[Newsletter Signup Script Small Problem]

Well, for one thing, the correct syntax for SQL INSERT is:

INSERT INTO table_name VALUES (value1, value2,....)
not
INSERT INTO table_name SET column = value, column = value, ....

I don't know much about WordPress, so I'll leave it at that.

[Site Logging...]

You could use the referer variable in php, $_SERVER['HTTP_REFERER'], but just remember that the referer can be spoofed very easily.

[Adding submitted and current values together]

[!--quoteo(post=352665:date=Mar 7 2006, 06:04 PM:name=master82)--][div class=\'quotetop\']QUOTE(master82 @ Mar 7 2006, 06:04 PM) [snapback]352665[/snapback][/div][div class=\'quotemain\'][!--quotec--]$sql = mysql_query("SELECT * FROM Users");

$finalnumber = intval($_POST['sales']) * (0.75);
$finalnum = ($finalnumber + intval($row['salestodate']));
[/quote]

In this line:
$finalnum = ($finalnumber + intval($row['salestodate']));
you haven't fetched the results from:
$sql = mysql_query("SELECT * FROM Users");

You need to use:
[code]$sql = mysql_query("SELECT * FROM Users");
$row = mysql_fetch_assoc($sql);

$finalnumber = intval($_POST['sales']) * (0.75);
$finalnum = ($finalnumber + intval($row['salestodate']));[/code]

[Having troubles - image does not show up.]

[!--quoteo(post=352656:date=Mar 7 2006, 05:33 PM:name=sjones)--][div class=\'quotetop\']QUOTE(sjones @ Mar 7 2006, 05:33 PM) [snapback]352656[/snapback][/div][div class=\'quotemain\'][!--quotec--]
$record = mysql_fetch_row($result);
//...
<td><img src='http://uswebproducts.com/country_savings/uploads/$record'>
[/quote]
mysql_fetch_row returns an array, that's why the image displayed is:
"http://uswebproducts.com/country_savings/uploads/Array"
Use $record[0...1...2...etc] according to field order

It would be better to use mysql_fetch_array or mysql_fetch_assoc in your case.

[Updating a single field in MYSQL]

You could:

[code]$sql = mysql_query("SELECT * FROM Users WHERE user_id = '$userid'"); //I'm just guessing about the user_id.
$row = mysql_fetch_assoc($sql);

$finalnumber = intval($_POST['sales']) * (0.75);
$finalnumber = $finalnumber + intval($row['num_field']);

$sql = mysql_query("UPDATE Users SET num_field = '$finalnumber'");
if($sql) {
  echo "Success!";
} else {
  echo "Failed...";
}[/code]
Of course, replace 'num_field' with the field in your database.

[Limiting # of Search Results Charactors]

You could use substr() to limit the characters to 50, 100, etc.

[code]string substr ( string string, int start [, int length] )[/code]
So, I'm guessing in your case it would be:

[code]$message = substr($message, 0, 50);[/code]
For viewing only text, you could use the strip_tags() function

[code]string strip_tags ( string str [, string allowable_tags] )[/code]

[PHP form not dropping data????]

Should be that you are checking for 'submit' to be empty, while your submit button's name is 'vendor_id':

[code]if(empty($_POST['submit'])) { //If the submit button has not been set we will echo a welcome statement[/code]

[code]<input name='vendor_id' type='submit' value='Remove'>[/code]
So.. it is [b]always[/b] trying to remove the vendor_id 'Remove'...
It should be trying to remove $_POST['cat_id']

[PHP form not dropping data????]

Your SQL syntax for DELETE is wrong, the correct syntax is:

[code]DELETE FROM table WHERE column = 'value';[/code]

[Simple form]

Well, first, are you sure your server has PHP installed? You might want to check that before anything.
And, second, are you using $_POST to get the values before this code?
Thirdly, you don't need to do $var = "$var", that's just wasting milliseconds :P

[PHP Newbie Question]

If you are running community.php then why do you want to check if you're running community.php?

[using ereg_replace(with variable stored in string)]

Hi there,
In order for PHP to parse (add the actual value of the var) you need double quotes around it:

$theData = ereg_replace('<b>', "<b> $person_name", $theData);
echo $theData;

OR:

$theData = ereg_replace('<b>', '<b> '.$person_name, $theData);
echo $theData;

[Need help with Dynamic list menu??]

You copied the else two times, use this:

[code]include ('../connections/mysql_connect.php');
if(empty($_POST['submit'])) {
    echo "<span class='content_blank'><strong>Welcome....</strong><BR></span>
          <span class='content_blank'>You can enter a new Vendor by selecting the category that they belong<BR></span>
          <span class='content_blank'>and entering the name of the new vendor</span><BR><BR>";
}else{
    if(!empty($_POST['cat_id']) & !empty($_POST['name'])) {
        $qry = mysql_query("INSERT INTO vendors (cat_id,vendor_name) VALUES ('".$_POST['cat_id']."','".$_POST['name']."');");
        if($qry) { echo "<span class='content_blank'><strong>New vendor has been added</strong></span>"; }
        else { echo "<span class='content_blank'><strong>Error inserting new vendor</strong></span>"; }
    } else {
        echo "<span class='content_blank'><strong>Error: something was left blank</strong></span>";
    }
}

echo "<table width='100%'  border='0' cellspacing='0' cellpadding='0'>
<tr>
<td>
   <span class='content_blank'>Please choose a category and type in the name for the new vendor.</span><br><br>
   <form method='post' action='csmag_admin.php?c=add_vendor'>
   <select name='cat_id'>
   <option>- Please select one -</option>\n";
   $qry = mysql_query("SELECT * FROM categories;");
   while($rows = mysql_fetch_assoc($qry)) {
       echo "<option value='".$rows['cat_id']."'>".$rows['cat_name']."</option>\n";
   }
   echo "</select>
   <input name='name' type='text' size=20>
   <input name='submit' type='submit' value='Submit'>
   </form>
</td>
</tr>
</table>";[/code]

[Need help with Dynamic list menu??]

The <select>'s options have the cat_id as the value when you choose them, therefore you don't need to get it out of the database when you are inserting it. What you actually see in the select is the cat_name.
When you insert the new vendor, yes it will auto increment the vendor_id.

PS: the form needs to be submitted to [b]'cs_mag.php?c=add_vendor'[/b] not [b]'cs_mag?c=add_vendor.php'[/b]

[Need help with Dynamic list menu??]

Um, well something like this?:

[code]if(empty($_POST['submit'])) {
    echo "<span class='content_blank'><strong>Welcome....</strong></span>";
} else {
    if(!empty($_POST['vendorid']) & !empty($_POST['name'])) {
        $qry = mysql_query("INSERT INTO vendors (cat_id,vendor_name) VALUES ('".$_POST['vendorid']."','".$_POST['name']."');");
        if($qry) { echo "<span class='content_blank'><strong>New vendor has been added</strong></span>"; }
        else { echo "<span class='content_blank'><strong>Error inserting new vendor</strong></span>"; }
    } else {
        echo "<span class='content_blank'><strong>Error: something was left blank</strong></span>";
    }
}

    } else {
        echo "<span class='content_blank'><strong>Error: something was left blank</strong></span>";
echo "<table width='100%'  border='0' cellspacing='0' cellpadding='0'>
<tr>
<td>
   <span class='content_blank'>Please choose a category and type in the name for the new vendor.</span><br><br>
   <form method='post' action='csmag_admin.php?c=add_vendor'>
   <select name='vendorid'>
   <option>- Please select one -</option>\n";
   $qry = mysql_query("SELECT * FROM categories;");
   while($rows = mysql_fetch_assoc($qry)) {
       echo "<option value='".$rows['cat_id']."'>".$rows['cat_name']."</option>\n";
   }
   echo "</select>
   <input name='name' type='text' size=20>
   <input name='submit' type='submit' value='Submit'>
   </form>
</td>
</tr>
</table>";[/code]