jbegreen

June 10, 2013

Thank you. Got it working with the EOD though, used your technique of not using an echo, didn't display as php but seems to work so THANKS

June 10, 2013

Hey can anyone help me.

Im trying to pass some for variables to an EOD statement as a confirmation slash profile preview page. However when i try to insert the variables surrounded with the standard <? ?> tags i get a message about unexpected white space.

here is the end o0f my processing script and the confirmation EOD which is echoes at the end of the script.

GetSQLValueString($insert_upload1, "text"),

GetSQLValueString($insert_upload2, "text"),

GetSQLValueString($insert_upload3, "text"),

GetSQLValueString($insert_upload4, "text"),

GetSQLValueString($insert_upload5, "text"));

mysql_select_db($database_w2w, $w2w);

$Result1 = mysql_query($insertSQL, $w2w) or die(mysql_error()); }

?>

<?

$confirmation = <<<EOD

// I WANT TO INSERT VARIABLES HERE FROM $_POST TO DISPLAY THEIR INFORMATION AFTER REGISTRATION AS A SORT OF PREVIEW PAGE

// THIS IS WHAT I HAVE TRIED

$password_entry = $_POST['password_entry'];

$business_name = $_POST['business_name'];

$address1 = $_POST['address1'];

$address2 = $_POST['address2'];

$address3 = $_POST['address3'];

$postcode = $_POST['postcode'];

$email = $_POST['email'];

$website = $_POST['website'];

$facebook = $_POST['facebook'];

$twitter = $_POST['twitter'];

$linkedin = $_POST['linkedin'];

$openfrom1 = $_POST['openfrom1'];

$openfrom2 = $_POST['openfrom2'];

$openfrom3 = $_POST['openfrom3'];

$openfrom4 = $_POST['openfrom4'];

$openfrom5 = $_POST['openfrom5'];

$openfrom6 = $_POST['openfrom6'];

$openfrom7 = $_POST['openfrom7'];

$opento1 = $_POST['opento1'];

$opento2 = $_POST['opento2'];

$opento3 = $_POST['opento3'];

$opento4 = $_POST['opento4'];

$opento5 = $_POST['opento5'];

$opento6 = $_POST['opento6'];

$opento7 = $_POST['opento7'];

$business_description = $_POST['business_description'];

$managers_message = $_POST['managers_message'];

echo $username_entry; ?>

<head>

<title>Registration complete! Thank you.</title>

but it seems i am unable to do this. Can anybody shed some light? Im sure its something silly.

Thanks

June 6, 2013

Can you explain please.

$query = mysql_query ("SELECT * FROM ofnissenisub321 WHERE username = $username_entry ");

while($rows = mysql_fetch_array("$query"));

Newbie

Thanks.

June 6, 2013

Hi Guys

Can anyone help me with this

<?php require_once('Connections/w2w.php'); ?>

<?

$username_entry = 'jack';

$query = mysql_query ("SELECT * FROM ofnissenisub321 WHERE username = $username_entry ");

$rows = mysql_fetch_array($query);

$username = $rows['username'];

$password = $rows['password'];

$business_name = $rows['business_name'];

$address1 = $rows['address1'];

$address2 = $rows['address2'];

$address3 = $rows['address3'];

$postcode = $rows['postcode'];

$email = $rows['email'];

$website = $rows['website'];

$facebook = $rows['facebook'];

$twitter = $rows['twitter'];

$linkedin = $rows['linkedin'];

$openfrom1 = $rows['openfrom1'];

$openfrom2 = $rows['openfrom2'];

$openfrom3 = $rows['openfrom3'];

$openfrom4 = $rows['openfrom4'];

$openfrom5 = $rows['openfrom5'];

$openfrom6 = $rows['openfrom6'];

$openfrom7 = $rows['openfrom7'];

$opento1 = $rows['opento1'];

$opento2 = $rows['opento2'];

$opento3 = $rows['opento3'];

$opento4 = $rows['opento4'];

$opento5 = $rows['opento5'];

$opento6 = $rows['opento6'];

$opento7 = $rows['opento7'];

$business_type = $rows['business_type'];

$business_description = $rows['business_description'];

$managers_message = $rows['managers_message'];

$image1 = $rows['image1'];

$image2 = $rows['image2'];

$image3 = $rows['image3'];

$image4 = $rows['image4'];

$image5 = $rows['image5'];

echo "$password";

?>

I get the following error Warning mysql_fetch_array is not a valid SQL result resource at........

Why is this happening??

Hope somebody can help

Thanks

May 31, 2013

solved it a different way.

You were right, by using a print_r i could get to the bottom of it

$image_upload1 = ($files['name'][0]);

thanks.

May 31, 2013

just says submit

May 31, 2013

The array displays fine, I'm trying to assign the 0 elements to a variable.

May 31, 2013

Hey Guys,

I have an Array Count On file uploads (Upto 5 Files) Called Upload[0];

Is this right to use an individual filename from the array as a variable?

$image_upload1 = $_POST['upload'[0];

As i want to use is in this...

$insert_upload1 = "user_images/$username_entry/$image_upload1";

As $insert_upload1 is the variable for an INSERT value on my Database.

I think its something in the display of the 'upload[ ] array.

Can anyone shed any light

Thanks

May 31, 2013

Is there anyway you could go into more detail with the PDO. Not very familiar with it. THanks so much.

J

May 31, 2013

Yes! HaHa! I knew it was something like that! I appreciate your time.

May 31, 2013

for the array count. Is there any easy way of selecting the file path (with variables) for each count and inserting into separate fields in a MySQL database?

Just thought i would throw that out there!

Cheers

Jack

May 31, 2013

Thanks DaveyK it was the curly braces! Although i didn't use these last time.

Thanks for your help!

May 31, 2013

Array(    [upload] => Array        (            [name] => Array                (                    [0] => test.jpg                    [1] => facebook_cover_small copy.jpg                    [2] =>                     [3] =>                     [4] =>                 )            [type] => Array                (                    [0] => image/jpeg                    [1] => image/jpeg                    [2] =>                     [3] =>                     [4] =>                 )            [tmp_name] => Array                (                    [0] => C:\WINDOWS\Temp\php93B.tmp                    [1] => C:\WINDOWS\Temp\php93C.tmp                    [2] =>                     [3] =>                     [4] =>                 )            [error] => Array                (                    [0] => 0                    [1] => 0                    [2] => 4                    [3] => 4                    [4] => 4                )            [size] => Array                (                    [0] => 595377                    [1] => 60665                    [2] => 0                    [3] => 0                    [4] => 0                )        ))

There are 5 files possible to upload. Used 2 to demonstrate print r.

Thanks for your help.

May 31, 2013

Hey Wondered if anyone could help me with this. I had this right a couple of days ago but for some reason it wasn't in my code yesterday and since rebuilding, has ceased to work.

I have a multi file upload that creates a directory as the name of the username variable ($username_entry).

When using this variable in the file path along with the variable for the name of the file ($name) the file path simply puts the variable text "$name" as some random blank file within the newly created directory.

Im sure it something pretty small that needs editing to use 2 variables in 1 path.

Anyway here's the code i need tweaking. Thanks to anyone that can help me

<?php

if (isset($_FILES['upload']) === true) {

$files = $_FILES['upload'];

for($x = 0; $x < count($files['name']); $x++) {

$name = $files['name'][$x];

$tmp_name = $files['tmp_name'][$x];

move_uploaded_file($tmp_name, 'user_images/'.$username_entry.'/ . $name');

}

?>

Its the move_uploaded_file($tmp_name, 'user_images/'.$username_entry.'/ . $name'); where i think the problem lies.......any takers?

Thanks.

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jbegreen

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