My PHP code to POST to my SQL on Godaddy's hosting is not working for some reason.. I added debug statements but I'm just not sure why it's not working. It's driving me crazy.
Here's my file named "homepage.php":
<?php
if (isset($_POST['submitted'])) {
include('mysql_connection.php');
$entry = $_POST['entry'];
$sql = "INSERT INTO posts (typed) VALUES ('$entry')";
if (!mysqli_query($dbcon, $sql)) {
die('Error inserting text.');
}
$newentry = "One entry added to the database.";
}
?>
HTML
<html>
<head> </head>
<body>
<center>
<form method="post" action="homepage.php">
<input type="hidden" name="submitted" value="true" />
<input type="text" name="entry" maxlength="200" />
<br></br>
<input type="submit" value="insert" />
</form>
</center>
<?php echo $newentry?>
</body>
</html>
And my database is named "subpost-db" with the table "posts" and a column named "typed" with VARCHAR values.
My SQL connection file is named "mysql-connection.php" and here's the code:
<?php
DEFINE ('DB_USER', 'xxxxxxxxxx');
DEFINE ('DB_PSWD', 'xxxxxxxxxx');
DEFINE ('DB_HOST', 'localhost');
DEFINE ('DB_NAME', 'subpost-db');
$dbcon = mysqli_connect(DB_HOST, DB_USER, DB_PSWD, DB_NAME);
if (!$dbcon) {
die('Error connecting to the requested database. ');
}
?>
By the way, when I go to mysql-connection.php on my website, the debug message does not pop up.
When I click "insert" after typing something on my form, it reloads homepage.php, where my form is, and only displays the text that says "Error inserting text." but I'm not sure what the problem is.
You are getting an error from your query so add the correct reporting to the test after the code that issues the query to the database and it will tell you what is wrong.
<?php
// from Fred-ii- comment
error_reporting(E_ALL);
ini_set('display_errors', 1);
if (isset($_POST['submitted'])) {
include('mysql_connection.php');
$entry = $_POST['entry'];
$sql = "INSERT INTO posts (typed) VALUES ('$entry')";
if (!mysqli_query($dbcon, $sql)) {
die('Error inserting text. ' . mysqli_error($dbcon) ); //<-- changed line
}
$newentry = "One entry added to the database.";
}
?>
Related
Whenever I tried to insert data to the table of database it keeps echoing "Connection successful" from connection.phpThe connection successful in my insert.php file
insert.php code
<?php
include_once("connection.php");
if(isset($_POST['Add'])){
$productCat = $_POST['category'];
$insertQuery = $pdo->prepare("INSERT INTO products_tbl (productCategory) VALUES (:ucategory)");
$insertQuery->bindParam(':ucategory',$productCat);
$insertQuery->execute();
echo "<script>alert('Successfully Register')</script>";
echo "<script>window.open('index.php','_self')</script>";
}
?>
connection code
<?php
$host = "localhost";
$username = "root";
$password = "";
$dbname = "Ecommercedb";
try{
$pdo = new PDO("mysql:host=$host;dbname=$dbname", $username, $password);
$pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
echo "Connection success";
} catch(PDOException $e){
echo "Connection failed: " . $e->getMessage();
}
?>
forms
<form action="insert.php" method="post">
<label for="">Product Category</label>
<input type="text" name="category" id="" placeholder="Category">
<input type="submit" value="Submit" class="btn" name="Add">
</form>
I tried removing the echo enter image description heresince it wasn't needed but it only lead to
HTTP:500 error.
I have also checked the connection and it is successful.
I expected there would be an alert saying it is successfully inserted but instead I keep getting the connection success from my connection.php
It looks like your code is working as intended, and the issue is likely related to the way you are submitting your form. The "Connection successful" message is being printed because your connection to the database is working, but the form data is not being passed to the insert.php script.
Make sure that you are correctly specifying the action attribute of the form element, and that the method is set to "post". Also, check that the name attributes of the form elements match the names of the variables in the insert.php script.
Additionally, you can try adding some debugging statements to the insert.php script, such as var_dump($_POST) or echo $productCat to check if the values are being passed to the script correctly.
Hello guys i need some help.I connected to database from server and can insert some info like $sql = "INSERT INTO Posts (Text_Post) VALUES ('Sample Text')";. Now I want to save on click text from <input type="text" /> to database. Can you tell me what i am doing wrong.
<?php
$servername = "google.com";
$username = "google";
$password = "google";
$dbname = "google";
// 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'])) {
$sql = "INSERT INTO Posts (Text_Post) VALUES ('".$_POST['text']."')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
}
?>
<!DOCTYPE html>
<html>
<head>
<title>anonim</title>
</head>
<body>
<form name="form" action="" method="post">
<input type="text" name="text" id="text" value="Salut" /=>
<input type="submit" id="Submit" />
</form>
</body>
</html>
You're missing the name tag on your submit. When data is POST'ed to the server, it uses the name tag.
<input type="submit" id="submit" name="Submit">
Remember to watch your Capitals also - (since you're checking if Submit is SET then you need to POST the submit).
You could just do:
if(isset($_POST['text'])) {
Also, going off the comments: I'd suggest taking a look at this link because you're prone to SQL Injections.
when we are going to post a form using POST or GET. we should always give name to all our fieds so we get get them just using $_POST['name'] or $_GET['name']. In Your case just give a name to your submit tag and check whether data is submitted or not.
replace
<input type="submit" id="Submit" />
with
<input type="submit" id="submit" name="submit">
and check it like
if(isset($_POST['submit'])) {// it will only check where form is posted or not
// your code...
}
I have two different directories on my wampserver, this code works on one, but not on the other, I don't understand why.
PHP
<?php
error_reporting(E_ALL & ~E_NOTICE);
session_start();
$msg = "";
if (!isset($_SESSION['username'])) {
header('Location: login.php');
die();
}
if(isset($_SESSION['username'])) {
if (isset($_POST['submit']))
{
$className = $_POST['className'];
$classColour = $_POST['classColour'];
include_once("connection.php");
$sql = "INSERT INTO class (className, classColour) VALUE ('$className', '$classColour')";
mysqli_query($dbConnection, $sql);
$msg = "New class '" . $className . "' added.";
} else {
$msg = "No class added yet.";
}
}
?>
HTML
<form method="post" action="add_class.php">
<input type="text" name="className" placeholder="Class" />
<input type="text" name="classColour" placeholder="Colour" />
<div><input type="submit" name="submit" value="Add" class="btn butn-orange"/></div>
</form>
This is in the file "add_class.php" and I've tried many different things, putting single quotes (`) around the table columns in the $sql but still, it won't work. I've tried adjusting the names in the table, which had underscores but now have camelCasing, still made no difference. This code works perfectly in another directory, can someone please tell me why this is happening and possibly propose a solution? Thank you in advance.
P.S My connection works because I inserted a new row via phpmyadmin and looped through the database printing every existing "className" and it worked, I just can't insert from the php script.
connection.php
<?php
$dbConnection = mysqli_connect("localhost","root", "", "main");
if(mysqli_connect_errno())
{
echo "Failed to connect" . mysqli_connect_error();
}
?>
When asked if you've had assigned the sessions, you replied I assigned this when the user logs in. Moving forward from that, let's assume the assigned session as one written below:
<?php
error_reporting(E_ALL & ~E_NOTICE);
session_start();
$_SESSION['username'] = "HawqasKaPujaari";
$msg = "";
// this fails, as session is already set.
if (!isset($_SESSION['username'])) {
header('Location: login.php');
die();
}
if(isset($_SESSION['username'])) {
if (isset($_POST['submit']))
{
$className = $_POST['className'];
$classColour = $_POST['classColour'];
include_once("connection.php");
$sql = "INSERT INTO class (className, classColour) VALUES ('$className', '$classColour')";
mysqli_query($dbConnection, $sql);
echo $msg = "New class '" . $className . "' added.";
} else {
$msg = "No class added yet.";
}
}
?>
<form method="post" action="">
<input type="text" name="className" placeholder="Class" />
<input type="text" name="classColour" placeholder="Colour" />
<div><input type="submit" name="submit" value="Add" class="btn butn-orange"/></div>
</form>
Note: The only changes I made were to change the action="" empty and changed VALUE to VALUES in your query.
connection.php:
<?php
$dbConnection = mysqli_connect("localhost","root", "", "main");
if(mysqli_connect_errno())
{
echo "Failed to connect" . mysqli_connect_error();
}
?>
As the above posted code seemed to be correct and was bugging me so I thought of testing it myself by creating the database/tables and it seemed to work properly without any errors. I have posted the relevant pictures below:
Note: Make sure you have the connection.php file in the same
directory as the add_class.php.
I'm learning PHP and I've hit a wall.
When the user submits to the form, it adds to the database.
I am also trying to display all items in the database on the same page as the form.
However,this only works if the form has just been submitted. If the form has not been submitted (but there is still content in the database), nothing is shown.
How can I always show what is in the current database?
<form action="" method="post">
<input type="text" name="todo_content" id="">
<input type="submit" value="Submit">
</form>
<?php
if ( isset($_POST['todo_content'])) {
$latest_content = $_POST['todo_content'];
} else {
die();
}
$todo_db = mysqli_connect('localhost', 'root', 'root');
mysqli_select_db($todo_db, 'todo_list');
mysqli_query(
$todo_db,
"INSERT INTO todo_items (item_content) VALUES ('$latest_content')"
);
$all_todos = mysqli_query( $todo_db, "SELECT item_content FROM todo_items" );
$all_todos_result = mysqli_fetch_array($all_todos);
var_dump($all_todos_result); // these show nothing
var_dump($all_todos); // these show nothing
?>
The die() is your problem:
if ( isset($_POST['todo_content'])) {
echo 'is set';
$latest_content = $_POST['todo_content'];
} else {
die();
}
If $_POST is not set, you will never reach the part where you start printing your items. And since the form is not submitted, $_POST is empty.
EDIT
You could do it like this:
<form action="" method="post">
<input type="text" name="todo_content" id="">
<input type="submit" value="Submit">
</form>
<?php
$todo_db = mysqli_connect('localhost', 'root', 'root');
mysqli_select_db($todo_db, 'todo_list');
if ( isset($_POST['todo_content'])) {
$latest_content = $_POST['todo_content'];
mysqli_query($todo_db, "INSERT INTO todo_items (item_content) VALUES ('$latest_content')");
}
$all_todos = mysqli_query( $todo_db, "SELECT item_content FROM todo_items" );
$all_todos_result = mysqli_fetch_array($all_todos);
var_dump($all_todos_result); // these show nothing
var_dump($all_todos); // these show nothing
?>
You need to move your insert query inside your if condition and you will need not to die if the form is not submitted but print the form and your query results
$todo_db = mysqli_connect('localhost', 'root', 'root');
mysqli_select_db($todo_db, 'todo_list');
if ( isset($_POST['todo_content'])) {
echo 'is set';
$latest_content = $_POST['todo_content'];
mysqli_query($todo_db, "INSERT INTO todo_items (item_content) VALUES ('$latest_content')");
} else {
?>
<form action="" method="post">
<input type="text" name="todo_content" id="">
<input type="submit" value="Submit">
</form>
<?php
$all_todos = mysqli_query( $todo_db, "SELECT item_content FROM todo_items" );
$all_todos_result = mysqli_fetch_array($all_todos);
var_dump($all_todos_result);
var_dump($all_todos);
}
As side note i'd say you are at high risk of mysql injection. You should use prepaed statments and not inserting $_POST data inside your database directly
I am experimenting with PHP and Mysql. I have created a database and table at mu localhost using xampp. I have also created a file that suppose to populate my table by executing a query, but the strange thing is that i get no errors but at the same time no DATA has been inserted into my DataBase:
CODE:
register.php:
<?php
session_start();
if(isset($_POST['submitted'])){
include('connectDB.php');
$UserN = $_POST['username'];
$Upass = $_POST['password'];
$Ufn = $_POST['first_name'];
$Uln = $_POST['last_name'];
$Uemail = $_POST['email'];
$NewAccountQuery = "INSERT INTO users (user_id,username, password, first_name, last_name, emial) VALUES ('$UserN','$Upass', '$Ufn', '$Uln', '$Uemail')";
if(!mysql_query($NewAccountQuery)){
die(mysql_error());
}//end of nested if statment
$newrecord = "1 record added to the database";
}//end of if statment
?>
<html>
<head>
<title>Home Page</title>
<meta http-equiv="content-type" content="text/html; charset=iso-8859-1" />
<link href="style.css" rel="stylesheet" type="text/css" />
</head>
<body>
<div id="wrapper">
<header><h1>E-Shop</h1></header>
<article>
<h1>Welcome</h1>
<h1>Create Account</h1>
<div id="login">
<ul id="login">
<form method="post" action="register.php" >
<fieldset>
<legend>Fill in the form</legend>
<label>Select Username : <input type="text" name="username" /></label>
<label>Password : <input type="password" name="password" /></label>
<label>Enter First Name : <input type="text" name="first_name" /></label>
<label>Enter Last Name : <input type="text" name="last_name" /></label>
<label>Enter E-mail Address: <input type="text" name="email" /></label>
</fieldset>
<br />
<input type="submit" submit="submit" value="Create Account" class="button">
</form>
</div>
<form action="index.php" method="post">
<div id="login">
<ul id="login">
<li>
<input type="submit" value="Cancel" onclick="index.php" class="button">
</li>
</ul>
</div>
</article>
<aside>
</aside>
<div id="footer">This is my site i Made coppyrights 2013 Tomazi</div>
</div>
</body>
</html>
I have also one include file which is connectDB:
<?php
session_start();
$con = mysql_connect("127.0.0.1", "root", "");
if(!$con)
die('Could not connect: ' . mysql_error());
mysql_select_db("eshop", $con) or die("Cannot select DB");
?>
Database structure:
database Name: eshop;
only one table in DB : users;
users table consists of:
user_id: A_I , PK
username
password
first_name
last_name
email
I spend a substantial amount of time to work this out did research and looked at some tutorials but with no luck
Can anyone spot what is the root of my problem...?
It is because if(isset($_POST['submitted'])){
you dont have input field with name submitted give the submit button name to submitted
<input name="submitted" type="submit" submit="submit" value="Create Account" class="button">
Check your insert query you have more fields than your values
Change :
$NewAccountQuery = "INSERT INTO users (user_id,username, password, first_name, last_name, email) VALUES ('$UserN','$Upass', '$Ufn', '$Uln', '$Uemail')";
to :
$NewAccountQuery = "INSERT INTO users (user_id,username, password, first_name, last_name, email) VALUES ('','$UserN','$Upass', '$Ufn', '$Uln', '$Uemail')";
Considering user_id is auto increment field.
Your email in query is written wrongly as emial.
Is error reporting turned on?
Put this on the top of your screen:
error_reporting(E_ALL);
ini_set('display_errors', '1');
Some good answers above, but I would also suggest you make use of newer MySQLi / PDO instead of outdated 2002 MySQL API.
Some examples: (i will use mysqli since you wrote your original example in procedural code)
connectDB.php
<?php
$db = mysqli_connect('host', 'user', 'password', 'database');
if (mysqli_connect_errno())
die(mysqli_connect_error());
?>
register.php -- i'll just write out an example php part and let you do the rest
<?php
//i'll always check if session was already started, if it was then
//there is no need to start it again
if (!isset($_SESSION)) {
session_start();
}
//no need to include again if it was already included before
include_once('connectDB.php');
//get all posted values
$username = $_POST['username'];
$userpass = $_POST['password'];
$usermail = $_POST['usermail'];
//and some more
//run checks here for if fields are empty etc?
//example check if username was empty
if($username == NULL) {
echo 'No username entered, try again';
mysqli_close($db);
exit();
} else {
//if username field is filled we will insert values into $db
//build query
$sql_query_string = "INSERT INTO _tablename_(username,userpassword,useremail) VALUES('$username','$userpass','$usermail')";
if(mysqli_query($db,$sql_query_string)) {
echo 'Record was entered into DB successfully';
mysqli_close($db);
} else {
echo 'Ooops - something went wrong.';
mysqli_close($db);
}
}
?>
this should work quite nicely and all you need to add is your proper posted values and build the form to post it, that's all.
<?php
$db = mysqli_connect('host', 'user', 'password', 'database');
if (mysqli_connect_errno())
die(mysqli_connect_error());
?>
register.php -- i'll just write out an example php part and let you do the rest
<?php
//i'll always check if session was already started, if it was then
//there is no need to start it again
if (!isset($_SESSION)) {
session_start();
}
//no need to include again if it was already included before
include_once('connectDB.php');
//get all posted values
$username = $_POST['username'];
$userpass = $_POST['password'];
$usermail = $_POST['usermail'];
//and some more
//run checks here for if fields are empty etc?
//example check if username was empty
if($username == NULL) {
echo 'No username entered, try again';
mysqli_close($db);
exit();
} else {
//if username field is filled we will insert values into $db
//build query
$sql_query_string = "INSERT INTO _tablename_(username,userpassword,useremail) VALUES('$username','$userpass','$usermail')";
if(mysqli_query($db,$sql_query_string)) {
echo 'Record was entered into DB successfully';
mysqli_close($db);`enter code here`
} else {
echo 'Ooops - something went wrong.';
mysqli_close($db);
}
}
?>