SpikeZ

Pretty well what the error message said! Foreach iterates through an array, if you dont give it an array then it cant iterate and throws the error.

in your form simply use the action attribute: <form name="contactForm" id="contactForm" action="http://myPHPserver.com/formProcess.php">

You are welcome

Darren, you need to JOIN the tables on their common fields and alias them. This probably wont run as is because I dont know your field names but it should give you an idea. select t1.product_id as t1_p_id , t1.product_name as product_name , t2.product_id as t2_p_id , t2.category_id as t2_cat_id , t3.category_id as t3_cat_id , t3.category_name as t3_cat_name from table_1 as t1 left join table_2 as t2 on ti_p_id = t2_p_id left join table_3 as t3 on t2_cat_id = t3_cat_id

If you have access to an different php enabled server you could direct the form action to that server instead. Then process the form and redirect back to the original site. Bit of a hack but it works.

Hi, you cant start a class definition with a number so .64x64 wont parse properly. Change it to .div64x64 and amend your HTML code accordingly and it will work fine

http://uk3.php.net/uniqid read the user submitted notes.

TBH I did the same! Wasnt until I set it up on a testbed that I noticed.

hey Drew Answered this on SitePoint but will post here too. You have: <input name="userfile[]" type="file" /> remove the [] as it creates a multidimentional array when the form is subitted. That kills the error checking.

phpDesigner 2008 for me, there are 2 editions - personal which is FREE and Pro which has a small fee. Tons of features if you want them as well as project management, file management and auto complete.

Hi again DFT, The problem stems from the location of the first $result variable. It is inside the first if() clause and so will only exist IF the form has been submitted. You then call the variable outside of the if() before the form has been submitted and so the variable hasnt been created yet - giving you the error. I would guess that moving the { on line 10 down to where the else is on line 16 would do the trick. <?php include("config.php"); if (isset($_POST["submit"])) { $txt = $_POST['txt']; $txt = mysql_real_escape_string($_POST['txt']); $result = mysql_query("UPDATE content SET txt='$txt' WHERE id='1'", $db); if ($result) { echo "<meta http-equiv=Refresh content=0;url=index.php>"; } } else { $result = mysql_query("SELECT * from content WHERE id='1'", $db); while ($row=mysql_fetch_assoc($result)){ require('fckeditor/fckeditor.php'); $oFCKeditor = new FCKeditor('txt'); $oFCKeditor->BasePath = 'fckeditor/'; $oFCKeditor->Value = $row['txt']; $oFCKeditor->Height = '300'; $oFCKeditor->Create(); } } ?>

Hi again DFT, The problem stems from the location of the first $result variable. It is inside the first if() clause and so will only exist IF the form has been submitted. You then call the variable outside of the if() before the form has been submitted and so the variable hasnt been created yet - giving you the error. I would guess that moving the { on line 10 down to where the else is on line 16 would do the trick. <?php include("config.php"); if (isset($_POST["submit"])) { $txt = $_POST['txt']; $txt = mysql_real_escape_string($_POST['txt']); $result = mysql_query("UPDATE content SET txt='$txt' WHERE id='1'", $db); if ($result) { echo "<meta http-equiv=Refresh content=0;url=index.php>"; } } else { $result = mysql_query("SELECT * from content WHERE id='1'", $db); while ($row=mysql_fetch_assoc($result)){ require('fckeditor/fckeditor.php'); $oFCKeditor = new FCKeditor('txt'); $oFCKeditor->BasePath = 'fckeditor/'; $oFCKeditor->Value = $row['txt']; $oFCKeditor->Height = '300'; $oFCKeditor->Create(); } } ?>

Have a look at http://uk.php.net/htmlspecialchars(). it converts charcters to their html entities

That seems a more than logical idea

You are most welcome DFT