[Resolved] Variable test to load content

rcoursey

OK,
Here I go again:

I am appending 2 variables to the URL on the previous page. One is the mode (add, edit, delete) and the other is a record number for the edit/delete.

When I run the attached code the initial test loads the first form whether the $mode variable == "add, edit, or delete". I wanted the test to display the form filled out only for edit/delete.

There is another form (blank) for $mode == 'add'

Where am I missing this logic?

code:
<?php // Home Content Add
// Begin Edit / Delete
echo $id;
echo $mode;

/ This is to make sure the above variables are assigned on this
Page ( I don't know why..just wanted to be sure since the test below does not work/
$mode = $mode;
$id = $id;
//******* REMEMBER TO DELETE ABOVE BEFORE PUBLISHING

if (!$mode == 'add' || $mode == 'edit' || $mode == 'delete');
{

// Display Record to edit or delete. The mode and id comes from URL on prior page

mysql_connect("localhost", "root", "") or die("There was a problem connecting to the database");
if (!msql_connect)
{
echo "There is a problem connecting to the server.";
exit;
}
mysql_select_db("dunhill");

$query = "SELECT * FROM homecontent WHERE hpcid='$id'";
$result = mysql_query($query);
if (!$result)
{
echo "The query failed.";
exit;
}
$row = mysql_fetch_array($result);
$num_rows = mysql_num_rows($result);
echo $num_rows;
// Test to see the result variables below
while ($row = mysql_fetch_array($result))
{
echo $row["hpcid"];
echo $row["headline"];

echo '\n';
}

// Display from

?>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<title>Home Page Management</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>

<body>
<h3>Edit or Delete Home Page Content</h3>
<form action="home_action.php" method="post">
snip...
</form>
snip
</body>
</html>
<?
}
// End Edit / Delete

*****the next form starts below

if ($mode == 'add')
{ // This form is dispalyed if edit or delete

?>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
snip.....

sdjensen

Change the lines

$mode = $mode; 
$id = $id;

$mode = $_POST['mode']; 
$id = $_POST['id'];

sfullman

this code:

mysql_connect("localhost", "root", "") or die("There was a problem connecting to the database");
if (!msql_connect)
{
echo "There is a problem connecting to the server.";
exit;
}

.. is worthless, msql_connect is wrong, try this:

$db_cnx=mysql_connect("localhost", "root", "") or die(mysql_error() . ":" . mysql_errno());

//it'll die automatically so you don't need the exit statement

============================

I found the following also clumsy:
$query = "SELECT * FROM homecontent WHERE hpcid='$id'";
$result = mysql_query($query);
if (!$result)
{
echo "The query failed.";
exit;
}

Try this:
$query = "SELECT * FROM homecontent WHERE hpcid='$id'";
$result = mysql_query($query) or die(mysql_errno() . ":" . mysql_error());

does the same thing

rcoursey

Thanks for the input. I tried it but it did not solve my problem.

The reason have the "echo $mode" statement at the beginning of the program is to see if the variable comes from the prior page. I see the correct value for the two variables at the top of the form when I echo the values of $mode and $id prior to testing the values. Yet the "if " statement allows a $mode with the value of "add" to meet the condition and display the first form for edit/delete instead of the form at the end of the script for $mode == 'add'

Any ideas on why the initial "IF" statement does not correctly prevent the $mode == 'add' from utilizing the block of code?

sfullman

Your logic is squirrely, you don't need:

if (!$mode == 'add' || $mode == 'edit' || $mode == 'delete');
{

// Display Record to edit or delete. The mode and id comes from URL on prior page

if your only options are add edit delete, just:

if($mode=='add' || $mode=='edit'){
....

}

your logic is faulty below too:

if ($mode == 'add')
{ // This form is dispalyed if edit or delete

Shouldn't it be:
if($mode !=='add'){
..

I don't remember but you are able to echo $mode right? Obviously none of this works until $mode is passed.

Sincerley,
Sam Fullman

rcoursey

You are correct. I can see the variables at the topof the page.

The last form should be displayed as a result of $mode == 'add' . I left the NOT out of the comment.

I had the if statement without the 'add' portion but the script would display the Edit/Delete form regardless of what the mode was. That is why I added the add condition.

BTW, I liked your suggestion on connections!

rcoursey

OK here we go:

This is the complete logic code for this script at this time.

<?php // Home Content Add Products
// Begin Edit / Delete
echo $id;
echo $mode; //******* REMEMBER TO DELETE THIS BEFORE PRODUCTION

// ********The above echo statements are giving the correct info from a known record

if ($mode=='edit' || $mode=='delete');
{

// Display Record to edit or delete

$db_cnx=mysql_connect("localhost", "root", "") or die(mysql_error() . ":" .
mysql_errno());

mysql_select_db("dunhill");

$query = "SELECT * FROM homecontent WHERE hpcid='$id'";
$result = mysql_query($query) or die(mysql_errno() . ":" . mysql_error());

$row = mysql_fetch_array($result);
// Result echo test
// *****This is not coming thru when I call the echo
if ($result)
{
echo $row["hpcid"] . " This rec_id \n ";
echo $row["headline"] . " This Headline \n ";

}

// Display from
// End Edit / Delete form
//test if add is passed in $mode
if ($mode=='add')
{ // This form is dispalyed if edit or delete is not passed
//begin add record form

My problem is;

Regardless what the value of $mode is I get the Edit/Delete Form.
I can see from the original "echo $mode" the value of the variable.
If $mode == add i should see the from designed for adding a record.

I feel like I am going in circles

sfullman

You do a good thing by echoing mode before your first logic. So you now know that that works.

Try echoing $mode again before your next if statement:

echo $mode;
if ($mode=='add')
{ // This form is dispalyed if edit or delete is not passed
//begin add record form

maybe something is changing the value of $mode. Maybe you have $mode in $_COOKIE somewhere.

I suggest if mode is coming in through a querystring, you use $GET['mode'], or $POST['mode'] if via a form post. Beyond that I can't help you that much.

Sincerely,
Sam Fullman

rcoursey

This works!!!

<?php // Home Content Add Products
// Begin Edit / Delete
echo $id;
echo $mode; //******* REMEMBER TO DELETE THIS BEFORE PRODUCTION

// ********The above echo statements are giving the correct info from a known record

if ($mode !== 'add')// ******added this
{
if ($mode=='edit' || $mode=='delete');
{
// ******added this to catch $mode
?> <br> The mode after the test is:<br>
<?
echo $mode;?><br><?

// Display Record to edit or delete

$db_cnx=mysql_connect("localhost", "root", "") or die(mysql_error() . ":" .
mysql_errno());

mysql_select_db("dunhill");

$query = "SELECT * FROM homecontent WHERE hpcid='$id'";
$result = mysql_query($query) or die(mysql_errno() . ":" . mysql_error());

$row = mysql_fetch_array($result);
$num_rows = mysql_num_rows($result);
// Result echo test
// *****This is not coming thru when I call the echo
if ($num_rows > 0)
{
echo $row["hpcid"] . " This rec_id \n ";
echo $row["headline"] . " This Headline \n ";

}

// Display from

I just don't know why the other way did not.

Thanks for all the input. I hope I am able to give back to the community soon.

Richard

tomhath

Remember to do what I always do when I make an error like this; pound your head repeatedly on the table. Look at the end of this line, notice the ";", that's what's causing the problem:

if ($mode=='edit' || $mode=='delete');

rcoursey

My forehead is starting to bleed....

when should I stop?