We have an assignment for school and I've tried to build the application, however some text that I want to have inserted into a database doesn't get submitted.
I've tried different things, but the page does not show an error either.
This is the code of my insert page
<head>
</head>
<body>
<form action="index.php" method="post">
ID: <input type="text" name="id"><br/>
Server: <input type="text" name="Server"><br/>
Student: <input type="text" name="Student"><br/>
Docent: <input type="text" name="Docent"><br/>
Project: <input type="text" name="Project"><br/>
Startdatum: <input type="text" name="Startdatum"><br/>
Einddatum: <input type="text" name="Einddatum"><br/>
<input type="submit" name="submit">
</form>
<?php
if(isset($_POST['submit'])) {
$con = mysqli_connect("localhost", "root", "usbw", "serverruimte");
if(!$con) {
die(mysqli_connect_error());
}
$sql = "INSERT INTO serverruimte (id,Server,Student,Docent,Project,startdatum,einddatum) VALUES ('$_POST[id]','$_POST[Server]','$_POST[Student]','$_POST[Docent]','$_POST[Project]','$_POST[startdatum]','$_POST[einddatum]')";
$result = mysqli_query($con, $sql);
if($result) {
echo "Opslaan voltooid!";
} else {
echo mysqli_error($con);
}
mysqli_close($con);
}
?>
</body>
</html>
Basically, what happens is: https://i.imgur.com/aUOx5yj.mp4
Does anyone know what the problem is and why the inserted data does not show up on the index page? The data does show on the page when I submit it directly into the MYSQL database.
Warning: You are wide open to SQL Injections and should use parameterized prepared statements instead of manually building your queries. They are provided by PDO or by MySQLi. Never trust any kind of input! Even when your queries are executed only by trusted users, you are still in risk of corrupting your data. Escaping is not enough!
When working with MySQLi you should enable automatic error reporting instead of checking for errors manually. Checking for errors manually is a terrible practice, very error prone and should be avoided at all costs. Let MySQLi throw exceptions and do not catch them. See How to get the error message in MySQLi?
When opening MySQLi connection you must specify the correct charset. The recommended one is utf8mb4.
if (isset($_POST['submit'])) {
// Enable automatic error reporting
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
// Create new instance of MySQLi class
$con = new mysqli("localhost", "root", "usbw", "serverruimte");
// Set correct charset. Important!
$con->set_charset('utf8mb4');
$stmt = $con->prepare('INSERT INTO serverruimte (id,Server,Student,Docent,Project,startdatum,einddatum) VALUES (?,?,?,?,?,?,?)');
$stmt->bind_param('sssssss', $_POST['id'], $_POST['Server'], $_POST['Student'], $_POST['Docent'], $_POST['Project'], $_POST['startdatum'], $_POST['einddatum']);
$stmt->execute();
echo "Opslaan voltooid!";
mysqli_close($con);
}
Change this line:
$sql = "INSERT INTO serverruimte (id,Server,Student,Docent,Project,startdatum,einddatum) VALUES ('$_POST[id]','$_POST[Server]','$_POST[Student]','$_POST[Docent]','$_POST[Project]','$_POST[startdatum]','$_POST[einddatum]')";
to:
$sql = "INSERT INTO serverruimte (id,Server,Student,Docent,Project,startdatum,einddatum) VALUES ('".$_POST['id']."','".$_POST['Server']."','".$_POST[Student]."','".$_POST['Docent']."','".$_POST['Project']."','".$_POST['Startdatum']."','".$_POST['Einddatum']."')";
Reason behind this change is because your query is wrong for the following reasons:
You were using strings instead of concatenating your real values coming from $_POST
Some of your indexes in $_POST were misspelled. For example:
$_POST[einddatum] should be $_POST['Einddatum']
Also, consider that this code is vulnerable to SQL Injection
Related
I got a little form:
<form id="plannerform" action="save.php" method="post">
<input id="plannername" placeholder=" " type="text" autocomplete="off" name="plannername">
<input id="plannersubmit" type="submit" value="eintragen">
</form>
As you can see there is the action="save.php" and method="post" on the text-input there is name="plannername".
And thats my php:
$con = mysql_connect("myHost","myUser","myPW");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("myDB", $con);
$sql="INSERT INTO anmeldungen (FR_PM)
VALUES ('$_POST[plannername]')";
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}
echo "1 record added";
The FR_PM is one column of my table. But when I press submit, not even a new row gets created. Nothing happens.
But when I call my php with "mywebsite.com/save.php" it adds a new row in my table (with no value at "FR_PM", what's pretty obvious)
What do I do wrong?
one of the things that you need to learn if you are a beginner, you should try by all means to stay away from using mysql_* function this is depreciated and its no longer supported in php. instead use mysqli_* with prepared statements, or use PDO prepared statements.
prepared statments make you code looks clean and its easy to debug.
this is you example with prepared statements.
<form id="plannerform" action="save.php" method="post">
<input id="plannername" placeholder=" " type="text" autocomplete="off" name="plannername">
<input id="plannersubmit" type="submit" value="eintragen" name="submit">
</form>
save.php
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if (isset($_POST['submit'])) {
if (empty($_POST['plannername'])) {
die("Enter plannername");
} else {
// prepare and bind
$stmt = $conn->prepare("INSERT INTO anmeldungen (FR_PM) VALUES (?)");
$stmt->bind_param("s", $_POST['plannername']);
if ($stmt->execute()) {
echo "New records created successfully";
} else {
echo "Could not insert record";
}
$stmt->close();
}
}
?>
The reason I used prepared statements :
Prepared statements reduces parsing time as the preparation on the
query is done only once (although the statement is executed multiple
times)
Bound parameters minimize bandwidth to the server as you need send
only the parameters each time, and not the whole query
Prepared statements are very useful against SQL injections, because
parameter values, which are transmitted later using a different
protocol, need not be correctly escaped. If the original statement
template is not derived from external input, SQL injection cannot
occur.
But when I call my php with "mywebsite.com/save.php" it adds a new row
in my table (with no value at "FR_PM", what's pretty obvious)
What do I do wrong?
Well do prevent that from happening you need to check if the form was submitted before you can actual process any thing.
Note: If we want to insert any data from external sources (like user input from a form ), it is very important that the data is sanitized
and validated. always treat input from a form as if its from a very
dangerous hacker
change your insert query:
$sql="INSERT INTO anmeldungen (FR_PM) VALUES ('".$_POST["plannername"]."')";
Or
$plannername = $_POST["plannername"];
$sql="INSERT INTO anmeldungen (FR_PM) VALUES ('".$plannername."')";
Also, use "name"= and not "id"= in the HTML form.
This is usually misleading when working with forms and HTTP POST method.
you may try
$value = $_POST['plannername'];
$sql="INSERT INTO anmeldungen (FR_PM) VALUES ('{$value}')";
When trying to fetch data from a Mysql database using PHP, the following code gets a message:
Getting error:
undefined variable result.
<?php
require_once 'login.php';
$conn = new mysqli($hn, $un, $pw, $db);
if ($conn->connect_error) die($conn->connect_error);
echo <<<_END
<form action="fetchdata.php" method="post"><pre>
Enter Country <input type="text" name="field">
<input type="submit" value="Display Records">
</pre></form>
_END;
if (isset($_POST['field'])) {
$field=$_post($conn,'field');
$query="SELECT * FROM customers WHERE Country = '$field'";
$result=$conn->query($query);
if (!$result) die($conn->error);
}
$rows = $result->num_rows;
Instead of
$field=$_post($conn,'field');
Maybe you mean
$field=$_POST['field'];
Additionally you use $result at the end, even when it is not defined:
$rows = $result->num_rows;
Also you never output any data and print it. You just store it in variables.
In any case: What you are doing there by writing form data directly into a query string, is dangerous. I recommend you to use PDO together with Named Parameters. Also maybe read up about SQL injections.
Here is another stackoverflow question with a nice answer, regarding SQL injections. It includes both PDO and mysqli: https://stackoverflow.com/a/60496/6637731
Without telling how request to this code is done, one guess is that isset($_POST['field']) returns false, hence variable $result is never defined but you use it anyway below in $result->num_rows.
You made a mistake in the below line. It should be
$field=$_POST['field'];
not
$field=$_post($conn,'field');
I was checking my webpages for SQL Injection, when the main pages didn't responded to it, I created a test script:
<?
$a = $_POST["a"];
$username="...";
$password="...";
$database="...";
mysql_connect ('...',$username,$password);
mysql_select_db($database) or die( "Unable to select database");
$ress=mysql_query("SELECT username FROM userinfo WHERE id='$a'");
$row = mysql_fetch_array($ress);
print $row[0];
?>
<form name="form" action="hackMe.php" method="POST">
<input id="a" name="a" size="150">
<input name="Submit" type="submit" value="Submit">
</form>
But when I try this line:
'; UPDATE userinfo SET email = 'steve#unixwiz.net' WHERE email = 'testusr#gmail.com
I just get an error, and no change in the database.
Any ideas why?
Quote from the manual:
mysql_query() sends a unique query (multiple queries are not supported) to the currently active database on the server that's associated with the specified
Highlighting by me. mysql_query() only allows a single query query per call, the second query behind the ; is ignored.
To test SQL injection you have to use a query that doesn't need a second one to do harm.
Edit:
It IS possible to allow multiple queries, but you have to explicitly state this in the mysql_connect() call.
mysql_connect($host, $username, $password, false, 65536);
// defined by MySQL:
// #define CLIENT_MULTI_STATEMENTS 65536 /* Enable/disable multi-stmt support */
first of all i am pretty new with mysql and php and for now i just want to insert some data in a mysql database form two text box using php.
here the database name is "info" and table name is "students" having three columns like id(primary key, auto increment activated), name and dept. There are two text boxes txtName and txtDept. I want that when i press the enter button the data form the text boxes will be inserted into the mysql database. I have tried the following code but data is not being inserted in the table....
<html>
<form mehtod="post" action="home.php">
<input type="text" name="txtName" />
<input type="text" name="txtDept" />
<input type="submit" value="Enter"/>
</form>
</html>
<?php
$con = mysqli_connect("localhost","root","","info");
if($_POST){
$name = $_POST['txtName'];
$dept = $_POST['txtDept'];
echo $name;
mysqli_query($con,"INSERT INTO students(name,dept) VALUES($name,$dept);");
}
?>
There are a few things wrong with your posted code.
mehtod="post" it should be method="post" - typo.
Plus, quote your VALUES
VALUES('$name','$dept')
DO use prepared statements, or PDO with prepared statements.
because your present code is open to SQL injection
and add error reporting
error_reporting(E_ALL);
ini_set('display_errors', 1);
You should also check for DB errors.
$con = mysqli_connect("localhost","root","","info")
or die("Error " . mysqli_error($con));
as well as or die(mysqli_error($con)) to mysqli_query()
Sidenote/suggestion:
If your entire code is inside the same file (which appears to be), consider wrapping your PHP/SQL inside a conditional statement using the submit button named attribute, otherwise, you may get an Undefined index... warning.
Naming your submit button <input type="submit" name="submit" value="Enter"/>
and doing
if(isset($_POST['submit'])){ code to execute }
Just doing if($_POST){ may give unexpected results when error reporting is set.
Rewrite: with some added security using mysqli_real_escape_string() and stripslashes()
<html>
<form method="post" action="home.php">
<input type="text" name="txtName" />
<input type="text" name="txtDept" />
<input type="submit" name="submit" value="Enter"/>
</form>
</html>
<?php
$con = mysqli_connect("localhost","root","","info")
or die("Error " . mysqli_error($con));
if(isset($_POST['submit'])){
$name = stripslashes($_POST['txtName']);
$name = mysqli_real_escape_string($con,$_POST['txtName']);
$dept = stripslashes($_POST['txtDept']);
$dept = mysqli_real_escape_string($con,$_POST['txtDept']);
echo $name;
mysqli_query($con,"INSERT INTO `students` (`name`, `dept`) VALUES ('$name','$dept')")
or die(mysqli_error($con));
}
?>
As per the manual: http://php.net/manual/en/mysqli.connect-error.php and if you wish to use the following method where a comment has been given to that effect:
<?php
$link = #mysqli_connect('localhost', 'fake_user', 'my_password', 'my_db');
if (!$link) {
die('Connect Error: ' . mysqli_connect_error());
}
?>
God save us all...
Use PDO class instead :). By using PDO you can additionally make prepared statement on client side and use named parameters. More over if you ever have to change your database driver PDO support around 12 different drivers (eighteen different databases!) where MySQLi supports only one driver (MySQL). :(
In term of performance MySQLi is around 2,5% faster however this is not a big difference at all. My choice is PDO anyway :).
My html code:
<form action="send_post.php" method="post">
<input type="submit" value="Login" />
</form>
PHP code:
<?php
$con = mysqli_connect("","","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
else
{
// echo('Connected with Mysql');
}
#mysql_select_db("a", $con);// connect to database db_name
if (isset($_POST['Submit']))
{
$email=$_POST['email'];
$pass=$_POST['pass'];
$sql_query="INSERT INTO formdata (email, pass) VALUES('$email', '$pass')";}
?>
Database name: mysql
Table name: formdata
Why it is not working? in second line I used 'local host' first but I was receiving error so I removed it.
You use the mysqli_ API to connect to your database and then test for errors and try to select a database with the mysql_ API. Pick one and stick to it. (Don't pick mysql_ it is deprecated).
You only run the form handling code if there is a Submit data item in the submitted data. You have no form control with name="Submit" so that will never happen.
Your form handling code expects there to be email and pass data in the submitted data but your form does not have fields with those names.
Constructing an SQL query as a string and storing it in a variable is insufficient to do anything to your database. You have to actually sent it to the database server. That would use mysqli_query with your current approach, but you should switch to prepared statements