INSERT INTO MYSQL - Can´t insert data

Question

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.

Answer 1

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.

Answer 2

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);

}

Answer 3

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

1. 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.
2. 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')
3. 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.

Answer 4

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.