NME
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Everything posted by NME
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? what do u mean? am i making 2d arrays incorrectly? im under the impression that to make 2d arrays in javascript, one must do the following: (ie for a 2x2 array). var stack = new Array(2); stack[1] = new Array(2); stack[2] = new Array(2); stack[1,1]="hello"; stack[2,1]="world"; document.write(stack[1,1] + " " + stack[2,1]);
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stack[1]=new Array(2); stack[2]=new Array(2); stack[1,1]="hello"; stack[2,1]="world"; document.write(stack[1,1] + " " + stack[2,1]); quick question.. why does the above write out the following: "world world"; i need it to write out: "hello world"; any ideas. thanks.
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ok i figured that might be a problem. so i put this in: ini_set("upload_max_filesize", "8M"); does that seem alright? i just want to increase the max_filesize to 8MB from 2 MB standard.
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haha crap i dont think thats what i meant. my problem is that i added a php.ini file to my server, but now when i do phpinfo(); it still shows information from the default php.ini that my webhost uses.
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ok i kinda get what ur talking about. when i make this element.php i should basically put in something like include ' php.ini'; ? or something else. sorry i just dont understand how to literally link the php.ini page.
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simple question. i made a php.ini, how do i get all my pages to link to it? do i just put it in the root folder of the server?
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any suggestions would be really appreciated.
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hey guys, thanks for all the help so far. but the problem is still there. here is the updated code: <html> <body> <!-- The data encoding type, enctype, MUST be specified as below --> <form enctype="multipart/form-data" action="uploadtest_script.php" method="POST"> <!-- MAX_FILE_SIZE must precede the file input field --> <input type="hidden" name="MAX_FILE_SIZE" value="30000" /> <!-- Name of input element determines name in $_FILES array --> Send this file: <input name="userfile" id="userfile" type="file" /> <input type="submit" value="Send File" /> </form> </body> </html> and its script: <? ini_set("memory_limit", "32M"); ini_set("max_execution_time", 0); error_reporting(E_ALL & ~E_NOTICE); print "size:". $_FILES['userfile']['size'] . " "; print "name:". $_FILES['userfile']['name'] . " "; print "type:". $_FILES['userfile']['type'] . " "; print "tempname:" . $_FILES['userfile']['tmp_name'] . ""; ?> and its result: size:0 name:2.mp3 type: tempname:
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in my php.ini i set the post_max_size to 8M. is there something else that needs to change? the mp3 i uploaded was 3.7M
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Hey guys, im trying to upload mp3's onto my site. i know how to do it with images but I'm getting some problems when i select an mp3 file. I cut the problem down to the following and made it very easy to read: <html> <body> <form action="uploadtest_script.php" method="post" enctype="multipart/form-data"> <label for="file">Filename:</label> <input type="file" name="file" id="file" /> <br /> <input type="submit" name="submit" value="Submit" /> </form> </body> </html> and the script is here. its a simple print out of all the FILES variables: <? print "name:". $_FILES['file']['name'] . " "; print "type:". $_FILES['file']['type'] . " "; print "size:". $_FILES['file']['size'] . " "; print "tempname:" . $_FILES['file']['tmp_name'] . ""; ?> and these are the results im getting???: size:0 name:2.mp3 type: tempname: Basically, the variables are not being set to the FILES array. any suggestions? thanks.
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Thanks a ton man. Your awesome
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Hey guys, I was just wondering if there was any command to jump to a different page without clicking a hyperlink. An example is suppose the user clicks a link that leads to a file where there are a few mysql commands, and then i want to go back to the initial page immediately after those commands are completed. I already know about include, but for some reason its not working when i do something like... include 'www.blah.com/blah.php?blah=$blahid'.
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any help would be really appreciate guys
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hmm im still getting an error: You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near ''Info_User_19' ( 'index' INT( 25 ) NOT NULL AUTO_INCREMENT , by the way, thanks so much for all your time poirot.
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Ok, this is what is says, but i still cant figure it out: You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near 'int(25) NOT NULL auto_increment, freinds int(25) NOT NULL,
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Thanks, but it stil doesnt create me a table.
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I am trying to dynamically create a table with a different name everytime. The following code is used, but it never seems to actually create a Table when i look at the database. [code]$dbhost = '<myhost>'; $dbusername = '<myusername>'; $dbpasswd = '<mypassword>'; $database_name = '<mydatabasename>'; $connection = mysql_pconnect("$dbhost","$dbusername","$dbpasswd") or die ("Couldn't connect to server."); $db = mysql_select_db("$database_name", $connection) or die("Couldn't select database."); $sql = mysql_query ( "CREATE TABLE user_" . $userid . " ( index int(25) NOT NULL auto_increment, count int(25) NOT NULL, PRIMARY KEY (index) )"); [/code] Any ideas why nothing is happening?
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ah ofcourse, im an idiot. thanks so much. should've been SELECT * FROM ...
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This is my code: [code]$query= "SELECT userid, first_name FROM users2 WHERE username='$username' AND password='$password'"; $result = @mysql_query($query); $row = mysql_fetch_array($result, MYSQL_NUM); //returns the row with that users information in the users2 table. if($row) { $_SESSION['userid']=$row[0]; $_SESSION['first_name']=$row[1]; $_SESSION['last_name']=$row[2];[/code] for some reason $row[2] returns nothing (meaning $_SESSION['last_name'] is nothing as well), although $row[1] and $row[0] return the values they intend from my SQL table. why won't row[3] or any subsequent rows show? thanks.
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Ok i got the solution, thanks for the help guys turns out it was a session problem: session_register('list1'); $_SESSION['list1']=$list1; i commented out the $_SESSION line... i didnt think i needed it, guess i do. thanks
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[!--quoteo(post=375327:date=May 19 2006, 04:09 PM:name=Crayon Violent)--][div class=\'quotetop\']QUOTE(Crayon Violent @ May 19 2006, 04:09 PM) [snapback]375327[/snapback][/div][div class=\'quotemain\'][!--quotec--] [code] $sql3 = mysql_query("INSERT INTO '$list1' (user_id, fullname) VALUES ('$userid', '$name')") or die (mysql_error()); [/code] try putting quotes around $list1 [/quote] I already tried that. I get this error: You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near ''' (user_id, fullname) VALUES ('1121', 'Jeff Gordon')' thanks tho
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My code looks like this : [code]$sql3 = mysql_query("INSERT INTO $list1 (user_id, fullname) VALUES ('$userid', '$name')") or die (mysql_error());[/code] However, it does not work and I get the following error: You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near '(user_id, fullname) VALUES ('1121', 'Jeff Gordon')' can anyone tell me what I'm doing wrong. Any help would be greatly appreciated. Thanks