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Needed help with altering code in mysql


mallika

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Hi,
I am really new to mysql, i am having a problem with a code.I have a form that can be seen here:
[url=http://hunstem.uhd.edu/savephp/form1.html]http://hunstem.uhd.edu/savephp/form1.html[/url]
In the last section where it says Judging references, it does not allow me to save without filling all the cells.I need to save without entering some or all cells.Please help me.
Thanks,
Mallika.
The code for the script to save is here:
[code]<?
include "db.php";

$lname = $_POST['lname'];
$fname = $_POST['fname'];
$cmpaff = $_POST['cmpaff'];
$title = $_POST['title'];
$street = $_POST['street'];
$city = $_POST['city'];
$zip = $_POST['zip'];
$ophc = $_POST['ophc'];
$oph = $_POST['oph'];
$hphc = $_POST['hphc'];
$hph = $_POST['hph'];
$faxc = $_POST['faxc'];
$fax = $_POST['fax'];
$email = $_POST['email'];
$degree = $_POST['degree'];
$ya = $_POST['ya'];
$disc = $_POST['disc'];
$sefhdisc = $_POST['sefhdisc'];
$yj = $_POST['yj'];
$jpip1 = $_POST['jpip1'];
$jpip2 = $_POST['jpip2'];
$jpip3 = $_POST['jpip3'];
$jpip4 = $_POST['jpip4'];
$jpip5 = $_POST['jpip5'];
$jpip6 = $_POST['jpip6'];
$jpip7 = $_POST['jpip7'];
$jpip8 = $_POST['jpip8'];
$jpip9 = $_POST['jpip9'];
$jpip10 = $_POST['jpip10'];
$jpip11 = $_POST['jpip11'];
$jpip12 = $_POST['jpip12'];
$jptp1 = $_POST['jptp1'];
$jptp2 = $_POST['jptp2'];
$jptp3 = $_POST['jptp3'];
$jptp4 = $_POST['jptp4'];

if ($_POST['phase1']) {
$phase1 = "on";
} else {
  $phase1 = "off";
}
if ($_POST['phase2']) {
$phase2 = "on";
} else {
  $phase2 = "off";
}
if ($_POST['intd']) {
$intd = "on";
} else {
  $intd = "off";
}
   
$Query = "INSERT INTO form1 VALUES ('0', '$lname', '$fname', '$cmpaff', '$title', '$street', '$city', '$zip', '$ophc', '$oph', '$hphc', '$hph', '$faxc', '$fax', '$email', '$degree', '$ya', '$disc', '$sefhdisc', '$yj', '$phase1', '$phase2', $jpip1, $jpip2, $jpip3, $jpip4, $jpip5, $jpip6, $jpip7, $jpip8, $jpip9, $jpip10, $jpip11, $jpip1, $jptp1, $jptp2, $jptp3, $jptp4, '$intd')";

$Result = mysql_db_query ($DBName, $Query);
mysql_close  ($Link);

if ($Result) {
  print "Data saved correctly";
} else {
  print "Error Reported on pform1.php<br>";
print "SQL: <b>$Query</b><br>";
print "$DBName, $Link, $Host<br>";
}
?>[/code]
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https://forums.phpfreaks.com/topic/22495-needed-help-with-altering-code-in-mysql/
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I think a quick and dirty way to fix it would be to do something like this:

[code]
if(isset($_POST['jpip1'])){
  $jpip1 = $_POST['jpip1'];
} else {
  $jpip1 = NULL;
}

.
.
.

if(isset($_POST['jptp1'])){
  $jptp1 = $_POST['jptp1'];
} else {
  $jptp1 = NULL;
}

.
.
.
[/code]

This should save NULL values in the database where the user didn't fill a cell on the form...so long as the database isn't set up such that it won't allow null values in those fields.

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