INSERT INTO MYSQL - Can´t insert data - mysql

After a couple of week´s trying to insert data in variables from form in MYSQL database, i´m asking here. I found a lot of example codes of INSERT INTO and also my provider checked my skript. He said I have a problem in my $sql=.
I tryed a lot of, but i can´t see any data in phpMyAdmin after click submit, but i receive the mail, that works fine.
Maybe anybody can see an issue in my script.
<?php
if(isset($_POST["sendcopy"])){
mail($mailToCC, $subject, $textCC, $from);
}
if(isset($_POST["submit"])){
$host_name = "database.myprovider";
$database = "db_name";
$user_name = "user_name";
$password = "*****************";
$connect = mysqli_connect($host_name, $user_name, $password, $database);
if (mysqli_connect_errno())
{
echo "KEINE VERBINDUNG MÖGLICH! " . mysqli_connect_error();
}
else
{
echo "ERFOLGREICHE VERBINDUNG ZUR DATENBANK!";}
$sql= 'INSERT INTO "offen"("id", "vorname", "nachname", "email", "telefon", "geburtsjahr", "postleitzahl", "datum", "stunde", "minute", "personen", "bereich", "nachricht")
VALUES ($id, $vorname, $nachname, $email, $tel, $geburtsjahr, $plz, $datum, $stunde, $minute, $personen, $bereich, $nachricht)';
mysql_close($connect);
}
include ("reservtrue.php");
exit;
?>
The "submit" comes from a <form> below.
<form method="post" action="mailer.php" onsubmit="return chkFormular()" name="Formular" id="formTemplate">
<table id="reservtable">
.
.
.
.
<input type="submit" name="submit" value="Reservieren" id="submit">
</td>
</tr>
</table>
</form>
I hope it´s no problem to use german words as variables, here in stackoverflow.
Thank´s for help.
EDIT
Thank you for you´r suggestions. I still can´t insert data from form to MYSQL. I changed my Code a bit. And if I paste the code into phpMyAdmin - SQL, it work´s! But not if I load my script to server and test my form in web.
This is my new Code:
<?php
error_reporting(E_ALL ^ E_NOTICE);
if(isset($_POST["submit"])){
$vorname = $_POST["Vorname"];
$nachname = $_POST["Nachname"];
$email = $_POST["Mailadresse"];
$tel = $_POST["Telefonnummer"];
$geburtsjahr = $_POST["Geburtsjahr"];
$plz = $_POST["PLZ"];
$datum = $_POST["Datum"];
$stunde = $_POST["Stunde"];
$minute = $_POST["Minute"];
$personen = $_POST["Personen"];
$bereich = $_POST["Bereich"];
$nachricht = $_POST["Nachricht"];
$host_name = "database.myprovider";
$database = "db_name";
$user_name = "user_name";
$password = "*****************";
$connect = mysqli_connect($host_name, $user_name, $password, $database);
if (mysqli_connect_errno())
{
echo "KEINE VERBINDUNG MÖGLICH! " . mysqli_connect_error();
}
else
{
echo "ERFOLGREICHE VERBINDUNG ZUR DATENBANK!";}
$insert = ("INSERT INTO offen(vorname, nachname, email, telefon, geburtsjahr, postleitzahl, datum, stunde, minute, personen, bereich, nachricht)
VALUES ('".$id."', '".$vorname."', '".$nachname."', '".$email."', '".$tel."', '".$geburtsjahr."', '".$plz."', '".$datum."', '".$stunde."', '".$minute."', '".$personen."', '".$bereich2."', '".$nachricht."')");
mysqli_query($insert, $sql);
}
include ("reservtrue.php");
exit;
?>
Problem solved
Problem is solved with following code:
$insert = "INSERT INTO `offene`
(
`id`, `vorname`, `nachname`, `email`, `telefon`, `geburtsjahr`, `postleitzahl`, `datum`, `stunde`, `minute`, `personen`, `bereich`, `nachricht`
)
VALUES
(
NULL, '$vorname', '$nachname', '$email', '$tel', '$geburtsjahr', '$plz', '$datum', '$stunde', '$minute', '$personen', '$bereich', '$nachricht');";
mysqli_query($connect, $insert);
Thank you guy´s for information, inspriation and tips! I learnd a lot in the last 2 days.

Where is your mysql_query($sql); ? Add this after your $sql. Your $sql query require mysql_query to run it. If you added that and it still doesn't work, I suggest your go to phpmyadmin's sql tab. Paste your query in with some random VALUES. That's how I check if my query is working or not.

echo "ERFOLGREICHE VERBINDUNG ZUR DATENBANK!";}
$sql= 'INSERT INTO offen (id, vorname, nachname, email, telefon, geburtsjahr, postleitzahl, datum, stunde, minute, personen, bereich, nachricht)
VALUES ('$id', '$vorname', '$nachname', '$email', '$tel', '$geburtsjahr', '$plz', '$datum', '$stunde', '$minute', '$personen', '$bereich', '$nachricht')';
mysqli_query($connect,$sql); //this line was missing from your code.
mysqli_close($connect); //updated to make it MySQLi
The mysqli_query($connect,$sql) actually does the work of applying the SQL you define in the $sql query.
NOTE
You have both MySQLi and MySQL functions in your script, you must stick with just one function, ALL your SQL functions must be MySQL i .
I would recommend that you change your SQL query to use single quotes and backticks. The table name offen and the column names do not need to be in quotes. The variables you insert do need to be in quotes, as I have illustrated.
You do not (usually) need to have mysqli_close because the SQL connection automatically closes once the PHP reaches the end of the page.

The variable names don't need quotes but the parametes do. Also you are missing the code to execute the SQL statement.
echo "ERFOLGREICHE VERBINDUNG ZUR DATENBANK!";}
$sql= "INSERT INTO offen(id, vorname, nachname, email, telefon, geburtsjahr, postleitzahl, datum, stunde, minute, personen, bereich, nachricht)
VALUES ($id, '$vorname', '$nachname', '$email', '$tel', '$geburtsjahr', '$plz', '$datum', '$stunde', '$minute', '$personen', '$bereich', '$nachricht')";
if ($connect->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $connect->error;
}
mysql_close($connect);
}

It appears you have some understandable confusion about quoting of the variables in $insert. You likely have a problem with how you are handling your datatypes.
Here are a few things to look out for
You have too many quotes around your variables. Wrap the query in "" but save the '' for individual variables (where necessary, seen below). For instance, just wrap the variable once like so: .'$vorname'.' if the variable you are inputting is a string. If it is an integer (INT variable type), leave the quotes off, e.g.$number`. In other words, input strings as strings, and INT as int.
If the variable you wish to input is set as auto-increment in your database (i.e. it is the primary key), you probably don't want to be inputting it at all. For example, in your case $id appears to be one. If this is true, you have the syntax backward. Just input it as NULL or leave it off the insert, like so: insert into tablename (id, vorname) values (NULL, '$vorname')
You need to also sanitize your ALL of your variables if you to prevent SQL Injection (very possible with your code). You can do this with mysqli_real_escape_string()
OR instead of going through all of that, you could use prepared statements, which would both handle that information in a cleaner way, but also would protect your code against SQL injection.
Here's how to do this using mysqli:
Connect as an object:
$connect = new mysqli($host_name, $user_name, $password, $database);
And then feed your query into the object and bind the parameters. In this section, "s" is a string, "i" is an integer.
$insert = $connect->prepare("INSERT INTO offen(id, vorname, nachname, email)
VALUES (NULL, ?,?,?");
$insert->bind_param("sss", $vorname, $nachname, $email);
$insert->execute();
$insert->close();
If you name your variables correctly, this should work.

Related

SQL insert is not working

Please can someone help? The code below seems to work and gives no errors, but when I check the database, it hasn't added anything. Tearing my hair out!
<?php
$bcode = $_GET['barcode'];
$businessid = $_GET['businessid'];
$servername = "---------";
$username = "-------";
$password = "-------";
// Create connection
$conn = new mysqli($servername, $username, $password);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO 'db741921215'.'Scans' (Barcode, Success, Business) VALUES ('".$barcode."', 'Y', '".$businessid."')";
$con->close();
} ?>
The columns in table 'Scans' is as below:
Help is very, very much appreciated!!!
I see a couple things - use backticks "`" around your table name definitions, not single quotes. Also, save yourself some eye strain and use the fact that PHP interpolates variables just fine within doublequoted strings.
$sql = "INSERT INTO `db741921215`.`Scans`
(Barcode, Success, Business)
VALUES
('$barcode', 'Y', '$businessid')";
Also - you never actual execute the query, do you?
$conn->query($sql);
You seem to be missing the step where you execute the sql statement
I can see where you define it but I don't see where it's executed. i.e.
$conn->query($sql);
Also, you seem to be missing a letter when closing the connection: $con->close() should be $conn->close();

Cannot push data to mysql

I can't seem to push data from my form to the database, I have checked the error_log and there's no error. Please check my codes below, thank you!
<?php
$con=mysqli_connect("localhost","admineventus","J7!ren;3") or Die ("Cannot connect");
mysqli_select_db($con,"eventus");
if(isset($_POST['submitpublish']))
{
mysqli_query($con,"insert into events values('$_POST[Category]','$_POST[Name]','$_POST[Location]','$_POST[Sdate]','$_POST[Edate]','$_POST[Etime]','$_POST[Fee]','$_POST[Free]','$_POST[About]')");
}
?>
Try this
if(isset($_POST['submitpublish']))
{
$sql = "INSERT INTO events (col1, col1, .. , colx)
VALUES ('".$_POST["Category"]."','".$_POST["Name"]."','".$_POST["Location"]."','".$_POST["Sdate"]."','".$_POST["Sdate"]."','".$_POST["Edate"]."','".$_POST["Etime"]."','".$_POST["Fee"]."','".$_POST["Free"]."','".$_POST["About"]."')";
$result = mysqli_query($conn,$sql);
}
Make sure your code enters in IF statement

Can't get form posts to be stored in the database

I can't figure out why the posts from my form won't save in the database.
This is the html:
<button type="button" class="write">Write Reviews</button>
<form class="writeForm"method="post">
<input type="text" required name="monicker" placeholder="Name">
<textarea name="review" required maxlength="5000" placeholder="Leave your review here (max 25,000 charachters)"></textarea>
<input type="submit" value="Submit">
</form>
And the php:
<?php
$dbc = mysqli::real_connect('localhost', 'user_name', 'not_password');
mysqli::select_db('db_name',$dbc);
$monicker = mysqli::real_escape_string ($_POST['monicker']);
$review = mysqli::real_escape_string ($_POST['review']);
if(isset($_POST['Submit'])) {
$query = "INSERT INTO reviews
(id, monicker, review, date)
VALUES (DEFAULT,'$monicker', '$review', 'CURDATE()');";
mysqlI::query($dbc, $query);
}
mysqlI::close();
?>
As #maku said, you need to execute the query. But there is also more to it.
You have added $mysql_query, this is incorrect.
It is the wrong API (mysql vs. mysqli).
You also need to include your db connection.
If your id column is auto_increment, there is no need to insert it. Happens automatically.
There is no reason to assign a variable to the query because you are not selecting anything.
I would recommend using prepared statements instead, then you don't need to use mysqli::real_escape_string.
if(isset($_POST['submit'])) {
$query = "INSERT INTO reviews (id, monicker, review, date)
VALUES (DEFAULT,'$monicker', '$review', 'CURDATE()')";
mysqli_query($dbc, $query) or die(mysqli_error($dbc);
}
mysqli::close();
You didnt execute your query :
Add this line.
$result = $mysqli::query($query);
I modified your code, please check this code :
<?php
$dbc = mysql_connect('localhost', 'communi3_root', 'typeset');
mysql_select_db('communi3_cfds',$dbc);
$monicker = isset($_POST['monicker']) ? mysql_real_escape_string ($_POST['monicker']) : '';
$review = isset($_POST['review']) ? mysql_real_escape_string ($_POST['review']) : '';
if(isset($_POST['submit'])){
$current_date = date('Y-m-d H:i:s');
$query = "INSERT INTO reviews (id, monicker, review, date) VALUES (DEFAULT,'$monicker', '$review', '$current_date');";
$result = mysql_query($query)or die(mysql_error($dbc));
}
mysql_close();
?>
I found an issues in your previous code :
1. No checking on your POST variables. It causes error because you used it directly without checking if it is set or not.
2. CURDATE() is an undefined function, use PHP date function instead.

Admin panel - Creating an edit users button

I've created an admin panel on my website so when the admin logs in he can edit users. I'm trying to get it to create a table that displays a list of all the users on the database, however, when I run it I get the error:
No database selected
Here is the code in my editusers.php:
<?php
include 'adminpage.php';
include 'connection.php';
$sql = "SELECT * FROM Users";
$result = mysql_query($sql)or die(mysql_error());
echo "<table>";
echo "<tr><th>UserID</th><th>First Name</th><th>Last Name</th><th>Email</th><th>D-O-B</th></tr>Username</th><th>Password</th><th>";
while($row = mysql_fetch_array($result)){
$userid = $row['UserID'];
$firstname = $row['FirstName'];
$lastname = $row['LastName'];
$email = $row['Email'];
$dob = $row['DateofBirth'];
$username = $row['Username'];
$password = $row['Password'];
// Now for each looped row
echo "<tr><td style='width: 200px;'>".$userid."</td><td style='width: 200px;'>".$firstname."</td><td>".$scale."</td><td>".$lastname."</td><td>".$email."</td></tr>".$dob."</td></tr>".$username."</td></tr>".$password."</td></tr>";
} // End our while loop
echo "</table>"
?>
First of all it looks like you are using mysql which isn't a wise move. This is because Mysql is actually deprecated and was improved to mysqli. Your problem may be to do with your database connection. You also haven't set a database. Like I said you can set an active database in your connection script. It should or could look something like this.
<?php
$conn = mysqli_connect("localhost", "root", "password", "database");
// Evaluate the connection
if (mysqli_connect_errno()) {
echo mysqli_connect_error();
exit();
}
?>
After that, your sql query is correct by selecting all from you table 'users' but in order to proceed I recommend creating a query where you use mysqli_query an select the $sql and $conn as parameters. In all honesty it is much advised to stop and continue once you have adapted to mysqli. Alternatively you can use PDO which in some cases can be seen as better to use rather than mysqli but the choice is yours. I personally would get to grips with mysqli and then look at some answers on Stack Overflow to decide whether you should use PDO or not. Visit the PHP manual here. Enter all the mysql functions you know and it will show you how to use the new mysqli version of the functions. Don't think that it is as simple as just adding and 'i' to the end of a mysql function. That's what I initially thought but there is alot to do with extra parameters etc. Hope this helps :-)

mysql query run only for numbers

$username = $_POST['username'];
$password = $_POST['password'];
$passworda = $_POST['passworda'];
$email = $_POST['email'];
if ($password == $passworda){
echo 'this is the user name to be chacked:' .$username .'<br>';
$query = "SELECT username FROM mishta where username=$username";
$queryrun = mysql_query($query);
echo $queryrun .'<br>';
echo mysql_num_rows($queryrun) .'<br>';
if (mysql_num_rows($queryrun)!=1){
mysql_query("INSERT INTO mishta(id, username, password, email) VALUES ('', '$username','$password','$email')");
}
else{
echo 'user already exist';
}
Hello there,
I'm trying to make a simple registration form, I get useername from html form and I want to make sure it doesn't already exist in my database.
the thing is that whenever I run the username as a number (1,2,3...) it runs smoothly.
but, when I try to write anything with letter, like Bob or Dora I get the following error:
Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in...
Yet the information is still being stored my my data base, means it got false on the last if statement, and it will continue to get false (It will just insert duplicates if I try again).
Any help would be appreciated.
You forgot the quotes around the username:
$query = "SELECT username FROM mishta where username='$username'"
Quotes are used as string delimiter. Numbers are not string and it works if you use a number.