Worstof Posted February 12, 2008 Share Posted February 12, 2008 Hello...just started yesterday trying out PHP and mysql. I have created a database and connect to it and add records through a simple html form. However, whenever I try to pass the input I type in via the form to either the form itself (via $_SERVER['PHP_SELF'] or calling another php file) nothing gets passed. When I try to use the $_SERVER method, I get a HTTP error 404, page not found add.phpaddrsvp=1 or when I pass to another php file, the code adds an empty record to the database. Right now, I'm on an iteration where I have two seperate files...one to enter the data and another to add to the database. I've included the code here. I'm at a loss...I've searched the forums with no success...I'm sure it is something very basic that I'm missing. Thanks for the help. -----This is the first file. Accepts the simple input and calls add.php ---------- <html> <head><title>User Admin Page : Add a User</title></head> <body bgcolor="#ffffff"> <table bgcolor=#000000 valign=top align=center border=0><tr><td bgcolor=#000000><table cellpadding=4 bgcolor=#ffffff cellspacing=2 border=0> <Tr><th>Add a User</th></tr><tr><td> <FORM METHOD="post" ACTION="add.php"> Last Name: <INPUT TYPE=text MAXLENGTH=70 NAME="lastname" SIZE=20><Br> First Name: <INPUT TYPE=text MAXLENGTH=70 NAME="firstname" SIZE=20><Br> <INPUT TYPE=submit VALUE="Add"> <INPUT type=reset VALUE="Reset Form"></form> </tr></td></table></tr></td></table> </body> </html> -----This is the second file...supposed to get the lastname and firstname from the first file and add to database but it doesn't------ <html> <body> <?php $db = mysql_connect("localhost","myusername","mypassword"); mysql_select_db (mydbtable); $result = mysql_query ("INSERT INTO mydbtable(lastname, firstname) VALUES ('$lastname', '$firstname')"); if(!$result) { echo "<b>User not added:</b> ", mysql_error(); exit; } if($result) { mysql_close($db); print "User <b>$lastname</b> added sucessfully!"; } else { print ("Didn't work"); } ?> </body> </html> Link to comment https://forums.phpfreaks.com/topic/90703-new-to-php-need-help-input-data-into-a-record/ Share on other sites More sharing options...
soycharliente Posted February 12, 2008 Share Posted February 12, 2008 You need to grab the data from the POST array. I don't see where you are doing that. $lastname = $_POST["lastname"]; Something to that effect for each variable that you're trying to work with. Link to comment https://forums.phpfreaks.com/topic/90703-new-to-php-need-help-input-data-into-a-record/#findComment-464929 Share on other sites More sharing options...
Worstof Posted February 12, 2008 Author Share Posted February 12, 2008 Thanks a ton! That made it work. I still can't get the $_SERVER[php_SELF] to pass arguments back to itself. Any thoughts would be appreciated. <html> <head> <title></title> </head> <body bgcolor="#FFFFFF"> <?php if (isset($_GET['addrsvp'])): //add a RSVP ?> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <label>Last name: <input type="test" name="lastname" /></label><br /> <label>First name: <input type="test" name="firstname" /></label><br /> <input type="submit" value="SUBMIT" /> </form> <?php else: //Default page display //Connect to DB server $dbcnx=@mysql_connect('localhost', 'db', 'pswd'); if (!$dbcnx) { exit('<p>Unable to connect to server</p>');} //Select the rsvp database if (!@mysql_select_db('mydb')) { exit('<p> Unable to connect to database</p>'); } if (isset($_POST['lastname'])) { $lastname=$_POST['lastname']; $firstname=$_POST['firstname']; $sql="INSERT INTO rsvp SET lastname='$lastname', firstname='$firstname'"; // if (!@mysql_query($sql)) { // echo '<p>Did not add RSVP to database</p>'; // } } //echo '<p>Here are all the people registered.</p>'; //Request all the rsvps //$result=@mysql_query('SELECT lastname FROM rsvp'); //if (!$result) { // exit('<p>Error performing query: ' . mysql_error() . '</p>; //} //while ($row=mysql_fetch_array($result)) { // echo '<p> .$row['lastname'] . '</p>'; //} //When clicked this link will load this page with the people listed echo '<p><a href"' . $_SERVER['PHP_SELF'] . '?addrsvp=1">RSVP</a></p>'; ?> </body> </html> Link to comment https://forums.phpfreaks.com/topic/90703-new-to-php-need-help-input-data-into-a-record/#findComment-465186 Share on other sites More sharing options...
chronister Posted February 12, 2008 Share Posted February 12, 2008 Maybe it's just me, but please explain a little more clearly on what your trying to accomplish... also, instead of <?php echo '<p><a href"' . $_SERVER['PHP_SELF'] . '?addrsvp=1">RSVP<a></p>'; ?> Try, <?php echo '<p><a href=" ' . $_SERVER['PHP_SELF'] . '?addrsvp=1">RSVP</a></p>'; ?> Nate Link to comment https://forums.phpfreaks.com/topic/90703-new-to-php-need-help-input-data-into-a-record/#findComment-465193 Share on other sites More sharing options...
Worstof Posted February 13, 2008 Author Share Posted February 13, 2008 Thanks...I'll fix the /a piece. Think I've got this fixed. Thanks for the help. Link to comment https://forums.phpfreaks.com/topic/90703-new-to-php-need-help-input-data-into-a-record/#findComment-465519 Share on other sites More sharing options...
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