Why are 5 of my 7 Variables reporting as Undefined? I'm Lost......

Modernvox · August 29, 2010

My form


<form action="process.php" method="post">
<font face= "calibri" size= "4">
<table border= "1">

<tr style= "background: #cccccc">  <td><b>Item Description:</b></td>
<td><input type= "text" name= "itemDescription" size= "30" value=""/></td>
</tr>
<tr style= "background: #99ff00">   <td><b>Item Price:</b></td>
<td><input type= "text" name= "itemPrice" size= "5" value=""/></td>
</tr>
</tr style= "background: #A8E5FF">   <td><b>Winning Bidders:</b></td>
<td><input type="text" name= "winningBidder" size= "5" />
</td>
</tr>
<tr>   <td><b>How many deals?:</b></td>
<td><input type="text" name= "itemQty" size= "3" value= "1" /></td>
</tr>
</table>
<br/>
<input type="reset" value="Reset Form"><input type="submit" name="submit" value= "Save & Cont."
" />


</form></font>

My process script:

<?php
error_reporting(E_ALL);
ini_set("display_errors", 1);

$host= "";
$db_name= "";
$db_user= "";
$db_password= "";


$logUser= $_POST['logUser'];
$deleteBidder = $_POST['deleteBidder'];
$newBidder= $_POST['newBidder'];
$itemDescription= $_POST['itemDescription'];
$itemPrice= $_POST['itemPrice'];
$winningBidder= $_POST['winningBidder'];
$itemQty= $_POST['itemQty'];
ob_start();

if ($_POST['newBidder']) {
$bidderId= $newBidder;

mysql_connect("$host", "$db_user", "$db_password")or die("cannot connect to server");
mysql_select_db("$db_name")or die("cannot select DB");

$sql="SELECT * FROM bidders WHERE biddersId='$bidderId'";
$result=mysql_query($sql);

// Mysql_num_row is counting table row
$count=mysql_num_rows($result);
// If result matched $myusername and $mypassword, table row must be 1 row

if($count==0){
// Add $biddersId and redirect to anypage
mysql_Query("INSERT INTO bidders (biddersId)
VALUES ('$bidderId')");
header("Location: index.php");
exit();
  }
}

if ($_POST['deleteBidder'])
{

mysql_connect("$host", "$db_user", "$db_password")or die("cannot connect to server");
mysql_select_db("$db_name")or die("cannot select DB");

mysql_query("DELETE FROM bidders WHERE biddersId='$deleteBidder'");
header("Location: index.php");
exit();
}

if ($_POST['itemDescription'] AND $_POST['itemPrice'] AND $_POST['winningBidder'] AND $_POST['itemQty'])
{


mysql_connect("$host", "$db_user", "$db_password")or die("cannot connect to server");
mysql_select_db("$db_name")or die("cannot select DB");

$sql="SELECT * FROM transactions WHERE biddersId='$winningBidder'";
$result=mysql_query($sql);

$count=mysql_num_rows($result);
// If result matched, table row must be 1 row

if($count==0){
echo "That Bidder Number is NOT logged in, ";
echo "would you like to set this bidder as active?";
echo " Enter 1 for NO or 2 for YES";
echo "<form action= \"process.php\" method= \"POST\">";
echo "<input type =\"text\" name= \"logUser\"/>";
echo "<input type= \"submit\" value = \"Submit\"/>";
$logUser= $_POST['logUser'];
exit();
}
if ($logUser= 1) {
header("Location: inprogress.php");
exit();
}

else if ($logUser= 2){
// Add $biddersId and redirect to anypage
mysql_connect("$host", "$db_user", "$db_password")or die("cannot connect to server");
mysql_select_db("$db_name")or die("cannot select DB");

mysql_Query("INSERT INTO bidders (biddersId)
VALUES ('$winningBidder')");


mysql_query("INSERT INTO transactions (itemDescription, itemPrice, bidderId, itemQty , totalPrice) 
VALUES('$itemDescription', '$itemPrice','$winningBidder', '$itemQty', '$totalPrice')")
or die(mysql_error());  
$_SESSION['itemDescription']='$itemDescription'; 
header("Location: index.php");
exit();
   }
}

echo "<font color= \"red\" face=\"calibri\" size=\"4\">That bidder is already logged, Please press your browsers back button and try again.</font>";


ob_end_flush();

?>

DeleteBidder and newBidder are the only variables working?

Alex · August 29, 2010

Is that the right form? There's no field named deleteBidder on that form.

This isn't valid either:

<input type="submit" name="submit" value= "Save & Cont."
" />

Modernvox · August 29, 2010

Is that the right form? There's no field named deleteBidder on that form.

That's from another form on the another page Alex. I am using process.php to process all forms

Modernvox · August 29, 2010

Here's what my code id producing for an error

Notice: Undefined index: deleteBidder in C:\xampp\htdocs\process.php on line 44

Notice: Undefined index: newBidder in C:\xampp\htdocs\process.php on line 45

Notice: Undefined index: itemDescription in C:\xampp\htdocs\process.php on line 46

Notice: Undefined index: itemPrice in C:\xampp\htdocs\process.php on line 47

Notice: Undefined index: winningBidder in C:\xampp\htdocs\process.php on line 48

Notice: Undefined index: itemQty in C:\xampp\htdocs\process.php on line 49

Notice: Undefined index: newBidder in C:\xampp\htdocs\process.php on line 56

Notice: Undefined index: deleteBidder in C:\xampp\htdocs\process.php on line 78

Notice: Undefined index: itemDescription in C:\xampp\htdocs\process.php on line 89

That bidder is already logged, Please press your browsers back button and try again.

Modernvox · August 29, 2010

Is that the right form? There's no field named deleteBidder on that form.

This isn't valid either:
<input type="submit" name="submit" value= "Save & Cont."
" />

Yeah..Fixed the <input type="submit" name="submit" value= "Save & Cont."

" /> crap, but still producing same errors

Alex · August 29, 2010

Did you fix that HTML error I pointed out?

If you're using multiple forms and all of those variables won't always be set than you'll always get those errors with your current code. You should be doing something like this for all your variables:

$logUser= isset($_POST['logUser']) ? $_POST['logUser'] : '';

This probably isn't related to your current problem, but it is an issue with your code:

$_SESSION['itemDescription']='$itemDescription';

That will set $_SESSION['itemDescription'] to th literal string '$itemDescription' and not the value that the variable holds. Instead it should be:

$_SESSION['itemDescription'] = $itemDescription;

Modernvox · August 29, 2010

These 2 forms work as expected:

<tr>    <td><font face= "calibri" size= "4"> Let's begin by adding bidders...</font></td>
</tr>
<Table border= "1">
        <form action="process.php" method="post">
        <tr style="background: #99cc33;">    <td><strong><font face= "calibri" style= "bold"> Bidders Id: </font></strong></td>
        <td><input type= "text" name= "newBidder"/></td>
</tr>
<tr>   <td><input type="Submit"  value="submit"></td>
</tr>
</table>

<tr>    <td><font face= "calibri" size= "4"> Here we can </font><font color= "red"> delete bidders.</font></td>
</tr>
<Table border= "1">
        <form action="process.php" method="post">
        <tr style="background: #ff6600;">    <td><strong><font face= "calibri" style= "bold"> Bidders Id: </font></strong></td>
        <td><input type= "text" name= "deleteBidder"/></td>
</tr>
<tr>   <td><input type="Submit"  value="submit"></td>
</tr>
</table>

Modernvox · August 29, 2010

Did you fix that HTML error I pointed out?

If you're using multiple forms and all of those variables won't always be set than you'll always get those errors with your current code. You should be doing something like this for all your variables:
$logUser= isset($_POST['logUser']) ? $_POST['logUser'] : '';
This probably isn't related to your current problem, but it is an issue with your code:
$_SESSION['itemDescription']='$itemDescription'; 
That will set $_SESSION['itemDescription'] to th literal string '$itemDescription' and not the value that the variable holds. Instead it should be:
$_SESSION['itemDescription'] = $itemDescription;

So I should be using isset instead of just if $_POST['whatever'] ?

Alex · August 29, 2010

You need to check if the variable is set or not before you use it using isset.

Modernvox · August 29, 2010

You need to check if the variable is set or not before you use it using isset.

Understood ALex, I thank you. How is that suppose to be written

if (isset(firstvar) && (second var) or is it isset($_POST['firstvar'] && ??

Sorry, multiple conditions confuse my little brain..

Alex · August 29, 2010

When using it to the set the value of variables like this:

$logUser= isset($_POST['logUser']) ? $_POST['logUser'] : '';

You wouldn't use multiple conditions. The purpose of that is to set $logUser to $_POST['logUser'] if it's set, otherwise set it equal to an empty string, ''.

For using isset in other contexts it does support multiple parameters and returns true if they're all set.

if(isset($_POST['something'], $_POST['somethingelse'])) {
    // ...
}

Modernvox · August 29, 2010

When using it to the set the value of variables like this:
$logUser= isset($_POST['logUser']) ? $_POST['logUser'] : '';
You wouldn't use multiple conditions. The purpose of that is to set $logUser to $_POST['logUser'] if it's set, otherwise set it equal to an empty string, ''.

For using isset in other contexts it does support multiple parameters and returns true if they're all set.
if(isset($_POST['something'], $_POST['somethingelse'])) {
    // ...
}

Ok. So, i must do this for each of my variables I suppose> $logUser= isset($_POST['logUser']) ? $_POST['logUser'] : ''; ?

Alex · August 29, 2010

For all your variables coming from outsides sources that are not guaranteed to be set, yes.

Modernvox · August 29, 2010

Ok. SO, what would my if statements look like if my variables are being set as $logUser= isset($_POST['logUser']) ? $_POST['logUser'] : '';

if ($logUser)

{

Conditions here

}

??

Alex · August 29, 2010

You could still use isset on the $_POST variables.

if ($_POST['itemDescription'] AND $_POST['itemPrice'] AND $_POST['winningBidder'] AND $_POST['itemQty'])

Could be:

if (isset($_POST['itemDescription'], $_POST['itemPrice'], $_POST['winningBidder'], $_POST['itemQty']))

Optimally you shouldn't define all the variables at the top if you're not going to be using them every time the script runs. You should have them defined within the if statements that determine if they're going to be needed or not.

Modernvox · August 29, 2010

You could still use isset on the $_POST variables.
if ($_POST['itemDescription'] AND $_POST['itemPrice'] AND $_POST['winningBidder'] AND $_POST['itemQty'])
Could be:
if (isset($_POST['itemDescription'], $_POST['itemPrice'], $_POST['winningBidder'], $_POST['itemQty']))
Optimally you shouldn't define all the variables at the top if you're not going to be using them every time the script runs. You should have them defined within the if statements that determine if they're going to be needed or not.

Even with all these fixes, I still get the same errors in that all my variables are undefined from that particular form..

I need a cheeseburger, i'm getting upset

Alex · August 29, 2010

If variables from the form that you're submitting are not set then it's most likely a problem with your HTML. Make sure that is all correct. I noticed that in the two forms you posted you weren't closing them off, but I figured that you probably closed it the line after what you posted.

Modernvox · August 29, 2010

After all fixes here's what we got for errors (smaller list:-)

Notice: Undefined index: deleteBidder in C:\xampp\htdocs\process.php on line 43

Notice: Undefined index: newBidder in C:\xampp\htdocs\process.php on line 44

Notice: Undefined index: newBidder in C:\xampp\htdocs\process.php on line 52

Notice: Undefined index: deleteBidder in C:\xampp\htdocs\process.php on line 74

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\process.php on line 98

Modernvox · August 29, 2010

If variables from the form that you're submitting are not set then it's most likely a problem with your HTML. Make sure that is all correct. I noticed that in the two forms you posted you weren't closing them off, but I figured that you probably closed it the line after what you posted.

Yeah the two forms I didn't close off are the ONLY 2 working:-)

Alex · August 29, 2010

Well they should be closed off.. did you close them off now?

Modernvox · August 29, 2010

Well they should be closed off.. did you close them off now?

Yes sir and here's what I have as of now for errors

Notice: Undefined index: newBidder in C:\xampp\htdocs\process.php on line 52

Notice: Undefined index: deleteBidder in C:\xampp\htdocs\process.php on line 75

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\process.php on line 99

That Bidder Number is NOT logged in, would you like to set this bidder as active? Enter 1 for NO or 2 for YES

Alex · August 29, 2010

Can you post your full code as it stands now? (Both HTML and PHP)

Modernvox · August 29, 2010

Ok this is the form that adds item description, price, qty and biddersid number.


<form action="process.php" method="post">
<font face= "calibri" size= "4">
<table border= "1">

<tr style= "background: #cccccc">  <td><b>Item Description:</b></td>
<td><input type= "text" name= "itemDescription" size= "30" value=""/></td>
</tr>
<tr style= "background: #99ff00">   <td><b>Item Price:</b></td>
<td><input type= "text" name= "itemPrice" size= "5" value=""/></td>
</tr>
</tr style= "background: #A8E5FF">   <td><b>Winning Bidders:</b></td>
<td><input type="text" name= "winningBidder" size= "5" /></td>
</tr>
<tr>   <td><b>How many deals?:</b></td>
<td><input type="text" name= "itemQty" size= "3" value= "1" /></td>
</tr>

<br/>
<input type="reset" value="Reset Form"><input type="submit" name="submit" value= "submit" />
</font>
</table>
</form>

These are the foms that add or delete a bidders ID located on a different page other than the of the itemdescription , etc form.

<tr>    <td><font face= "calibri" size= "4"> Let's begin by adding bidders...</font></td>
</tr>
<Table border= "1">
        <form action="process.php" method="post">
        <tr style="background: #99cc33;">    <td><strong><font face= "calibri" style= "bold"> Bidders Id: </font></strong></td>
        <td><input type= "text" name= "newBidder"/></td>
</tr>
<tr>   <td><input type="Submit"  value="submit"></td>
</tr>
</table>
</form>


<tr>    <td><font face= "calibri" size= "4"> Here we can </font><font color= "red"> delete bidders.</font></td>
</tr>
<Table border= "1">
        <form action="process.php" method="post">
        <tr style="background: #ff6600;">    <td><strong><font face= "calibri" style= "bold"> Bidders Id: </font></strong></td>
        <td><input type= "text" name= "deleteBidder"/></td>
</tr>
<tr>   <td><input type="Submit"  value="submit"></td>
</tr>
</table>
</form>

Still don't understand why the script tells me delete and newbidder are undefined when I am not submitting those forms??

Modernvox · August 29, 2010

Oops and here's the process.php page for all the forms.

<?php
error_reporting(E_ALL);
ini_set("display_errors", 1);

$host= "";
$db_name= "";
$db_user= "";
$db_password= "";

ob_start();
if ($_POST['newBidder']) {
$newBidder= isset($_POST['newBidder']) ? $_POST['newBidder'] : '';
$bidderId= $newBidder;

mysql_connect("$host", "$db_user", "$db_password")or die("cannot connect to server");
mysql_select_db("$db_name")or die("cannot select DB");

$sql="SELECT * FROM bidders WHERE biddersId='$bidderId'";
$result=mysql_query($sql);

// Mysql_num_row is counting table row
$count=mysql_num_rows($result);
// If result matched $myusername and $mypassword, table row must be 1 row

if($count==0){
// Add $biddersId and redirect to anypage
mysql_Query("INSERT INTO bidders (biddersId)
VALUES ('$bidderId')");
header("Location: index.php");
exit();
  }
}

if ($_POST['deleteBidder'])
{
$deleteBidder= isset($_POST['deleteBidder']) ? $_POST['deleteBidder'] : '';
mysql_connect("$host", "$db_user", "$db_password")or die("cannot connect to server");
mysql_select_db("$db_name")or die("cannot select DB");

mysql_query("DELETE FROM bidders WHERE biddersId='$deleteBidder'");
header("Location: index.php");
exit();
}

if (isset($_POST['itemDescription'], $_POST['itemPrice'], $_POST['winningBidder'], $_POST['itemQty']))
{
$itemDescription= isset($_POST['itemDescription']) ? $_POST['itemDescription'] : '';
$itemPrice= isset($_POST['itemPrice']) ? $_POST['itemPrice'] : '';
$winningBidder= isset($_POST['winningBidder']) ? $_POST['winningBidder'] : '';
$itemQty= isset($_POST['itemQty']) ? $_POST['itemQty'] : '';

mysql_connect("$host", "$db_user", "$db_password")or die("cannot connect to server");
mysql_select_db("$db_name")or die("cannot select DB");

$sql="SELECT * FROM transactions WHERE biddersId='$winningBidder'";
$result=mysql_query($sql);

$count=mysql_num_rows($result);
// If result matched, table row must be 1 row

if($count==0){
echo "That Bidder Number is NOT logged in, ";
echo "would you like to set this bidder as active?";
echo " Enter 1 for NO or 2 for YES";
echo "<form action= \"process.php\" method= \"POST\">";
echo "<input type =\"text\" name= \"logUser\"/>";
echo "<input type= \"submit\" value = \"Submit\"/>";
$logUser= isset($_POST['logUser']) ? $_POST['logUser'] : '';
exit();
}
if ($logUser= 1) {
header("Location: inprogress.php");
exit();
}

else if ($logUser= 2){
// Add $biddersId and redirect to anypage
mysql_connect("$host", "$db_user", "$db_password")or die("cannot connect to server");
mysql_select_db("$db_name")or die("cannot select DB");

mysql_Query("INSERT INTO bidders (biddersId)
VALUES ('$winningBidder')");


mysql_query("INSERT INTO transactions (itemDescription, itemPrice, bidderId, itemQty , totalPrice) 
VALUES('$itemDescription', '$itemPrice','$winningBidder', '$itemQty', '$totalPrice')")
or die(mysql_error());   
header("Location: index.php");
exit();
   }
}

echo "<font color= \"red\" face=\"calibri\" size=\"4\">That bidder is already logged, Please press your browsers back button and try again.</font>";


ob_end_flush();

?>

Alex · August 29, 2010

Still don't understand why the script tells me delete and newbidder are undefined when I am not submitting those forms??

Because even though you're not submitting that form the code is getting a point where it's using a variable that is not defined (obviously). Just make sure that before you use a variable from an outside source that might not be set you make sure that it is using isset.

You must've edited that code because the errors you posted can't be happening on those lines in the code provided.

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Why are 5 of my 7 Variables reporting as Undefined? I'm Lost......

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