jbegreen
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Trying to insert variables inside an EOD statement
jbegreen replied to jbegreen's topic in PHP Coding Help
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 -
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> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> <title>Registration complete! Thank you.</title> <style type="text/css"> but it seems i am unable to do this. Can anybody shed some light? Im sure its something silly. Thanks
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Can you explain please. $query = mysql_query ("SELECT * FROM ofnissenisub321 WHERE username = $username_entry "); while($rows = mysql_fetch_array("$query")); Newbie Thanks.
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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
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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.
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just says submit
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The array displays fine, I'm trying to assign the 0 elements to a variable.
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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 $insert_upload1 is the variable for an INSERT value on my Database. I think its something in the display of the 'upload[ ] array. Can anyone shed any light Thanks
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Is there anyway you could go into more detail with the PDO. Not very familiar with it. THanks so much. J
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Yes! HaHa! I knew it was something like that! I appreciate your time.
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for the array count. Is there any easy way of selecting the file path (with variables) for each count and inserting into separate fields in a MySQL database? Just thought i would throw that out there! Cheers Jack
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Thanks DaveyK it was the curly braces! Although i didn't use these last time. Thanks for your help!
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Array( [upload] => Array ( [name] => Array ( [0] => test.jpg [1] => facebook_cover_small copy.jpg [2] => [3] => [4] => ) [type] => Array ( [0] => image/jpeg [1] => image/jpeg [2] => [3] => [4] => ) [tmp_name] => Array ( [0] => C:\WINDOWS\Temp\php93B.tmp [1] => C:\WINDOWS\Temp\php93C.tmp [2] => [3] => [4] => ) [error] => Array ( [0] => 0 [1] => 0 [2] => 4 [3] => 4 [4] => 4 ) [size] => Array ( [0] => 595377 [1] => 60665 [2] => 0 [3] => 0 [4] => 0 ) )) There are 5 files possible to upload. Used 2 to demonstrate print r. Thanks for your help.
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Hey Wondered if anyone could help me with this. I had this right a couple of days ago but for some reason it wasn't in my code yesterday and since rebuilding, has ceased to work. I have a multi file upload that creates a directory as the name of the username variable ($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.