($sql); $result = mysql_db_query($edge_database,$q,$mysql); ?>

Make sure your connection information in your header file is correct,
$edge_database = "databaseserver"
$q = "username"
$mysql = "password"

Then choose one of the above, but not both ... mysql_query is the prevered one ... the rest of the code looks fine to me

You'll be a true php coder soon :wink:

Anonymous · Oct 29, 2002

Thanks..... that was dumb on my part.
Ok now I got it showing me information!!!!!!!!! So I decided to finish it up and make it update the database and guess what... It didn't work, do this is what i got..

Code:

<?PHP
$db = mysql_connect("localhost", "*****", "******") or die ("Could not connect to localhost");
mysql_select_db("edge",$db) or die ("Could not find edge");

$db_name ="people";
    


if(!$submit)
{
	 // query the DB
	$query = "SELECT * FROM $db_name";
	$result = mysql_query($query);	


		echo "<form method='post' action=$PHP_SELF?>" ;
        	echo "<table><tr><td width=100><b>User Name</b></td><td><b>Account Balance</b></td></tr>"; 
      
	while ($row = mysql_fetch_array($result)) {
    	   echo "<tr><td>" . $row['login'] . "</td>";
	    	  echo "<td><input type='Text' name='account' value=" . $row['likes'] . "></tr>";
    
    		  }
    		  echo "<input type='Submit' name='submit' value='Enter information'>";
	    	  echo "</form>";
}
if($submit)
    {
        $query = "UPDATE people SET likes='$likes' WHERE login=$login";
        $result = mysql_query($query);
        echo "Thank you! Information updated.";
    }

    ?>

Also how would i go about making it alphabitize the userid's?
~Carl

Anonymous · Oct 30, 2002

So so so very wrong, here's the things you need to correct

In your form output, set the name of the input as an array (name='account[]'). If you want you can set it to a particular index (as long as it's unique) to match it up to the client or whatever.

Next, when the form is submitted, you'll get an array of values. You will then need to perform a separate update query for each and every record that needs updating, at the moment you're treating them as if it's one!

That should see you sorted!

Anonymous · Oct 30, 2002

I am not to sure I know how to do what you are saying

Anonymous · Oct 30, 2002

I read over your post and attempted to do that.. well once again it didn' work. Here it is

Code:

<?PHP
      $db = mysql_connect("localhost", "smmborg", "") or die ("Could not connect to localhost");
      mysql_select_db("edge",$db) or die ("Could not find edge");
      
      $db_name ="people";
          
      
      
      if(!$submit)
      {
      	 // query the DB
      	$query = "SELECT * FROM $db_name";
      	$result = mysql_query($query);	
      $id = $myrow["id"];
  
      $login = $myrow["login"];
  
      $likes = $myrow["likes"];
  
     
      		echo "<form method='post' action=$PHP_SELF?>" ;
              	echo "<table><tr><td width=100><b>User Name</b></td><td><b>Account Balance</b></td></tr>"; 
            
      	while ($row = mysql_fetch_array($result)) {
          	  
      	   	#  echo "<input type=hidden name='login' value=" . $row['login'] . " >;                 
         	  echo "<tr><td>" . echo $login . "</td>";
      	    	  echo "<td><input type='Text' name='account' value=" . echo $likes . "></tr>";
          
          		  }
          		  echo "<input type='Submit' name='submit' value='Enter information'>";
      	    	  echo "</form>";
      }
      if($submit)
          {
              $query = "UPDATE people SET likes='$likes' WHERE login=$login";
              $result = mysql_query($query);
              echo "Thank you! Information updated.";
          }
      
          ?>

Also how when it is working, how would I make it alphabitize it by their login and how would I make it only show rows that contain a value in a certain column? Ex.. Only show records that in $row['AccountHolder'] it says yes. Thanks,

~Carl
...A day will come when I wont be a newbie...it will come...

Anonymous · Oct 30, 2002

Say you have 10 people and for 5 of them you change theis balance, how is MySQl to know what to change and where if you only run this query once

Code:

$query = "UPDATE people SET likes='$likes' WHERE login=$login";

To solve this Jay means do

Code:

............='Text' name='account[]' value=" .........

insteead of

Code:

........='Text' name='account' value=" .........

This way the variable $account is an array of all different accounts and theyre (new) value. Now you can use foreach http://www.php.net/manual/en/control-structures.foreach.php to traverse the array and insert the new values one by one

eg.
if($submit){
foreach($account as $value){
... INSERT STUFF .......
}
}

Hope this helps ... you can buy us all a couple of beers once you get all of this up and runnig huh

Anonymous · Oct 30, 2002

I appoligize again for my newbie-ness but I can't help it, just think how bad I would be If it ws not for you guys. But I dont quite understand what you where telling me about the foreach thing, I understand your point but I dont know how to do it. Also in my attempt to use variables in my code instead of how I did before I no longer get a list of all members and info I used to get. Oh ya , the beer is on me when this is over.

Code:

<?PHP
         $db = mysql_connect("localhost", "*******", "******") or die ("Could not connect to localhost");
         mysql_select_db("edge",$db) or die ("Could not find edge");
         
         $db_name ="people";
             
         
         
         if(!$submit)
         {
         	 // query the DB
         	$query = "SELECT * FROM $db_name";
         	$result = mysql_query($query);	
         $id = $row["id"];
     
         $login = $row["login"];
     
         $likes = $row["likes"];
     
        
         		echo "<form method='post' action=$PHP_SELF?>" ;
                 	echo "<table><tr><td width=100><b>User Name</b></td><td><b>Account Balance</b></td></tr>"; 
               
         	while ($row = mysql_fetch_array($result)) {
             	  
         	               	  echo "<tr><td> $login </td>";
         	    	  echo "<td><input type='Text' name='account[]' value='$likes'></tr>";
             
             		  }
             		  echo "<input type='Submit' name='submit' value='Enter information'>";
         	    	  echo "</form>";
         }
         if($submit)
             {
              foreach($account as $value){   
                 $query = "UPDATE people SET likes='$likes' WHERE login=$login";
                 $result = mysql_query($query);
                 echo "Thank you! Information updated.";
             }
         }
?>

Carl

Anonymous · Oct 30, 2002

Your code has got sooooo many errors, I've corrected most of them, but left a couple of obvious errors that you'll have to hunt down on your own. If you find them, then you'll understand how the script works and why you couldn't get it to work

Code:

$db = mysql_connect("localhost", "*******", "******") or die ("Could not connect to localhost");
mysql_select_db("edge",$db) or die ("Could not find edge");

$db_name ="people";

if(!$submit) {
	// query the DB
	$query = "SELECT * FROM $db_name";
	$result = mysql_query($query);
	
	$id = $row["id"];
	$login = $row["login"];
	$likes = $row["likes"];
	
	echo "<form method='post' action=".$PHP_SELF.">";
	echo "<table><tr><td width=100><b>User Name</b></td><td><b>Account Balance</b></td></tr>";
	
	while ($row = mysql_fetch_array($result)) {
		echo "<tr><td> $login </td>";
		echo "<td><input type='Text' name='account[$id]' value='$likes'></tr>";
	}
	print "</table>\n";
	echo "<input type='Submit' name='submit' value='Enter information'>";
	echo "</form>";
}
if($submit) {
	foreach($account as $id => $value){
		$query = "UPDATE people SET likes='$value' WHERE login=$id";
		$result = mysql_query($query);
		echo "Thank you! Information updated.";
	}
}

Anonymous · Oct 30, 2002

Ok I must be totally clueless.. just when I open it up.. it showes this

Its odd to me at least that it creates the right amount of rows for the number of entered data but has no information. Maybe this is the obvious thing I am supposed to notice. Another thing is I dont get is the $value in the foreach function. I found the error in the update part

Code:

login=$id
to
id=$id

Thanks
~Carl[/code]

Anonymous · Oct 30, 2002

Well, have a look at what information you were expecting to be filled in, and then where they are defined.

The correct amount of rows are generated, because it generates a row every time the code loops. But the variables it's been assigned are outside the loop are they not?

Code:

   $id = $row["id"]; 
   $login = $row["login"]; 
   $likes = $row["likes"];

What is $row defined as? Figure that one out, and you'll be one step closer. And until you get the form to display right, there's no point checking the update part of it.

Most likely this is simple

Anonymous

Guest

Anonymous

Guest

Anonymous

Guest

Anonymous

Guest

Anonymous

Guest

Anonymous

Guest

Anonymous

Guest

Anonymous

Guest

Anonymous

Guest

Anonymous

Guest

Anonymous

Guest

Anonymous

Guest

Anonymous

Guest

Anonymous

Guest

Anonymous

Guest

Anonymous

Guest

Anonymous

Guest

Anonymous

Guest

Anonymous

Guest

Anonymous

Guest