here is my code.....
after i press the save button cant save my data into mysql
and that page became blank
<?php $nama_pengguna =$_GET['nama_pengguna']; ?>
<?php include('../Connections/connection_db.php'); ?>
<?php
$Nama = $_POST['Nama'];
$No_KP = $_POST['No_KP'];
$NoNDP = $_POST['NoNDP'];
$Semester = $_POST['Semester'];
$kursus = $_POST['kursus'];
$NoTelPelajar = $_POST['NoTelPelajar'];
$alamat_1 = $_POST['alamat_1'];
$alamat_2 = $_POST['alamat_2'];
$Poskod = $_POST['Poskod'];
$Bandar = $_POST['Bandar'];
$Negeri = $_POST['Negeri'];
$namabp = $_POST['namabp'];
$No_Tel_bp = $_POST['No_Tel_bp'];
$alamat_1_bp = $_POST['alamat_1_bp'];
$alamat_2_bp = $_POST['alamat_2_bp'];
$poskod_bp = $_POST['poskod_bp'];
$Bandar_bp = $_POST['Bandar_bp'];
$Negeri_bp = $_POST['Negeri_bp'];
$data =mysql_query ("INSERT INTO maklumat_pelajar (No_KP, Nama, No.NDP, Semester, Kursus, No_Tel_Pelajar, Alamat, Alamat_2, Poskod, Bandar, Negeri, Nama_BP, No_Tel_BP, Alamat_BP, Alamat_BP_2, Poskod_BP, Bandar_BP, Negeri_BP) VALUES ('".$_POST['No_KP']. "','".$_POST['Nama']. "','".$_POST['NoNDP']. "','".$_POST['Semester']. "','".$_POST['kursus']. "','".$_POST['NoTelPelajar']. "','".$_POST['alamat_1']. "','".$_POST['alamat_2']. "','".$_POST['Poskod']. "','".$_POST['Bandar']. "','".$_POST['Negeri']. "','".$_POST['namabp']. "','".$_POST['No_Tel_bp']. "','".$_POST['alamat_1_bp']. "','".$_POST['alamat_2_bp']. "','".$_POST['poskod_bp']. "','".$_POST['Bandar_bp']. "','".$_POST['Negeri_bp']. "')");
//$data = mysql_query($query);
echo $data;
//header ("location:../bppl/register_pelajar.php?nama_pengguna=$nama_pengguna");
?>
can anyone got another way to do it?
change that
No.NDP
to
`No.NDP`
in your columns
you should escape your variables first before inserting them by
mysql_real_escape_string()
to see what error you have use mysql_error() like that
$data =mysql_query ("INSERT INTO ...........")or die (mysql_error());
Try this.
$sql=mysql_query("INSERT INTO maklumat_pelajar('$No_KP','$nama','$NoNDP',
'$smester','$kursus',
'$NoTelPelajar','$alamat_1','$alamat_2','$Poskod','$Bandar','$Negeri',
'$namabp','$No_Tel_bp','$alamat_1_bp','$alamat_2_bp','$poskod_bp',
'$Bandar_bp','$Negeri_bp')");
if(!$sql) {
die("Database query failed: " . mysql_error());
}
Related
final.php
Here I am trying to get the data from the url using GET method and trying to insert into the database. I was able to insert the data for first few rows after that the data is not inserted. Can anyone help me regarding this?
when I try to run the url: www.myii.com/app/final.php?name=123&glucose=3232...
the data is not inserting.
<?php
include("query_connect.php");
$name = $_GET['name'];
$glucose = $_GET['glucose'];
$temp = $_GET['temp'];
$battery = $_GET['battery'];
$tgs_a = $_GET['tgs_a'];
$tgs_g = $_GET['tgs_g'];
$heartrate = $_GET['heartrate'];
$spo2 = $_GET['spo2'];
$rr = $_GET['rr'];
$hb = $_GET['hb'];
$ina22 = $_GET['ina22'];
$accucheck = $_GET['accucheck'];
$isactive = $_GET['isactive'];
$address = $_GET['address'];
$deviceno = $_GET['deviceno'];
$sql_insert = "insert into query (name,glucose,temp,battery,tgs_a,tgs_g,heartrate,spo2,rr,hb,ina22,accucheck,isactive,address,deviceno) values ('$name','$glucose','$temp',$battery','$tgs_a','$tgs_g','$heartrate','$spo2','$rr','$hb','$ina22','$accucheck','$isactive','$address','$deviceno')";
mysqli_query($sql_insert);
if($sql_insert)
{
echo "Saving succeed";
//echo $date_time;
}
else{
echo "Error occured";
}
?>
Query_connect.php
This is my database config php file.
<?php
$user = "m33root";
$password = "me3i434";
$host = "localhost";
$connection = mysqli_connect($host,$user,$password);
$select = mysqli_select_db('miiyy',$connection);
if($connection)
{
echo "connection succesfull<br>";
}
else {
echo "Error";
}
?>
Make sure that all columns can contain NULL so that not filled fields will stay NULL instead of throwing an error.
Try below to see mysql error:
mysql_query($sql_insert);
echo mysql_error()
Try these
In SQL
Change the table name of query to some other name. Because "query" is reserved in SQL
In code
if (!mysqli_query($con,$sql_insert ))
{
echo("Error description: " . mysqli_error($con));
}
else
{
echo "Success";
}
Use mysqli_error() function
I keep getting a 'Resource id # 6' failure when submitting a script on my website. The code I'm using is the same type of code I use for registering for the website and that works but this script doesn't work at all. What my code does is send a booking request with the fields as shown to the database. I keep getting a Resource id#6 error , and I've googled what that is but I can't seem to figure out whats wrong. I am a beginner at php , so any tips on whats to look for to avoid a resource id # 6 error would be a lot of help
<?php
//$pattern="/^.+#.+/.com/";
//error_reporting(0);
if(isset($_POST["submit"])){
$Name_of_Person = $_POST['Name_of_Person'];
$Name_of_Group = $_POST['Name_of_Group'];
$room = $_POST['room'];
$How_Many_People = $_POST['How_Many_People'];
$Date_of_Booking = $_POST['Date_of_Booking'];
$End_time = $_POST['End_time'];
$Purpose = $_POST['Purpose'];
$Contact_Number = $_POST['Contact_Number'];
$Contact_Email = $_POST['Contact_Email'];
$Alcohol = $_POST['Alcohol'];
$Security = $_POST['Security'];
$Projector = $_POST['Projector'];
$Extra_Chairs = $_POST['Extra_Chairs'];
$Extra_Info = $_POST['Extra_Info'];
$Activated = '0';
$con = mysql_connect('localhost','root','test123') or die("couldn't connect");
mysql_select_db('bookerdb') or die("couldn't connect to DB");
//if(filter_var($email, FILTER_VALIDATE_EMAIL)){//(preg_match($pattern, $_POST['Contact_Email'])){
$query = mysql_query("SELECT * FROM `booking_table` WHERE Date_of_Booking='".$Date_of_Booking."' AND room='".$room."'");
$numrows = mysql_num_rows($query);
echo $query;
if($numrows==0){
$sql="INSERT INTO `booking_table` (Name_of_Person,Name_of_Group,room,How_Many_People,Date_of_Booking,End_time,Purpose,Contact_Number,Contact_Email,Alcohol,Security,Projector,Extra_Chairs,Extra_Info, Activated) VALUES ('$Name_of_Person','$Name_of_Group','$room','$How_Many_People','$Date_of_Booking','$End_time','$Purpose','$Contact_Number','$Alcohol','$Security','$Projector','$Extra_Chairs','$Extra_Info',$Activated)";
$result = mysql_query($sql);
if($result){
echo "Sent to be approved";
$redirect_page = '../ASC.php';
$redirect = true;
if($redirect==true){
header('Location: ' .$redirect_page);
}
}else{
echo "Failed";
}
}else{
echo"There is already a requested booking on that date & time";
$redirect_page = '../EAR.php';
$redirect = true;
if($redirect==true){
header('Location: ' .$redirect_page);
}
}
/*}else{
echo "error";
$redirect_page = '../EWF.php';
$redirect = true;
if($redirect==true){
header('Location: ' .$redirect_page);
}
}*/
}
?>
You have error in your second SQL query. You try to insert 14 values into 15 columns (in values you forgot $Contact_Email).
$sql="INSERT INTO `booking_table` (Name_of_Person,Name_of_Group,room,How_Many_People,Date_of_Booking,End_time,Purpose,Contact_Number,Contact_Email,Alcohol,Security,Projector,Extra_Chairs,Extra_Info, Activated) VALUES ('$Name_of_Person','$Name_of_Group','$room','$How_Many_People','$Date_of_Booking','$End_time','$Purpose','$Contact_Number','$Contact_Email','$Alcohol','$Security','$Projector','$Extra_Chairs','$Extra_Info',$Activated)";
Than remove echo $query from your code, line 30.
In $query isn't query, but mysql result object. You can't work with that by this way, you can't echo it.
Login.php
session_start();
<?php
$username = "root";
$password = "tiger";
$hostname = "localhost";
//connection to the database
$dbhandle = mysqli_connect($hostname, $username, $password)
or die("Unable to connect to MySQL");
//select a database to work with
/* #var $selected type */
$selected = mysqli_select_db($dbhandle,"sample")
or die("Could not select sample");
$name=(\filter_input(\INPUT_POST,'name'));
$phone=(\filter_input(\INPUT_POST,'phone'));
$email=(\filter_input(\INPUT_POST,'email'));
//$custno=(\filter_input(\INPUT_POST,'custno'));
if(!empty(\filter_input(\INPUT_POST,'continue')))
{
echo "<script type='text/javascript'>\n";
'check()';
echo "</script>";
$sql="insert into customersignin(name,phone,email)values('$name','$phone','$email')";
$result=mysqli_query($dbhandle,$sql) or die(\mysqli_error($dbhandle));
}
else
{
$sql1="insert into customersignin(custno)values(NULL)";
$result1=mysqli_query($dbhandle,$sql1) or die(\mysqli_error($dbhandle));
}
$sql2="select custno from customersignin";
$result2=mysqli_query($dbhandle,$sql2) or die (mysqli_error($dbhandle));
$row= mysqli_fetch_array($result2);
if(mysqli_num_rows($result2)>0)
{
echo "$_SESSION['custno']";
unset($_SESSION['custno'];
header('Location:customersvsoup.php');
}
mysqli_close($dbhandle);
$_SESSION[name]=(\filter_input(INPUT_POST,'name'));
customer.php
<body>
<?php session_start(); ?>
<input type="text" style="position: absolute;top:200px;" value="<?php echo $_SESSION["custno"]?>">
</body>
In the php file the customer log in is done,the custno is the auto generate field,i have 2 buttons called continue and skip,for both the auto generate works fine,after any of the button action is done,i need to display the custno in the text box of the next page using session.But the problem is the text box is empty when i run this code.But the session['name'] is working..Please help.
Your session_start(); should come at the beginning of the file in login.php. I see you using $_SESSION[custno] before it's called. That's why your textbox is empty.
Also it should be:
$_SESSION['custno']
$_SESSION['name']note the single quotes
Regarding your logical problem (in the comments) try:
$_SESSION['name'] = (filter_input(INPUT_POST, 'name'));
if (!empty(filter_input(INPUT_POST, 'continue')))
{
echo "<script type='text/javascript'>\n";
'check()';
echo "</script>";
$sql = "insert into customersignin(name,phone,email)values('$name','$phone','$email')";
$result = mysqli_query($dbhandle, $sql) or die(mysqli_error($dbhandle));
$sql2 = "select max(custno) as last_custno from customersignin";
$result2 = mysqli_query($dbhandle, $sql2) or die(mysqli_error($dbhandle));
if (mysqli_num_rows($result2) > 0)
{
$row = mysqli_fetch_assoc($result2);
$_SESSION['custno'] = $row['last_custno'];
header('Location:customersvsoup.php');
}
}
else
{
$sql1 = "insert into customersignin(custno)values(NULL)";
$result1 = mysqli_query($dbhandle, $sql1) or die(mysqli_error($dbhandle));
//since this bit of code is repeating,
//you could even use a function to shorten it
$sql2 = "select max(custno) as last_custno from customersignin";
$result2 = mysqli_query($dbhandle, $sql2) or die(mysqli_error($dbhandle));
if (mysqli_num_rows($result2) > 0)
{
$row = mysqli_fetch_assoc($result2);
$_SESSION['custno'] = $row['last_custno'];
header('Location:customersvsoup.php');
}
}
And please put the session_start(); inside after <?php. All php code should be within the PHP tags.
you have error in insert query:
$sql="insertintocustomersignin(name,phone,email)values('$name','$phone','$email')";
should be :
$sql="insert into customersignin(name,phone,email) values ('$name','$phone','$email')";
you should use quotes in array index :
$_SESSION[custno], $_SESSION[name] should be $_SESSION['custno'], $_SESSION['name']
Can anyone tell me why I am getting an Undefined Index error on my code here.
I have used this setup using the if(isset) condition in other parts of my project after
researching my original Undefined Index errors and ISSET fixed my problems. But it is not working here for some reason and I cannot see why.
This form is POSTING the input:
<form action="addAlbum_Processed.php" method="POST">
<p>Enter artistID of Artist<input type="number" name="artist_id" maxlength="2" size="2"></p>
<p>Enter name of Album to be created<input type="text" name="album_name" size="20"></p>
<input type="submit" name="submit" value="submit"></form>
and this page is processing the form input and updating the albums table in my database:
<?php
$connection = mysql_connect('localhost','root','')
or die(mysql_error());
echo "Connected to php Server <br>";
or die("Could not select assi2 database");
echo "Connected to assi2 database <br>";
if(isset($_POST['submit']))
{
$album_name = $_POST['album_name'];
$artist_id = $_POST['artist_id'];
}
$album_name = $_POST['album_name'];
$artist_id = $_POST['artist_id'];
$sqlQuery = "SELECT * FROM albums WHERE album = '{$album_name}'";
$result = mysql_query($sqlQuery, $connection) or die("Selection Query Failed !!!");
if (mysql_num_rows($result) != 0)
{
header ("Location: Album_Exists.html");
}
else
{
$sqlInsert = "INSERT INTO albums (ArtistID, Album, delete_marker)
VALUES ('{$artist_id}','{$album_name}','delete_marker = 0')";
$result = mysql_query($sqlInsert, $connection) or die("Selection Query Failed !!!");
header ("Location: addAlbum_Processed.php");
}
mysql_close($connection);
?>
I cannot see where I am going wrong. Regards, TW
This is a tiny example of your problem:
if(isset($_POST['submit']))
{
$album_name = $_POST['album_name'];
$artist_id = $_POST['artist_id'];
}
You check whether a submit form field was posted before using the other fields. So far, so good. (I would check for the fields that were going to be used, but at least you're checking something.)
But then:
$album_name = $_POST['album_name'];
$artist_id = $_POST['artist_id'];
You use the fields anyway.
What's more...you don't keep from trying to insert stuff if a form isn't being posted. So any time some rogue spider visits your page, you end up with a blank album in your database.
And that's not even mentioning the fact that you're still using mysql_query.
if(isset($_POST['submit']))
{
$album_name = $_POST['album_name'];
$artist_id = $_POST['artist_id'];
}
|__________________________| first
$album_name = $_POST['album_name'];
$artist_id = $_POST['artist_id'];
|_________________________| Repeated
you are fetching variables twice.only one that is if condition is enough.Also use isset for both the variables.
if(isset($_POST['submit']))
{
if isset($_POST['album_name'])
$album_name = $_POST['album_name'];
if isset($_POST['artist_id'])
$artist_id = $_POST['artist_id'];
}
Try something like in addalbam_process.php
<?php
$connection = mysql_connect('localhost','root','')
or die(mysql_error());
echo "Connected to php Server <br>";
or die("Could not select assi2 database");
echo "Connected to assi2 database <br>";
if(isset($_POST['submit']))
{
if(isset($_POST['albam_name']){$album_name = $_POST['album_name']};
if(isset($_POST['artist_id']){$artist_id = $_POST['artist_id']};
}
$sqlQuery = "SELECT * FROM albums WHERE album = '{$album_name}'";
$result = mysql_query($sqlQuery, $connection) or die("Selection Query Failed !!!");
if (mysql_num_rows($result) != 0)
{
header ("Location: Album_Exists.html");
}
else
{
$sqlInsert = "INSERT INTO albums (ArtistID, Album, delete_marker)
VALUES ('{$artist_id}','{$album_name}','delete_marker = 0')";
$result = mysql_query($sqlInsert, $connection) or die("Selection Query Failed !!!");
header ("Location: addAlbum_Processed.php");
}
mysql_close($connection);
Please, use MYSQLI or PDO to Prevent SQL INJECTION
here </form> is missing
and try something like this
if(isset($_POST['submit']))
{
$album_name = $_POST['album_name'];
$artist_id = $_POST['artist_id'];
}
A few things.
This line 'delete_marker = 0' should most probably read as
VALUES ('{$artist_id}','{$album_name}','0')
or VALUES ('{$artist_id}','{$album_name}',0)
As I read it 'delete_marker = 0' you are attempting to actually write this value inside the delete_marker column (ArtistID, Album, delete_marker)
Or, you're attempting to use a WHERE delete_marker = 0 clause, which can't be used in an INSERT INTO, but an UPDATE or SELECT rather.
And your if(isset($_POST['submit'])) conditional statement should be wrapping your entire code, instead of just your 2 form variables, because it's basically saying "Ok, assign these 2 variables, then ignore the rest if it's NOT set."
Plus, you're repeating those 2 input variables.
$album_name = $_POST['album_name'];
$artist_id = $_POST['artist_id'];
(I wrapped your entire code inside the if(isset($_POST['submit'])) conditional statement, btw.
Side note: If you're having a DB connection issue, use this instead:
$connection = mysql_connect('localhost', 'root', '');
if (!$connection) {
die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
PHP Give this a try:
Sidenote: If this line fails VALUES ('{$artist_id}','{$album_name}', 0) put quotes around the 0 as in '0'
<?php
$connection = mysql_connect('localhost','root','')
or die(mysql_error());
echo "Connected to php Server <br>";
or die("Could not select assi2 database");
echo "Connected to assi2 database <br>";
if(isset($_POST['submit']))
{
$album_name = $_POST['album_name'];
$artist_id = $_POST['artist_id'];
$sqlQuery = "SELECT * FROM albums WHERE album = '{$album_name}'";
$result = mysql_query($sqlQuery, $connection) or die("Selection Query Failed !!!");
if (mysql_num_rows($result) != 0)
{
header ("Location: Album_Exists.html");
}
else
{
$sqlInsert = "INSERT INTO albums (ArtistID, Album, delete_marker)
VALUES ('{$artist_id}','{$album_name}', 0)"; // or add quotes around the zero
$result = mysql_query($sqlInsert, $connection) or die("Selection Query Failed !!!");
header ("Location: addAlbum_Processed.php");
}
} // closing brace for if(isset($_POST['submit']))
mysql_close($connection);
?>
I am trying to set a session variable after first successful insert query. I have a javascript Confirm to check if the user wants to add more data. If user wants to add more Products then I don't have to insert into the first table so keep the session value. If not then clear out the session data. I am not sure this is is right way of doing it. Could you please let me know if there are any suggestions.
<?php
if ($_SERVER['REQUEST_METHOD'] == 'POST')
{
session_start();
if (empty($_SESSION['new_id']))
{
$first = $_POST["first"];
$first = mysql_real_escape_string($first);
$last = $_POST["last"];
$last = mysql_real_escape_string($last);
$insertsql = "INSERT INTO Table1(FirstName,LastName) VALUES ('".$first."', '" .$last. "')";
$result1 = mysql_query($insertsql) or die(mysql_error());
$new_id = MySql_Insert_Id();
$_SESSION['new_id'] = $new_id;
}
$prodName = $_POST["prodName"];
$prodName = mysql_real_escape_string($prodName);
$prod_details = $_POST["prod_details"];
$prod_details = mysql_real_escape_string($prod_details);
$insertSQl2 = "INSERT INTO Table2($_SESSION['new_id'], Product_Name, Product_Details) VALUES ('".$new_id."', '" .$prodName. "', '" .$prod_details. "')";
$Result2 = mysql_query($insertSQl2) or die(mysql_error());
}
?>
<form>
//controls here
</form>
<?php
if($Result2 == true)
{
echo "Data saved Successfully";
echo '<script type="text/javascript">' , 'if confirm("Would like to add more Products?") ' , '</script>';
}
?>