HTML/PHP form doesn't insert data to MySQL Database [duplicate] - php

This question already has answers here:
How to include a PHP variable inside a MySQL statement
(5 answers)
Closed 1 year ago.
I am new to PHP, and I have been trying to make a small Homework Organiser Application.
The idea is that you can input a Subject and Description and it will be added to a MySQL Database (that's it for now).
I have created a html form:
<form action="insert.php">
<label for="subj">Subject:</label>
<br>
<input type="text" name="subj">
<br>
<label for="desc">Description:</label>
<br>
<input type="text" name="desc">
<br>
<input type="submit" value="Submit" name="submit">
</form>
and some php code:
<?php
$subject = $_POST['subj'];
$description = $_POST['desc'];
$subject = mysql_real_escape_string($subject);
$description = mysql_real_escape_string($description);
$dbhost = ''; //These are filled in actually
$dbuser = ''; //These are filled in actually
$dbpass = ''; //These are filled in actually
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
die('Could not connect: ' . mysql_error());
}
$sql = 'INSERT INTO organiser '.
'(subject,description) '.
'VALUES ('$subject', '$description')';
mysql_select_db(''); //These are filled in actually
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
die('Could not enter data: ' . mysql_error());
}
echo "Entered data successfully\n";
mysql_close($conn);
?>
The problem is that the input from the Subject and Description boxes on the HTML form don't go into the MySQL Database.
However, If I set
'VALUES ('$subject', '$description')';
to
'VALUES ("test", "test")';
it works.
Any help is appreciated!
Thanks!

In addition to the answer already given in regards to missing dots for the concatenate:
Form method defaults to a GET method, if the method is omitted from the form tag.
You are using <form action="insert.php"> which is equivalent to doing
<form action="insert.php" method = "get"> which is not what you want nor required.
Change it to
<form action="insert.php" method="post">
since you are using POST variables.
That is the contributing reason why 'VALUES ("test", "test")'; works and not the other way, since both of these variables $subject - $description, are based on your POST variables:
$subject = $_POST['subj'];
$description = $_POST['desc'];
You can either do
$sql = "INSERT INTO organiser (subject,description) VALUES ('$subject', '$description')";
as stated in a comment.
or
$sql = "INSERT INTO organiser (subject,description) VALUES ('".$subject."', '".$description."')";
Add error reporting to the top of your file(s)
error_reporting(E_ALL);
ini_set('display_errors', 1);
which would have signaled errors found in your code.
http://php.net/manual/en/function.error-reporting.php
Yet, your method is open to SQL injection. Use mysqli_* with prepared statements, or PDO with prepared statements.
Use a mysqli prepared statements method; it's safer.
<?php
$DB_HOST = "xxx"; // replace with your own
$DB_NAME = "xxx";
$DB_USER = "xxx";
$DB_PASS = "xxx";
$conn = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME);
if($conn->connect_errno > 0) {
die('Connection failed [' . $conn->connect_error . ']');
}
// optional to check for empty fields
// if(isset($_POST['submit']) && !empty($_POST['subj']) && !empty($_POST['desc'])) {
if(isset($_POST['submit'])) {
$subject = stripslashes($_POST['subj']);
$description = stripslashes($_POST['desc']);
$stmt = $conn->prepare("INSERT INTO organiser (`subject`, `description`) VALUES (?, ?)");
$stmt->bind_param('ss', $subject, $description);
if (!$stmt->execute()) {
echo "Execute failed: (" . $stmt->errno . ") " . $stmt->error;
}
else{
echo "<h2>Success!</h2>";
}
$stmt->close(); // Statement
$conn->close(); // MySQLi
}
?>
Form: (new)
<form action = "insert.php" method = "post">
<label for="subj">Subject:</label>
<br>
<input type="text" name="subj">
<br>
<label for="desc">Description:</label>
<br>
<input type="text" name="desc">
<br>
<input type="submit" value="Submit" name="submit">
</form>

Your problem is because you are using single quotes in your $sql declaration,
$sql = 'INSERT INTO organiser '.'(subject,description) '.'VALUES ('$subject', '$description')';
When you use single quotes you are telling PHP that you would like everything within the quotes to be taken literally.
$age = 20;
$myAge = 'I am $age years';
echo $myAge;
This would echo I am $age years since you used single quotes.
However, if you used double quotes instead,
$age = 20;
$myAge = "I am $age years";
echo $myAge;
The output would be,
I am 20 years
Thus, your $sql statement should be (I write it on one line instead),
$sql = "INSERT INTO organiser (subject,description) VALUES ('$subject', '$description')";
^ ^
If you echo your $sql it would become something along the lines of,
INSERT INTO organiser (subject,description) VALUES ('Your subject', 'Your description')
Your use of single quotes within your SQL-statement is correct, you can read more about the subject here: When to use single quotes, double quotes and backticks

You forgot the dots around the variables and you can add some double quotes around that depending on the content of your variables :)
'VALUES ("'.$subject.'", "'.$description.'")';

Related

Trying to pass user input into database [duplicate]

This question already has answers here:
When to use single quotes, double quotes, and backticks in MySQL
(13 answers)
Closed 3 years ago.
So I'm working with XAMPP and I was following this tutorial on how to set up a website. Right now the webpage is takes a first name and last name. Once you submit it should place those into the database. I'm getting this error every time I try to test it: Connected "successfullyError: INSERT into 'user'('fname', 'lname') VALUES ('abc','xyz') You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near ''user'('fname', 'lname') VALUES ('abc','xyz')' at line 1"
Here's what the html looks like:
<!DOCTYPE html>
<html>
<body>
<form action="submit.php" method="post">
First Name:<br>
<input type="text" name="firstname">
<br>
Last Name:<br>
<input type="text" name="lastname">
<br><br>
<input type="submit" value="Submit">
</form>
</body>
</html>
And here is the file that has the php code to connect the page to the database. I am new to php and I tried to locate the syntax error, but to no avail.
<?php
$x = $_POST['firstname'];
$y = $_POST['lastname'];
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "db1";
//create connection
$conn = new mysqli($servername, $username, $password, $dbname);
//Check connection
if ($conn->connect_error)
{
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
$sql = "INSERT into 'user'('fname', 'lname') VALUES ('$x','$y')";
if($conn->query($sql) === TRUE)
{
echo "That's going on your permanent record loser";
}
else { echo "Error: " . $sql . "<br>" . $conn->error; }
$conn->close();
?>
The only problem that I see that you use a single quote ' to your table and column instead of using "`". The single quote are use as string delimiter.
Try changing this part
'user'('fname', 'lname')
into
`user`(`fname`, `lname`)

Validating email for a specific username using php and MYSQL [duplicate]

This question already has answers here:
When to use single quotes, double quotes, and backticks in MySQL
(13 answers)
Closed 6 years ago.
So, I am trying to validate whether username and email entered by a user in an html form are correct according to Database
I have a database called testing and a table called info which has two fields:
name and email.
this is the code
<html>
<body>
<form action="conn.php" method="post">
Name: <input type="text" name="name"><br>
E-mail: <input type="text" name="email"><br>
<input type="submit">
</form>
</body>
</html>
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname="testing";
$name=$_POST['name'];
$mail=$_POST['email'];
// Create connection
$conn1 = new mysqli($servername, $username, $password, $dbname);
$query='select email from info where name= "$name"';
$results = mysqli_query($conn1,$query);
$row= mysqli_fetch_assoc($results);
echo $row["email"];
$email=$row["email"];
if($mail==$email){
echo " correct values";
}
else
echo " incorrect";
//echo $name, $email;
// Check connection
//if ($conn1->connect_error) {
// die("Connection failed: " . $conn1->connect_error);
//}
//echo "Connected successfully";
$conn1->close();
?>
but the result is always incorrect. The values that i enter in the text boxes match the values in the database.
You should use placeholders, the code would be something like this:
$query = 'SELECT emal FROM info WHERE name = ?';
$stmt = mysqli_stmt_prepare($conn1, $query);
mysqli_stmt_bind_param($stmt, 's', $name);
mysqli_stmt_execute($stmt);
$results = mysqli_stmt_get_result($stmt);
For better readable code, read up on the object-oriented mysqli-interface, or better yet use PDO.
alternatively you can use string concatenation like
Update this line
$query='select email from info where name= "$name"';
with
$query='select email from info where name= "'.$name.'" ';
or with
$query="select email from info where name= '".$name."' ";

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 ')' at line 5

I can't get the data from my form to load into MySQL Database. Please help!
This is the error that i get:
Error: INSERT INTO add_review (name,email,details) VALUES ( 'Darron
Brown', 'blank#gmail.com', 'ldldjd', ) 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 ')' at line 5
<?php
// Connect to MySQL
// a. Variables
$host = "hostname";
$username = "user";
$password = "secretpassword";
$dbname = "myDatabase";
// b. Connection
$connection = mysqli_connect($host, $username, $password, $dbname);
// c. Check our connection
if(mysqli_connect_errno()) {
die("Database connection failed: " .
mysqli_connect_error() .
" (" . mysqli_connect_errno() . ")"
);
}
// Insert our data
$name = isset($_POST["name"]) ? $_POST["name"] : "";
$email = isset($_POST["email"]) ? $_POST["email"] : "";
$details = isset($_POST["details"]) ? $_POST["details"] : "";
$name = mysqli_real_escape_string($connection, $name);
$email = mysqli_real_escape_string($connection, $email);
$details = mysqli_real_escape_string($connection, $details);
$sql = "INSERT INTO add_review (name,email,details) VALUES (
'$name',
'$email',
'$details',
)";
// $insert = $connection->query($sql);
// Print response from MySQL
if (mysqli_query($connection, $sql)) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($connection);
}
// Close our connection
mysqli_close($connection);
?>
<div class = "section page">
<div class="wrapper">
<h1>Add a Review</h1>
<p>If you think there is something I am missing, let me know! Complete the form to send me an email.</p>
<form method="post" action="">
<table>
<tr>
<th><label for="name">Movie Name</label></th>
<td><input type="text" id="name" name="name" /></td>
</tr>
<tr>
<th><label for="email">Email</label></th>
<td><input type="text" id="email" name="email" /></td>
</tr>
<tr>
<th><label for="email">Suggest Movie Details</label></th>
<td><textarea name="details" id="details"></textarea></td>
</tr>
</table>
<input type="submit" value="Send" />
</form>
</div>
</div>
See the syntex for '$details', change your query and remove the extra comma after $details like:
$sql = "INSERT INTO add_review (name,email,details) VALUES (
'$name',
'$email',
'$details'
)";
NB : You are at risk of SQL Injections, Learn mysqli_prepared to prevent sql injections, you can learn here
Your code with prepared statements would look like this :
<?php
// prepare and bind
$sql = $conn->prepare("INSERT INTO add_review (name,email, details) VALUES (?, ?, ?)");
$sql->bind_param("sss", $name, $email, $details);
$sql->execute();
echo "New records created successfully";
$sq]->close();
$conn->close();
?>
Now explaining the function :
$sql->bind_param("sss", $name, $email, $details);
This function binds the parameters to the SQL query and tells the database what the parameters are. The "sss" argument lists the types of data that the parameters are. The s character tells mysql that the parameter is a string.
The argument may be one of four types:
i - integer
d - double
s - string
b - BLOB
We must have one of these for each parameter.
By telling mysqli what type of data to expect, we minimize the risk of SQL injections.
Important : When you insert any data from external sources (eg user
input from a form), it is very important that the data is sanitized and validated.

PHP not inserting into tables

I'm having trouble getting a practice signup form to submit data to my database ...
<!DOCTYPE HTML>
<html>
<head>
</head>
<body>
<?php
$name = $email = $password = "";
?>
<form method="post">
Name: <input type="text" name="name">
<br><br>
E-mail: <input type="text" name="email">
<br><br>
Password: <input type="text" name="password">
<br><br>
<input type="submit" value="Submit" name="submit">
</form>
<?php
if(isset($_POST['submit'])){
$name = fix_input($_POST["name"]);
$email = fix_input($_POST["email"]);
$password = fix_input($_POST["password"]);
mysqli_connect("localhost","username","password","dbname") or die(mysql_error());
mysql_query("INSERT INTO ('username','password') VALUES ('$name', md5('$password'))");
Print "You've been signed up successfully"; }
function fix_input($data)
{
$data = trim($data);
$data = stripslashes($data);
$data = htmlspecialchars($data);
return $data;
}
?>
</body>
</html>
In addition to Ugur's answer, you are mismatching mysqli commands and mysql commands. Here's how to do this in an object oriented fashion:
// create mysqli database object
$mysqli = new mysqli_connect("localhost","username","password","database");
// store your query in a variable. question marks are filled by variables
$sql = "INSERT INTO table_name ('username','password') VALUES (?,?)";
// prepare command uses $sql variable as query
$stmt = mysqli->prepare($sql);
// "ss" means your 2 variables are strings, then you pass your two variables.
$stmt->bind_param("ss",$name,md5($password));
// execute does as it seems, executes the query.
$stmt->execute();
// then print your success message here.
Using prepared statements removes the need to sanitize user input, as harmful input is not substituted into the query directly. For more reading:
http://www.php.net/manual/en/mysqli.quickstart.prepared-statements.php
There are some good tips for using prepared statements in many different scenarios, as well as towards the bottom, there is an explanation on how prepared statements prevent SQL injection.
Missing table name
mysql_query("INSERT INTO ...... ('username','password') VALUES ('$name', md5('$password'))");
You're mixing mysql_* with mysqli_* functions, i.e.: mysqli_connect and mysql_query and you're wrapping your column names in quotes, plus you're missing the table name to insert into.
Try the following, fixed code:
if(isset($_POST['submit'])){
$name = fix_input($_POST["name"]);
$email = fix_input($_POST["email"]);
$password = fix_input($_POST["password"]);
mysqli_connect("localhost","username","password","dbname") or die(mysql_error());
mysqli_query("INSERT INTO `your_table` (`username`,`password`) VALUES ('$name', md5('$password'))");
Print "You've been signed up successfully"; }
You're also using password storage technology that dates back to 1996. MD5 is no longer considered safe to use.
I suggest you look into PHP's password function: http://php.net/password
And if you're having problems with your fix_input() function, you should consider using the mysqli_real_escape_string() function.
then setting up a DB connection while passing a variable to it.
$DB_HOST = "xxx";
$DB_NAME = "xxx";
$DB_PASS = "xxx";
$DB_USER = "xxx";
$db = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME);
if($db->connect_errno > 0) {
die('Connection failed [' . $db->connect_error . ']');
}
and instead of using:
$name = fix_input($_POST["name"]);
use the following:
$name= mysqli_real_escape_string($db, $_POST['name']);
and do the same for the rest.
you don't have table name in your query! also do not use quotation in your column name :)
you have mixed up mysqli and mysql.
Change
mysql_query("INSERT INTO ('username','password') VALUES ('$name', md5('$password'))");
to
mysqli_query("INSERT INTO yoour_table(username',password) VALUES ('$name', md5('$password'))");

I cant get the form data to go into database. What am I doing wrong?

CODE UPDATED, STILL NOT WORKING.
I know I´m apparently using mysql function which will be outdated. But for now all I want is for this code to work. I want to know what I´m doing wrong:(
I´m very new to php and databases... I have been struggling to get simple html form data to go into the database table. And I just can´t get it to work:( Can anyone help and see what is wrong with my code? I´ve just done a simple table in the database with the fields ID, FIRSTNAME and SURNAME.
Here is the code:
<?php
//connect to database
$mysql_host = 'localhost';
$mysql_user = 'root';
$mysql_pass = '';
$mysql_db = 'test';
if (!mysql_connect ($mysql_host, $mysql_user, $mysql_pass)||!mysql_select_db ($mysql_db) ) {
die(mysql_error());
}
// Code
if (isset($_POST['firstname'])&&
isset($_POST['surname'])) {
$firstname = $_POST['firstname'];
$surname = $_POST['surname'];
if (!empty($username)&&!empty($password)) {
$query = "INSERT INTO `test`.`test_tabell`
VALUES ('', '" . mysql_real_escape_string($firstname) . "', '" . mysql_real_escape_string($surname) . "')";
/*$query = "INSERT INTO `test`.`test_tabell` VALUES (``, `.$firstname.`, `.$surname.`)"; */
$query_run = mysql_query($query);
if (!$query_run) echo mysql_error();
}
}
?>
<form action="add.php" method="POST">
Firstname:<br> <input type="text" name="firstname" value="<?php if (isset($firstname)) { echo $firstname; } ?>"><br><br>
Surname:<br> <input type="text" name="surname" value="<?php if (isset($surname)) { echo $surname; } ?>"><br><br>
<input type="submit" value="Submit">
</form>
Thank you!
Don't use mysql specific syntax, It's outdated and it begins to be annoying when you need to do some high level stuff, and you can't switch to sqlite or postgresql.
I recommend using PDO, you can do something like:
// Usage: $db = connectToDataBase($dbHost, $dbName, $dbUsername, $dbPassword);
// Pre: $dbHost is the database hostname,
// $dbName is the name of the database itself,
// $dbUsername is the username to access the database,
// $dbPassword is the password for the user of the database.
// Post: $db is an PDO connection to the database, based on the input parameters.
function connectToDataBase($dbHost, $dbName, $dbUsername, $dbPassword)
{
try
{
return new PDO("mysql:host=$dbHost;dbname=$dbName;charset=UTF-8", $dbUsername, $dbPassword);
}
catch(PDOException $PDOexception)
{
exit("<p>An error ocurred: Can't connect to database. </p><p>More preciesly: ". $PDOexception->getMessage(). "</p>");
}
}
And then init the variables (I think you forgot to define the name of the database);
$host = 'localhost';
$user = 'root';
$dataBaseName = 'databaseName';
$pass = '';
Now you can access your database via
$GLOBALS['db'] = connectToDataBase($host , $databaseName, $user, $pass);
Now you have an instance of a PDO database donnection.
One thing I want to point out is that you're vonurable to sql injections, you want to use prepared statements in your query, like:
$query = "INSERT INTO test(first_name, sur_name) VALUES (:firstname, :surname);";
Where we will execute two variables $firstName and $surName on the query, making them replace the values of :firstName and :surName, let me show you by first creating a simple insertion function:
function insertFunction($db, $query, $firstName, $surName)
{
$statement = $db->prepare($query);
return $statement->execute(array(":firstName" => $firstName, ":surName" => $surName));
}
So It's easy for you to do something like
$firstName = 'Smith';
$surName = 'John';
$db = $GLOBALS['db'];
$success = insertFunction($db, $query, $firstName, $surName);
Now you can check if it was successful or not, by checking whether $success is true or false.
If you want to see more advanced use of PDO (multiple rows etc) then you can check out one of my comments here: Javascript function as php?
(Not the top comment).
I hope this helps. Please comment if anything is odd.
Hard to tell without seeing your schema but try this:
$query = "INSERT INTO `test`.`test_tabell` VALUES ('', '$firstname', '$surname')";
$query_run = mysql_query($query);
You're using backticks instead of apostrophes. Also, you're trying to execute a query before defining what the query is.
Your insert query is wrong and also open to SQL injections. Here's how it should be:
$query = "INSERT INTO `test`.`test_tabell`
VALUES ('', '" . mysql_real_escape_string($firstname) . "', '" . mysql_real_escape_string($surname) . "')";
Notice the changing of all backticks to apostrophe.
Also, you're trying to execute the query before defining it.
EDIT
As per your information related to table definition, you can skip the id field from your table. The INSERT query will become:
$query = "INSERT INTO `test`.`test_tabell` (`FIRSTNAME`, `SURNAME`)
VALUES ('" . mysql_real_escape_string($firstname) . "', '" . mysql_real_escape_string($surname) . "')";
$query_run = mysql_query( $query );
As posted in the comments, you REALLY SHOULD NOT use/learn/practice using any function that starts with "mysql_" since it will NOT work as soon as PHP is updated. These functions are on their way out. Best of luck with learning to use PHP and SQL databases - just make sure you're learning something that will be useful in the future. Make sure to read up on Object Oriented Programming (OOP) in relation to PHP and both the PDO and mysqli_* functions.

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