What have I over looked?

EQtitan

Script basics

Check to see if any data was entered at all, if no data was entered show code.html again. If data was entered take data using the $_POST variable. Open a database connection. Look to see if the code has already been entered into the database, if it has not then enter it into 'codeDB' in table 'code', field name code_set.

I get no error messages after submitting the form and running do_auth.php, just a blank white page.

Thanks in advance...

<?
/* Check to see if data was submitted */
$chk_code = $_POST[dew_code]
if ($chk_code == '') {
	header("Location: http://127.0.0.1/code.html");
	exit;
}
/* Create a variable to hold database */
$db_name = "codeDB";
/* Create a variable to hold table within db */
$table_name = "code";
/* Connect  to db */
$connect = mysql_connect("localhost", "root", "qwerty") or die(mysql_error());
/* Look at all codes within code table */
$db = mysql_select_db($db_name, $connect) or die(mysql_error());
/* Try to find a duplicate */
$sql = "SELECT * FROM $table_name WHERE code = '$_POST[dew_code]'";
/* Variable to hold the results in */
$result = mysql_query($sql,$connection) or die(mysql_error());
/* If a result is found make it = $num */
$num = mysql_num_rows($result);
if ($num != 0) {
	echo = "I already have this code on file please try again";
} else {
	$sql = "INSERT INTO $table_name (dew_code) VALUES ('$_POST[dew_code]')");
		echo = "Code ($dew_code) has been added to $table_name";
}
?>

<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<title>Do Auth</title>
<style type="text/css">
<!--
body,td,th {
	color: #FFFFFF;
}
body {
	background-color: #000000;
}
-->
</style>
</head>

<body>

</body>
</html>

noimad1

Ok, something I would try is to use mysql_connect instead of mysql_db_select.....Also, I think you might have your result backwards. It hink it is mysql_db_query(name_of_database, query_variable)

mabye give this code a try:

$dbh=mysql_connect ($servername, $dbusername, $dbpassword);
$query ="SELECT * FROM $table_name WHERE code = '$_POST[dew_code]'";
$result = mysql_db_query ($dbname,$query);
if($result == false){
echo mysql_errno().":".mysql_error()."<BR>";
exit;
}

$num = mysql_num_rows($result);
if ($num != 0) {
	echo = "I already have this code on file please try again";
} else {
$sql = "INSERT INTO $table_name (dew_code) VALUES ('$_POST[dew_code]')");
		echo = "Code ($dew_code) has been added to $table_name";
}

Lastly, if you are getting a blank page, then you are having some problem with $num.....echo out $num after you assign the value to it. It needs to be set to something....

EQtitan

Tried your suggestion and I still get a white screen how do I echo 'out' the $num???

tekky

I would recommend using [man]mysql_query[/man] instead of mysql_db_query

and with mysql_query, you only need mysql_query($sql) no connection info if you already selected the DB

and mysql_query() NEVER takes a dbname it always takes a resource ID for the connection, IE $dbh

however if you wish to use mysql_db_query its done using

mysql_db_query('databasename', $sql, [optional $dbh]);

if you dont supply $dbh, it uses the last active SQL connection that use did use

and I would try something like...

$dbh=mysql_connect ($servername, $dbusername, $dbpassword); 
$dew_code = $_POST['dew_code'];
$query ="SELECT * FROM $table_name WHERE code = '$dew_code'"; 
$result = mysql_query ($query); 
if(!$result){ 
echo mysql_errno().":".mysql_error()."<BR>"; 
exit; 
} 

$num = mysql_num_rows($result); 
if ($num) { 
    echo = "I already have this code on file please try again"; 
} else { 
$sql = "INSERT INTO $table_name (dew_code) VALUES ('$dew_code')"); 
        echo = "Code ($dew_code) has been added to $table_name"; 
}

I would also recommend not using $_POST['values'] in strings w/o concatenating them....

EQtitan

Still not working with the changes. I'm still gettig the white screen.

/sigh oh well.

noimad1

Ok, to echo out the value of $num, do this

echo "Num = $num";

right after this line:
$num = mysql_num_rows($result);

That will tell you what the value of $num is.

Now that I look closer at your code, I think your echo statements are incorrect. Also, I don't think you need the () in your sql statement.

And lastly, I don't think anything is being put into your database because you are building the sql insert statement, but not actually executing it.

Try the code like this:

$dbh=mysql_connect ($servername, $dbusername, $dbpassword);
$query ="SELECT * FROM $table_name WHERE code = '$_POST[dew_code]'";
$result = mysql_db_query ($dbname,$query);
if($result == false){
echo mysql_errno().":".mysql_error()."<BR>";
exit;
}

$num = mysql_num_rows($result); 
//for error checking:
echo "<BR><BR>Num = $num<BR><BR>";
if ($num <> 0) { 
    echo "I already have this code on file please try again"; 
} else { 
mysql_select_db($dbname, $dbh);
$sql = "INSERT INTO $table_name dew_code VALUES '$dew_code'"; 
        echo "Code $dew_code has been added to $table_name"; 
$result = mysql_query($sql);
if ($result == false){
echo mysql_errno().":".mysql_error()."<BR>";
exit;
}

}

give that a try and tell me what happens.

EQtitan

Still getting a white screen and nothing added to db. Do I need to make a field within the table?

like so:

Database = codeDB
Table = code
Field = new_code <---- this is the place where data would be stored and queried...

noimad1

Originally posted by EQtitan
Still getting a white screen and nothing added to db. Do I need to make a field within the table?

like so:

Database = codeDB
Table = code
Field = new_code <---- this is the place where data would be stored and queried...

Yes. There has to be an existing field in order to add data to the table....

Also, it just doesn't make sense why you are getting a blank screen. By echo'ing out $num you should get something....

At minimum, with the code I gave you you should have seen at least "Num ="

Can you create a simple script and run it to see if it echo's out:

<?
echo "this is a test";
?>

Just to double check, your filename is with the .php file extention on it?

EQtitan

The echo works, and the extension is .php

here it the full code

Code.html

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

</style>
</head>

<body>

<form method="POST" action="do_auth.php">

<p class="style2">
<strong>DewU Code:</strong></p>
<span class="style2">

</span>
<p><input type="text" name="dew_code" length="10" maxlength="10"></p>

<p><input type="submit" name="submit" value="Enter Code"></p>

</body>
</html>

do_auth.php

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">

<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<title>Do Auth</title>
<style type="text/css">
<!--
body,td,th {
	color: #FFFFFF;
}
body {
	background-color: #000000;
}
-->
</style>
</head>

<body>

<?
/* Check to see if data was submitted */
$chk_code = $_POST[dew_code]
if ($chk_code == '') {
	header("Location: http://127.0.0.1/code.html");
	exit;
}
/* Create a variable to hold database */
$db_name = "codeDB";
/* Create a variable to hold table within db */
$table_name = "code";
/* Create a variable to hold filed within table */
$field_name = "new_code";
/* Connect  to db */
$dbh= mysql_connect (localhost, root, qwerty); 
$query = "SELECT * FROM $table_name WHERE $field_name = '$_POST[dew_code]'"; 
$result = mysql_db_query ($dbname,$query); 
if($result == false){ 
echo mysql_errno().":".mysql_error()."<BR>"; 
exit; 
} 

$num = mysql_num_rows($result);  

//for error checking: 
echo "<BR><BR>Num = $num<BR><BR>"; 
if ($num <> 0) {  

    echo "I already have this code on file please try again";  

} else {  

mysql_select_db($dbname, $dbh); 
$sql = "INSERT INTO $table_name dew_code VALUES '$dew_code'";  

        echo "Code $dew_code has been added to $table_name";  

$result = mysql_query($sql); 
if ($result == false){ 
echo mysql_errno().":".mysql_error()."<BR>"; 
exit; 
} 

}
?>
</body>
</html>

noimad1

what kind of server is this running on?

Is it your local machine, or something I might be able to see somewhere?

Everything in your code looks ok, and something should be getting echo'd out......

The only thing I can think of is that the first database query is failing, and exiting the code, but not actually printing your error...

noimad1

mabye you could set this at the beginning of your php code?

ini_set ('error_reporting', E_ALL);
display_errors(1);

I have a feeling if we can't see the errors, then we might have a problem finding out what is happening...

EQtitan

I'm using Apache2 on my PC, with MySQL, PHP4, anmd phpmyadmin.

Also I tried using that code of yours at the very beginning of the script and still i get a blank screen.

onion2k

Re your sig..

if ($php_level = "beginner") {

Thats an assignment operation and should always return true, so whatever $php_level is set to you'll get the stupid question message. You want == instead.

This is the second time I've had to post about the PHP in peoples sigs today. Sort it out folks.

noimad1

Originally posted by EQtitan
I'm using Apache2 on my PC, with MySQL, PHP4, anmd phpmyadmin.

Also I tried using that code of yours at the very beginning of the script and still i get a blank screen.

This one just has me stumped. If you are runnign your own Apache, do you have access to your php.ini file? Mabye we can turn error checking on there:

there should be a section that says
Error handling and logging
put this line in there:
error_reporting = E_ALL & ~E_NOTICE

can you echo something out at the beginning of the script to see if can at least do that:

echo "hello world 1" right at the top of your php script.

Then do that every so often in the script, just increase the 1 to 2, 3 etc. That way if it doesn't echo one of them out, we can see exactly where in the code it stops.....

noimad1

OK,

I took the script and tested it on my server. I added a temp database/table/field to try it out. I did come accross a lot of little errors. I cleaned it up and this code for sure works:

<?
/* Check to see if data was submitted */
$chk_code = $_POST[dew_code]
if ($chk_code == '') {
    header("Location: [url]http://127.0.0.1/code.html[/url]");
    exit;
}
/* Create a variable to hold database */
$db_name = "dmchenry_php";
/* Create a variable to hold table within db */
$table_name = "testing";
/* Create a variable to hold filed within table */
$field_name = "fly";
/* Connect  to db */
$dbh= mysql_connect (localhost, yourusername, yourpassword);
$query = "SELECT * FROM $table_name WHERE $field_name = '$chk_code'";
$result = mysql_db_query ($db_name,$query); 
if($result == false){ 
echo mysql_errno().":".mysql_error()." error 1<BR>"; 
exit; 
} 

$num = mysql_num_rows($result);  

//for error checking:
echo "<BR><BR>Num = $num<BR><BR>"; 
if ($num <> 0) {  

    echo "I already have this code on file please try again";  

} else {  

mysql_select_db($db_name, $dbh); 
$sql = "INSERT INTO $table_name ($field_name) VALUES ('$chk_code')";  

        echo "Code $chk_code has been added to $table_name";  

$result = mysql_query($sql); 
if ($result == false){ 
echo mysql_errno().":".mysql_error()."<BR>"; 
exit; 
} 

}
?>

so give that a try and let me know what happens. I found little things like the Select statement was a little off, a few places we were using $dbname instead of $db_name and such.

For some reason your php is not echoing out the errors. That makes it really hard to debug.

As soon as I set it up on my machine it gave me the errors and I was able to track them down in the code.....

EQtitan

My display errors was off I turned it on and debugged it. Ha, ha. Your going to kick yourself after I tell you what the problem was.

There was no Semicolon at the end of the function.

$chk_code = $_POST[dew_code];

So, far it looks to be running right with no errors.

Thanks, for all your help

PHP exp. ++1

noimad1

Always something small isn't it! Those dang semi-colons get me all the time.

I think I did catch that in my last code when I put it on my server...

EQtitan

Ok, I have the script all setup and running great, but now I have a few more questions.

Ok, currently the code is setup that if no data is entered into the form or if a code has already been entered to reshow the form again. Now my next question is how do I make a $msg to display a message on the code.html page.

For instance say no data was entered, how do I redirect it back to the code.html page with a message saying "No data was entered", or in the instance that the same code was found in the database to redirect back to code.html to display "Code has already been entered please try another". The redirects already work with the current script.

Could I just make the do_auth.php a block on the code.html page, or vise versa?

<?php
/* Check to see if data was submitted */ 
$chk_code = $_POST[dew_code];
if($chk_code == '') { 
    header("Location: http://127.0.0.1/code.html"); 
    exit; 
} 
/* Create a variable to hold database */ 
$db_name = "codedb"; 
/* Create a variable to hold table within db */ 
$table_name = "code"; 
/* Create a variable to hold filed within table */ 
$field_name = "new_code"; 
/* Connect  to db */ 
$dbh= mysql_connect (localhost, root, qwerty); 
$query = "SELECT * FROM $table_name WHERE $field_name = '$chk_code'"; 
$result = mysql_db_query ($db_name,$query);  

if($result == false){
echo mysql_errno().":".mysql_error()." error 1<BR>";  

exit;  

}  

$num = mysql_num_rows($result);   

//for error checking: 
echo "$num matches found in database<BR>";
if ($num <> 0) {   

	echo "I already have this code on file please try again";
	header("Location: http://127.0.0.1/points/code.html");
} else {   

mysql_select_db($db_name, $dbh);  

$sql = "INSERT INTO $table_name ($field_name) VALUES ('$chk_code')";   

        echo "Code $chk_code has been added to $table_name";
$result = mysql_query($sql);  

if ($result == false){  

echo mysql_errno().":".mysql_error()."<BR>";  

exit;  

}  

} 
?>

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

<body>

</body>
</html>

noimad1

Well you can do this a couple different ways. The easiest would probably be to have the do_auth.php file echo out to the user that either the code already exists or that they didn't enter anything. Then instruct them to use their back button to re-enter.

You could get a little fancier and have the do_auth.php file redirect back to the form.html page, passing a variable back to it stating what the error was. For example:

if($chk_code == '') { 
    header("Location: http://127.0.0.1/code.html?failed=nocode[/url]"); 
    exit; 
}

Then later on instead of echoing out to the user that the data was entered, just do:

f ($num <> 0) {   

header("Location: http://127.0.0.1/points/code.html?failed=duplicate[/url]");

(note, you don't need the [/url] in there...that is just the message forum putting that in there...)

Then you would need to change your code.html to code.php and in the middle (where you want the error message to come out) put something like:

<?php
$failed = $_REQUEST['failed'];
if ($failed == "nocode") {
echo "I'm sorry, you did not enter any code in.";
}
if (failed == "duplicate") {
echo "Im sorry, that cod eis already entered, please try again";
?>

In case you didn't know or not, you can embed that php right into the middle of your html page. As long as you file extention is php it will run the html code and the php code.

There is a trick with those header redirects though. There can not be any html, php echo or anything outputed to the screen before you do that redirect or it will give you an error.

You could get even fancier if you wanted to. This is how I usually do code like the one you have. I create one file called code.php which contains all of my form information. Then I have the form post to itself <form method="post" action="code.php>

Then what I do is put a hidden field in my form called "action" and make it equal to yes.

Then I embed all my php code inside the code.php file under one big if statement:

<?php
$action = $_POST['action'];

if ($action == "yes" {

do all my php

}

?>

Then you could just echo the errors out to the user and you don't have to re-direct them. It saves a little time not having to load one page with just php and then redirect them back to the same page.

And it looks a little more flawless to the end user as they don't get a blank screen...then wait...then go back...if you see what I mean.

Let me know if all that makes sense.

EQtitan

I get this error when using the the variable passing from the do_auth.php to the code.php(formly html)

Parse error: parse error, unexpected $end in F:\www\webroot\points\code.php on line 41

<html>
<head>
<?php 
$failed = $_REQUEST['failed']; 
if ($failed == "nocode") { 
echo "I'm sorry, you did not enter any code in."; 
} 
if (failed == "duplicate") { 
echo "Im sorry, that code is already entered, please try again"; 
?>

<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<title>Donate your DewU codes</title>
<style type="text/css">
<!--
body {
	background-color: #000000;
}
body,td,th {
	color: #009900;
}
.style2 {color: #009900}
-->
</style>
</head>

<body>

<form method="POST" action="do_auth.php">
<!-- Label for data field -->
<p class="style2">
<strong>DewU Code:</strong></p>
<span class="style2">
<!-- Make text_field for code entry -->
</span>
<p><input type="text" name="dew_code" length="10" maxlength="10"></p>
<!-- Make 'submit' button -->
<p><input type="submit" name="submit" value="Enter Code"></p>

</body>
</html>

Line 41 is

</html>