I am trying to get some informations from users such as name midterm final grades and post them into database which I am using the MySQL server.
The problem is when I press the Add to DD button nothing changes and the data does not go into my table in database.
here is my main code:
<html>
<body>
<table border ="1">
<tr>
<td>Name</td>
<td><input id="name"> </input></td>
</tr>
<tr>
<td>Midterm</td>
<td> <input id="midterm"> </input></td>
</tr>
<tr>
<td>Final</td>
<td> <input id="final"> </input></td>
</tr>
<tr>
<td>Grade</td>
<td> <input id="grade"> </input></td>
</tr>
<td><input type="button" onclick="calculate()" value="Calculate"></td>
<td> <input type="submit" value="Add to DB"> </td>
</tr>
</table>
</body>
<script type="text/javascript">
function calculate(){
var mid=document.getElementById('midterm').value;
var fin=document.getElementById('final').value;
var grade=mid*(0.3)+fin*(0.7);
document.getElementById('grade').value=grade;
}
</script>
</html>
and this is also the code that inserts datas :
<html>
<body>
<form method="get" action="grade.php">
<input type="text" value="Welcome to Student Grades Calculator">
<br>
<input type="submit" value="GO">
</form>
</body>
</html>
<?php
$name=$_POST['name'];
$midterm=$_POST['midterm'];
$final=$_POST['final'];
$grade=$_POST['grade'];
echo 'Hey';
$connect= mysql_connect('localhost','root','','test');
if(mysqli_connect_errno())
{
echo "Failed to connect to MySQL:".mysql_connect_errno();;
}
$s = "INSERT INTO(name,midterm,final,grade) VALUES('rr','33','33','33')";
$sql = "INSERT INTO(name,midterm,final,grade) VALUES ('$name','$midterm','$final','$grade')";
mysqli_query($connect , $sql);
mysqli_close($connect);
?>
You need to set your variables to $_GETnot $_POST
$name=$_GET['name'];
$midterm=$_GET['midterm'];
$final=$_GET['final'];
$grade=$_GET['grade'];
I also recommend using post instead of get just for security
Related
I can't seem to delete row in database by id in php
I think the the id is not passed to the $_POST['delete']
however, the popup "Your data is deleted" is displayed, but the data is not deleted.
So I'm not sure where is the error in this code.
I also try to delete the data by its id
for example: Delete book where no='4';
and the code seems to run fine because the data is deleted in the database.
<html>
<script>
function confirmDelete() {
return confirm('Are you sure?');
}
</script>
<!DOCTYPE html>
<head>
<form action="test.php" method="POST">
<br><br><br>
<table bordercolor="#FFCC66" align="center" bgcolor="#FFFFFF">
<tr>
<th>No</th>
<th>Title</th>
<th>Author</th>
<th>Year</th>
<th>Donor's Name</th>
<th>Call Number</th>
<th>Date Received</th>
<th>Handled By</th>
<th></th>
<th></th>
</tr>
<?php
include ('config.php');
$view=mysqli_query($conn,"SELECT * FROM book");
?>
<?php while($v=mysqli_fetch_array($view)){ ?>
<tr>
<td>
<?php echo $v["no"];?>
</td>
<td>
<?php echo $v["title"];?>
</td>
<td>
<?php echo $v["author"];?>
</td>
<td>
<?php echo $v["year"];?>
</td>
<td>
<?php echo $v["donorname"];?>
</td>
<td>
<?php echo $v["callnum"];?>
</td>
<td>
<?php echo $v["datereceived"];?>
</td>
<td>
<?php echo $v["handledby"];?>
</td>
<td><input type="submit" name="delete" value="Delete" onclick="return confirmDelete('Are you sure?');" /></td>
</tr>
<?php
} ?>
</tr>
</table>
<br><br>
</form>
</body>
</html>
<?php
if(isset($_POST['delete']))
{
include('config.php');
$no =$v["no"];
$d=mysqli_query($conn,"DELETE FROM `book` WHERE no='$no'");
if ($d)
{
echo "<script type='text/javascript'> alert('Your data is deleted!'); </script>";
echo"<meta http-equiv='Refresh' content='0' >";
}
else
{
echo "<script type='text/javascript'> alert('Your data cannot delete!'); </script>";
}
mysqli_close($conn);
}
?>
Change the submit element to
<td>
<input type="submit" name="delete" value="<?php echo $v['no'];?>" onclick="return confirmDelete('Are you sure?');" />
</td>
and
$no = $_POST["delete"];
Another solution si to add a hidden input with your value.
<td>
<?php echo $v["no"];?>
<input type="hidden" value="<?php echo $v["no"];?>" />
</td>
In your php you will find the value in $_POST['no']
This solution is better to pass multiple arguments in POST like a captcha or a confirmation (checkbox).
logic is not correct, while you press the delete button, all the data will be passed along with submitting because your tag is outside of the loop.
As my opinion, you should use ajax like functionality here, or follow this method.
<?php while($v=mysqli_fetch_array($view)){ ?>
<form action="test.php" method="POST">
<tr>
<td>
<?php echo $v["no"];?>
<input type="hidden" value="<?php echo $v["no"];?>" name="no" >
</td>
<td><input type="submit" name="delete" value="Delete" onclick="return confirmDelete('Are you sure?');" /></td>
</tr>
</form>
<?php } ?>
and in your post call use $no = $_POST['no']; instead of $no =$v["no"];
I've got two tables: Subjects and Careers :
"Subjects" includes (id, careers_id (is the foreign key to the column "id" of the table Careers) subject, description, hours)
"Careers" includes (id,name,description)
I put a button which allows me to add a new subject. So when I click on it another page open. I need to add a slider/select which shows me the careers available in the table career. Take a look,I need something like this :
Here is my code to add a new subject (it works,but i dont know how to make the slider/select which fetch data from the table careers :/)
<?php include('connect.php');
$error="";
if(isset($_POST['btnsave']))
{
$carreras_id=$_POST['txtcarreras_id'];
$subject=$_POST['txtsubject'];
$descripcion=$_POST['txtdescripcion'];
$carga_horaria=$_POST['txtcarga_horaria'];
if($_POST['txtid']=="0")
{
$a_sql=mysql_query("INSERT INTO subjects VALUES('','$carreras_id','$subject','$descripcion','$carga_horaria')");
if($a_sql)
{
header("location:index.php");
}
}else{
echo "Actualizar";
}
}
?>
<h2 align="center">ADD NEW SUBJECT</h2>
<form method="Post">
<table align="center">
<tr>
<td>Career:</td>
<td>
<input type='text' name='txtcarreras_id' />
<input type="hidden" name="txtid" value="0" />
</td>
</tr>
<tr>
<td>Subject:</td>
<td>
<input type='text' name='txtsubject' />
</td>
</tr>
<tr>
<td>Description:</td>
<td>
<input type='text' name='txtdescripcion' />
</td>
</tr>
<tr>
<td>Hours:</td>
<td>
<input type='text' name='txtcarga_horaria' />
</td>
</tr>
<tr>
<td></td>
<td>
<input type='submit' value=save name='btnsave' />
</td>
</tr>
</table>
</form>
I don't know what to do?
Hope you can help me!
Thanks!
To build your <option>s you would do something like this ->
$sql = mysql_query("SELECT * FROM Careers");
$options = "";
while($result = mysql_fetch_array($sql)){
$options .= "<option value='".$result['id']."'>".$result['name']."</option>";
}
(Note - you should update from mysql_ to MySQLi or PDO - MySQL: choosing an API )
Then you would change your <input> to a <select> with your $options ->
<tr>
<td>Career:</td>
<td><select name='txtcarreras_id'><?php echo $options; ?></select><input type="hidden" name="txtid" value="0" /></td>
</tr>
<?php if(isset($_POST['submit']))
{
$tadd=$_POST["tadd"]; //getting values
$pname=$_POST["pname"];
$date=$_POST["date"];
$result=mysql_query("insert into pannel(tadd,pname,date)values('$tadd','$pname','$date')");
echo "<script type='text/javascript'>
alert('Quotation Generated Successfully!')
</script>";
} ?>
<center>
<h1>Title</h1>
</center>
<form name="form" method="post" action="" onSubmit="submit;">
<center><table border="0" cellspacing="0" style="width:350px">
<tr> <td><b>To Address</td> <td><textarea name="tadd" rows="5"
cols="30"></textarea></td></tr>
<tr> <td><b>Project Name</td> <td><input type="text" name="pname" required></td></tr>
<tr> <td><b>Date</td> <td><input type="text" name="date"
id="datepicker" required></td></tr>
<tr> <td colspan="2" align="center"><input type="submit" name="submit"
value="submit"/></td> </tr></center> </table> </form>
I have one record in my database with
id tadd pname date
1 hello vvv 22/10/2014
if i insert values into database again it should data already inserted
please help me regarding this issue
You can achieve this using mysql_num_rows() which is one way to do this, which I believe the goal is to avoid duplicates.
Sidenote: You can also set your column(s) as UNIQUE to avoid duplicates.
N.B.: I used the pname column as an example. It's up to you to check which one will always be unique in regards to a username for instance.
$query = "SELECT * FROM pannel where pname = '".$pname."'";
$result = mysql_query($query);
if(mysql_num_rows($result) > 0){
echo "Already exists.";
}
else{
mysql_query("insert into pannel (tadd, pname,date) values ('$tadd','$pname','$date')");
}
Do sanitize your data:
$tadd = mysql_real_escape_string($_POST["tadd"]);
and do the same for the others.
Even better, use mysqli with prepared statements, or PDO with prepared statements.
They're much safer, because your present code is open to SQL injection.
Footnotes:
You should get rid of onSubmit="submit;" in your form. As outlined in comments, it's not going to do anything.
Edit:
<?php
// assuming DB connection has been made.
if(isset($_POST['submit'])) {
$tadd= mysql_real_escape_string($_POST["tadd"]);
$pname= mysql_real_escape_string($_POST["pname"]);
$date= mysql_real_escape_string($_POST["date"]);
$query = "SELECT * FROM pannel where pname = '".$pname."'";
$result = mysql_query($query);
if(mysql_num_rows($result) > 0){
echo "Already exists.";
exit;
}
else{
mysql_query("insert into pannel (tadd, pname,date) values ('$tadd','$pname','$date')");
echo "<script type='text/javascript'>alert('Quotation Generated Successfully!')</script>";
}
} // brace for if(isset($_POST['submit']))
?>
<!DOCTYPE html>
<head></head>
<body>
<center><h1>Title</h1></center>
<form method="post" action="">
<div align="center">
<center>
<table border="0" cellspacing="0" style="width:350px">
<tr> <td><b>To Address</td> <td><textarea name="tadd" rows="5" cols="30"></textarea></td></tr>
<tr> <td><b>Project Name</td> <td><input type="text" name="pname" required></td></tr>
<tr> <td><b>Date</td> <td><input type="text" name="date" id="datepicker" required></td></tr>
<tr> <td colspan="2" align="center">
<input type="submit" name="submit" value="submit"/>
</td> </tr>
</table>
</center>
</div>
</form>
</body>
</html>
It's been a little while with PHP, so please excuse my ignorance. I have this web page:
<?php
mysql_connect ("localhost", "user", "pass");
mysql_select_db ("expiration");
if (isset ($_REQUEST['new_expire']) && $_REQUEST['new_expire'] != "") {
$insert_query = "INSERT INTO files (path, expires) VALUES ('";
$insert_query .= $_REQUEST['new_path'];
$insert_query .= "', '";
$insert_query .= $_REQUEST['new_expire'];
$insert_query .= "');";
mysql_query ($insert_query);
}
?>
<html>
<head></head>
<body>
<?php echo print_r $_REQUEST; ?>
<form action="index.php" method="POST">
<p>Add New Expiration</p>
<table>
<tr>
<td align="right">
Select File:
</td>
<td>
<select id="new_path">
<?php $options = buildOptions (); echo $options; ?>
</select>
</td>
</tr>
<tr>
<td align="right">
MySQL Expire Time:
</td>
<td>
<input type="text" id="new_expire" />
</td>
</tr>
<tr>
<td></td>
<td>
<input type="submit" value="Save" />
</td>
</tr>
</table>
</form>
</body>
</html>
<?php
mysql_close ();
?>
When I load the page, the result of the print_r is an empty array. When I submit the form, it's still empty. I get no new record in the database. Any ideas?
Change all the places you have id to name, for example:
<input type="text" id="new_expire" /> --> <input type="text" name="new_expire" />
The _REQUEST (either _POST or _GET) is only from input elements with a name
I'm trying to get the emails corresponding to the checkbox using the following codes. But, I'm not getting the correct checked emails in the new variable. Can anyone please check ??
<?php
include("connection.php");
$username=$_SESSION['username'];
$query=mysql_query("SELECT * FROM contacts WHERE username='$username'");
$num=mysql_num_rows($query);
$info=mysql_fetch_array($query);
$i=0;
$msg='';
?>
<table width="672" border="0">
<?php
$i=0;
while($info)
{
?>
<form action="compose.php" method="post">
<tr style="font-size:14px;">
<td width="21" bgcolor="#f2f2f2"> <input type="checkbox" name="add" onSelect="<?php $msg=$msg.$info['email'].", ";?>"/> </td>
<td width="229" bgcolor="#f2f2f2"> <?php echo $info['email']; ?> </td>
<td width="408" bgcolor="#f2f2f2"> <?php echo $info['name']; ?> </td>
</tr>
<?php
$info=mysql_fetch_array($query);
$i++;
}
$_SESSION['contacts']=$msg;
?>
<tr><td></td><td></td><td><br />
<input class="new-button" type="submit" value="Insert & Compose" name="submit" /></td>
</tr>
</form>
</table>
To get any value back for checkboxes they must have a value=. In your case you probably would want the value to be the according email address.
One problem with your code is using onSelect= instead of value=, and second you didn't print the actual value into the page. Rewrite it to:
<td width="21" bgcolor="#f2f2f2">
<input type="checkbox" name="add"
value="<?php print $info['email']; ?>"/> </td>
If you need the $msg variable to do something, assemble it after the output.
<input type="checkbox" name="add" value="<?php echo $msg.$info['email'];?>"/>
checkbox does not have onSelect event probobly you got value in mind and in PHP code you should echo and what .", " is for?