0
Create table Cameras
	(CameraID varchar (35),
	 CheckOutDate varchar(10),
	 ReturnDate varchar(10),
	 primary key(CameraID),
	 foreign key(CameraId) references History(CameraID));


	Create table Borrowers
	(PsuID int not null,
	 BorrowerName varchar(30),
	 Email varchar(50),
	 Telephone varchar(12),
	 ReasonUse varchar(200),
	 primary key(PsuID),
	 foreign key(PsuID) references History(PsuID));




	Create table History
	(PsuID int not null,
	 CameraID varchar (35),
	 BorrowedFrequency int,
	 primary key(PsuID, CameraID),
	 foreign key(PsuID) references Borrowers(PsuID),
	 foreign key(CameraID) references Cameras(CameraID));


This is my table

I want to insert Cameras and Borrowers at the same time

<?php
$CameraID	= $_POST['cameraid'];
$CheckOutDate	= $_POST['checkoutdate'];
$ReturnDate	= $_POST['returndate'];
$PsuID = $_POST['psuid'];
$BorrowerName = $_POST['borrowername'];
$Email	= $_POST['email'];
$Telephone = $_POST['telephone'];
$ReasonUse	= $_POST['reasonuse'];


//Connect To Database
$hostname="localhost";
$username="your_dbusername";
$password="your_dbpassword";
$dbname="your_dbusername";

mysql_connect($hostname,$username, $password) or die ("ERROR: Unable to connect to database!");
mysql_select_db($dbname);






#Add new entry
	
$sql = "insert into Borrowers() values ()";
$result = mysql_query($sql, $conn) or die(mysql_error());

$sql = "insert into Cameras() values ()";
$result = mysql_query($sql, $conn) or die(mysql_error());

Hope I can get some helps,

Thanks

2
Contributors
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Views
6 Years
Discussion Span
Last Post by sumandy
0

Change

insert into Borrowers() values ()

to

insert into Borrowers values (NULL,'$BorrowerName','$Email','$Telephone','$ReasonUse')

Do likewise with the Camera table insert.

0

Thanks, I would put the attribute in, but I was wondering if that is legitimate

Let's say I want all three tables to be involved


$sql = "insert into Borrowers(contents here) values ($contents here)";
$result = mysql_query($sql, $conn) or die(mysql_error());

$sql1 = "insert into Cameras(contents here) values ($contents here)";
$result1 = mysql_query($sql, $conn) or die(mysql_error());

$sql2 = "insert into History(contents here) values ($contents here)";
$result2 = mysql_query($sql, $conn) or die(mysql_error());

Is this right?
Do I need any Where Clause code?

Thanks a lot

0

Insert does not need a where clause because you do not insert at a specific location other than the table. You just insert a row with some values.

0

This is my HTML code:

this is my php code

But it's not working still
it has errors like:

Notice: Undefined index: cameraid in /websites/RFID/checkout.php on line 4

Notice: Undefined index: checkoutdate in /websites/RFID/checkout.php on line 5

Notice: Undefined index: returndate in /websites/RFID/checkout.php on line 6

Notice: Undefined index: psuid in /websites/RFID/checkout.php on line 7

Notice: Undefined index: borrowername in /websites/RFID/checkout.php on line 8

Notice: Undefined index: email in /websites/RFID/checkout.php on line 9

Notice: Undefined index: telephone in /websites/RFID/checkout.php on line 10

Notice: Undefined index: reasonuse in /websites//RFID/checkout.php on line 11
... etc
Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /websites/RFID/checkout.php on line 31


What else can I do?


Thanks a Lot!

0

Sorry I made a mistake before

This is my html code

<html>

<head>
<title>Check Out</title>










<style type="text/css">
.style1 {
	font-weight: bold;
	text-decoration: underline;
}
</style>




</head>

<body background="wall.jpg" text="navy">

<center><img src="wblogo.png"></center>

<br>
<form method="post" action="checkout.php">

<center>
<table width="500" border ="1" cellpadding="4" cellspacing="0" BORDERCOLOR="navy">
<tr>
	<td colspan="2">
		<center style="width: 488px; height: 46px"><b><h1>Check Out </h1> </b></center>
	</td>
</tr>
<tr>
	<td style="height: 36px">
		<b> Camera ID: </b>
	</td>
	<td align="left" style="height: 36px">
		<input name="CameraID" size="50" style="width: 294px; height: 23px">
    </td>
</tr>
<tr>
	<td>
		<b> PSU ID: </b>
	</td>
	<td align="left">
		<input name="PsuID" size="50" style="width: 294px; height: 23px">
    </td>
</tr>
<tr>
	<td>
		<b> Borrower Name (First/Last): </b>
	</td>
	<td align="left">
		<input name="BorrowerName" size="50" style="width: 294px; height: 23px">
	</td>
</tr>


<tr>
	<td>
		<b> Phone: </b>
	</td>
	<td align="left">
		<input name="Telephone" size="50" style="width: 294px; height: 23px">
    </td>
</tr>
<tr>
	<td>
		<b> E-mail: </b>
	</td>
	<td align="left">
		<input name="Email" size="50" style="width: 294px; height: 23px"> 
	 </td>
</tr>
<tr>
	<td>
		<b> Date Borrowed: </b>
	</td>
	<td align="left">
		<input name="CheckOutDate" size="50" style="width: 294px; height: 23px"> 

	</td>
</tr>



<tr>
	<td>
		<b> Date to be Returned: </b>
	</td>
	<td align="left">
		<input name="ReturnDate" size="50" style="width: 294px; height: 23px"> 
    </td>
</tr>
<tr>
	<td class="style1">
		Reason for use: 
	</td>
	<td align="left">		
		<textarea name="ReasonUse" style="width: 295px; height: 246px;"></textarea>
		
		
	</td>
</tr>




<tr>
	<td align="left">
		<input type="reset" value="Clear" style="width: 80px; height: 26px">
	</td>
	<td align="right">
		<input type="submit" value ="Check Out">
	</td>	
</tr>
</table>
</center>

</form>













</body>

</html>

This is my php code

<?php
$CameraID	= $_POST['cameraid'];
$CheckOutDate	= $_POST['checkoutdate'];
$ReturnDate	= $_POST['returndate'];
$PsuID		 = $_POST['psuid'];
$BorrowerName	 = $_POST['borrowername'];
$Email		= $_POST['email'];
$Telephone 	= $_POST['telephone'];
$ReasonUse	= $_POST['reasonuse'];





#Connect To Database
$hostname="localhost";
$username="sxc455";
$password="P0tassium19";
$dbname="sxc455";

mysql_connect($hostname,$username, $password) or die ("ERROR: Unable to connect to database!");
mysql_select_db($dbname);

	
#Add new entry


$sql = "insert into Cameras(CameraID, CheckOutDate, ReturnDate) values ('$CameraID', '$CheckOutDate', '$ReturnDate')";
$result = mysql_query($sql, $conn) or die(mysql_error());

$sql1 = "insert into Borrowers(PsuID, BorrowerName, Email, Telephone, ReasonUse) values ($PsuID, '$BorrowerName', '$Email', '$Telephone', '$ReasonUse')";
$result1 = mysql_query($sql, $conn) or die(mysql_error());




print("
	
	<center>
	<table width='400' height='200' cellpadding='4' cellspacing='0'>
	  <tr align='left'>
	    <td>
	      The Entry <b>"); print($PsuID); print(" </b> was added to the database
	    </td>
	  </tr>
	</table>
	</center>");
	
?>

But it's not working still
it has errors like:

Notice: Undefined index: cameraid in /websites/RFID/checkout.php on line 4

Notice: Undefined index: checkoutdate in /websites/RFID/checkout.php on line 5

Notice: Undefined index: returndate in /websites/RFID/checkout.php on line 6

Notice: Undefined index: psuid in /websites/RFID/checkout.php on line 7

Notice: Undefined index: borrowername in /websites/RFID/checkout.php on line 8

Notice: Undefined index: email in /websites/RFID/checkout.php on line 9

Notice: Undefined index: telephone in /websites/RFID/checkout.php on line 10

Notice: Undefined index: reasonuse in /websites//RFID/checkout.php on line 11
... etc
Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /websites/RFID/checkout.php on line 31


What else can I do?


Thanks a Lot!

Edited by sumandy: n/a

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