PHP - Transfer Data from table to another within a database
Question
I'm still new to PHP programming, my goal here is to
-transfer data from Table to another Table
<html> <head> <title>Sending Application Form by Email</title> </head> <body>
<?PHP
include ("dbase.php");
if(isset($_COOKIE['ID_my_site']))
{
$username = $_COOKIE['ID_my_site'];
$query = mysql_query("SELECT * FROM employee_database WHERE Employee_Name= '$username'");
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$insert = mysql_query ("INSERT INTO `leave_requested`
(`Employee_Name`,
`Employee_ID`,
`Designation`,
`Department`,
`Days_on_Leave`,
`Leave_Start_On`,
`Leave_End_On`,
`Reason for Leave`)
SELECT `Employee_Name`,
`Employee_ID`,
`Designation`,
`Department`,
'$Days',
'$Start',
'$End',
'$Reason'
FROM `employee_database`
WHERE `Employee_Name` = '$username'");
/*
("INSERT INTO leave_requested (Employee_Name, Employee_ID, Designation, Department, Days_on_Leave, Leave_Start_On, Leave_End_On, Reason for Leave)
SELECT Employee_Name, Employee_ID, Designation, Department FROM employee_database
VALUES ('$row[Employee_Name]','$row[Employee_ID]','$row[Designation]','$row[Department]','$_POST['No_Days']','$_POST['StartDate']','$_POST['EndDate']','$_POST['Leave_Reason']')");
*/
}
}
mysql_close ($qry);
?>
<?php
require("C:\AppServ\www\Project\PHPMailer_5.2.4\class.phpmailer.php");
$mail = new PHPMailer();
$mail->IsSMTP();
//GMAIL config
$mail->SMTPAuth = true; // Enable SMTP authentication
$mail->SMTPSecure = "tls"; // Sets the prefix to the server
$mail->Host = "smtp.gmail.com"; // Sets GMAIL as the SMTP server
$mail->Port = '587'; // Sets the SMTP port for the GMAIL server
$mail->Username = ("my@gmail.com"); // SMTP UserName (Email Address)
$mail->Password = "******"; // SMTP PassWord (Email Address PassWord)
$mail->SMTPDebug = 2; // Enables SMTP debug information (for testing)## 1 = errors and messages ## 2 = messages only ##
$mail->From = $mail->Username; // Sender Email Account on Email
$mail->FromName = 'HR Management'; // Sender Name on Email
$mail->AddAddress('my@yahoo.com', 'JD'); // Add a recipient Email and Name
$mail->WordWrap = 50;
$mail->IsHTML(true);
$mail->Subject = '$username have added Leave Requested'; // Email Subject
$mail->Body = 'This is the HTML message body <b>in bold!</b>'; // HTML Plain Text
$mail->AltBody = 'This is the body in plain text for non-HTML mail clients'; // HTML Function
if(!$mail->Send())
{
echo 'Message could not be sent.';
echo 'Mailer Error: ' . $mail->ErrorInfo;
exit;
}
echo 'Message has been sent';
?>
</body>
</html>
I would be glad if anyone were to able to send an example of 2 sql query. Thanks
The error now I'm getting is an odd one, i haven't encounter this type of ERROR before. Do anyone have knowledge on this ERROR. Please suggest a solution to this ERROR. Thanks
Page 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 'Resource id #4' at line 1
3 answers
solution1
2 2014-01-02 13:05:41
$query = mysql_query("SELECT * FROM employee_database WHERE Employee_Name= '$username'");
$result = mysql_query($query) or die(mysql_error());
In the first line, mysql_query() returns a resource .
In the second line, you pass the return value back into another call to mysql_query() .
mysql_query() expects a SQL query string as parameter.
Easy fix:
$query = "SELECT * FROM employee_database WHERE Employee_Name= '$username'";
$result = mysql_query($query) or die(mysql_error());
Some further tips:
- google error messages before consulting on a forum
- read the documentation before consulting a forum
- don't use mysql extension in PHP, it is deprecated !
- use prepared statements - your code is prone to SQL injection attacks
solution2
0 2014-01-02 08:43:19
First I have to tell you about your query that it is prone to SQL injection. Second mysql_* functions are deprecated and no more will be supported.
You can do it with sinle query. You may use this query:
INSERT INTO `leave_requested` (`Employee_Name`,
`Employee_ID`,
`Designation`,
`Department`,
`Days_on_Leave`,
`Leave_Start_On`,
`Leave_End_On`,
`Reason for Leave`)
SELECT `Employee_Name`,
`Employee_ID`,
`Designation`,
`Department`,
'$_POST[\'No_Days\']',
'$_POST[\'StartDate\']',
'$_POST[\'EndDate\']',
'$_POST[\'Leave_Reason\']'
FROM `employee_database`
WHERE `Employee_Name` = '$username'
solution3
0 2014-01-02 08:43:26
There is such a thing as INSERT SELECT . (I've linked to MySQL documentation, but so far as I know it's standard SQL.)
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