I am using a simple php script to insert data into database but it's failing. The query just doesn't become successful without showing a single error which is why I am unable to figure out the problem. Some expert here help me please.
echo $name." ".$email." ".$pass." ".$phone." ".$area." ".$specialization." ".$city." ".$latitude." ".$longitude;
The result of echo is normal - without any null elements.
$query = mysqli_query($conn, "INSERT INTO users (name, email, pass, phone, area, specialization, hospital, city, latitude, longitude)
VALUES ('$name', '$email', '$pass', '$phone', '$area', '$specialization', '$hospital', '$city', '$latitude', '$longitude') ");
if ($query) {
echo "Status: Registeration Successful!";
// creating directory for user and storing dummy profile picture
//mkdir('../profiles/'.$email_trim, 0777);
//$result_copy = copy("img/dp.jpg.jpg", "../profiles/".$email_trim."/dp.jpg.jpg");
} else {
echo "Status: Err";
}
This "Status: Err" is always printed. I don't know why.
P.S I have double checked the database the field labels are fine.
UPDATE 1:
I added the
die(mysqli_error($conn));
statement and it says "DUPLICATE ENTRY '0' FOR KEY PRIMARY'.
PROBLEM AND SOLUTION:
The issue was that I had an 'id' field which was primary key of the table but it was not set to AUTO_INCREMENT. So, whenever I tried to insert a new record, I was actually inserting entries with duplicate PKs which was the issue. I change it to AUTO_INCREMENT and it solved the problem.
It seems you try to insert a new element with a PK = 0, but there is already a record with this key !
What is the primary key of your table ? Do you use an "id" field which is not shown in your insert statement ? Is this field AUTO_INCREMENT ?
It would be helpful to see the structure of your 'users' table.
Wild guess: looks like you may have defined an "id" column (or with whatever other name) which is primary key with default value "0", but it's not auto increment. That way you can insert 1 row and it will get "0" as "id" column's value, but you cannot insert another row because it will also try to use default value "0", which cannot happen as primary key has to be unique.
If that is the case, then please alter users table and make sure that the primary key column is also 'auto increment'.
Please check the proper error by adding below mentioned code inside else:
echo mysqli_errno($conn) . '----' . mysqli_error($conn);
<?php
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
?>
$query = mysqli_query($conn, "INSERT INTO `users` (`name`, `email`, `pass`, `phone`, `area`, `specialization`, `hospital`, `city`, `latitude`, `longitude`)
VALUES ('$name', '$email', '$pass', '$phone', '$area', '$specialization', '$hospital', '$city', '$latitude', '$longitude') ") or die(mysqli_error());
$query = mysqli_query($conn, "INSERT INTO `users` (`name`, `email`, `pass`, `phone`, `area`, `specialization`, `hospital`, `city`, `latitude`, `longitude`)
VALUES ('$name', '$email', '$pass', '$phone', '$area', '$specialization', '$hospital', '$city', '$latitude', '$longitude') ");
use ` Tick maybe because there's some reserved word in your fields.
I think you doubled your close parenthesis and do not put $conn inside the query..
$query = "INSERT INTO users (name, email, pass, phone, area, specialization, hospital, city, latitude, longitude)
VALUES ('$name', '$email', '$pass', '$phone', '$area', '$specialization', '$hospital', '$city', '$latitude', '$longitude')";
mysqli_query($query, $conn);
Something like this. I hope this helps
Related
$name = mysqli_real_escape_string($connection, $_POST["name"]);
$surname = mysqli_real_escape_string($connection, $_POST["surname"]);
$username = mysqli_real_escape_string($connection, $_POST["username"]);
$email = mysqli_real_escape_string($connection, $_POST["email"]);
$pw1 = mysqli_real_escape_string($connection, $_POST["pw1"]);
$query = "INSERT INTO 'users' ('id','name', 'surname', 'username', 'email', 'password') VALUES (NULL,'$name', '$surname', '$username', '$email', '$pw1')";
$result = mysqli_query($connection, $query);
if(!$result){
echo ("fail");
}
I test if the query has worked using if(!$result){ echo ("fail");} and it echoes fail every time and no data is inserted into the database every time! I have checked the syntax and i believe it is correct... could this be because of the database "collation"?
You should not use the single quote at the table or field name. You have to use a Backtick (like ``) which is located in under Esc key or left side of 1 Key or upper side of Tab key. It should looks like:
$query = "INSERT INTO `users` (`id`, `name`, `surname`, `username`, `email`,
`password`) VALUES ('null', '$name', '$surname', '$username', '$email', '$pw1')";
or
$query = "INSERT INTO users (id, name, surname, username, email,
password) VALUES ('null', '$name', '$surname', '$username', '$email', '$pw1')";
Note: If your id field is already set auto increment then you can remove id and value null. Because id value will automatically increment.
Hope it will helpful.
This php insert query is not working in MYSQL xampp. I couldn't find any error
QUERY:
$query = "INSERT INTO member (id, username,fname,lname,email, password, salt )
VALUES ( '$username', '$password', '$email', '$salt' )";
you are missing $fname, $lname in query also use NULL for id if auto incremented
$query = "INSERT INTO member (id, username,fname,lname,email, password, salt )
VALUES (NULL, '$username', '$fname', '$lname', '$password', '$email', '$salt' )";
you are passing wrong number of column names.your correct query should look like this:
$query = "INSERT INTO member (username,password,email,salt )
VALUES ( '$username', '$password', '$email', '$salt' )";
You are not inserting values to all the columns specified in the query.
In your query
$query = "INSERT INTO member (id, username,fname,lname,email, password, salt )
VALUES ( '$username', '$password', '$email', '$salt' )";
You are specifying 7 columns and only 4 values .So either add more values or remove unnecessary columns specified in the query like
$query = "INSERT INTO member (username, password,email, salt )
VALUES ( '$username', '$password', '$email', '$salt' )";
OR
$query = "INSERT INTO member (id, username,fname,lname,email, password, salt )
VALUES ('$id', '$username','$fname','$lname','$email', '$password', '$salt' )";
I'm trying to add $_SESSION['user_id'] into a database and, when I echo it in this function, it works fine. However, when I try to push it into my MySQL database, it adds the value 0. Really confused as to why. Thanks for any help!
function fill_team() {
$i = 1;
while ($i < 24) {
$first_name = first_name();
$last_name = second_name();
echo "<br>";
$add_names = mysql_query("INSERT INTO `players` (`first_name`, `last_name`, `user_id`) VALUES ('$first_name', '$last_name', '.$_SESSION['user_id']'.)");
$i++;
}
}
Replace echo $row2['first_name']; with return $row2['first_name'];.
If you want to get some value from the function, you should use return operator to pass the value back. It has nothing to do with printing the value with print or echo.
There's some messy mixed quoting inside the SQL statement:
mysql_query("INSERT INTO `players` (`first_name`, `last_name`, `user_id`)
VALUES ('$first_name', '$last_name', '.$_SESSION['user_id']'.)");
Off the top of my head, I have no idea what SQL this will produce, but this should be written as:
mysql_query("INSERT INTO `players` (`first_name`, `last_name`, `user_id`)
VALUES ('$first_name', '$last_name', '$_SESSION[user_id]')");
or
mysql_query("INSERT INTO `players` (`first_name`, `last_name`, `user_id`)
VALUES ('$first_name', '$last_name', '".$_SESSION['user_id']."')");
I've created a registration form and I'm adding the data with php but for some reason it will not let me add the data into the database. Can you see anything wrong with this code?
<?php
mysql_connect("localhost", "root", "password") or die("No Connection");;
mysql_select_db("music") or die("No Database");;
$username= mysql_real_escape_string($_REQUEST["username"]);
$password= mysql_real_escape_string($_REQUEST["password"]);
$email= mysql_real_escape_string($_REQUEST["email"]);
$hash = md5( rand(0,1000) );
mysql_query("INSERT INTO members (id, username, password, email, hash, active) VALUES('', '$username', '$password', '$email', '$hash', '')") or die("Can't Add");
if(mysql_affected_rows()>0){
echo "1";
}else{
echo "2";
}
?>
I Keep getting a Can't Add error indicating that there is a simple problem with the mysql_query row
Thank you
Can't add is not an error, just a catch all statement you added at the end.
To see your actual problem, change your code at the end of the mysql_query line to include the actual error retrieved from mysql_error().
mysql_query("INSERT INTO members (username, password, email, hash, active) VALUES('$username', '$password', '$email', '$hash', '')")
or die("Can't Add - " . mysql_error());
That will give you more details regarding the error, if you post that, I can update my answer with the reason why.
Note that I've also removed the insertion of id, it's not needed if your column is AUTO_INCREMENT, which it should be.
if id is your primary key then replane '' with null
Make id as auto-incremented primary key and remove it from insert query.
there may be unique constraints for any of the column
Either have
mysql_query("INSERT INTO members (id, username, password, email, hash, active)
VALUES(null, '$username', '$password', '$email', '$hash', '')") or die("Can't Add");
^^^^
OR
mysql_query("INSERT INTO members (username, password, email, hash, active)
VALUES( '$username', '$password', '$email', '$hash', '')") or die("Can't Add");
i have a db query in php that is not inserting into database. Have used this format lots of times but for some reason its not working now. any ideas please
$query = "INSERT INTO `databasename`.`member_users` (`id`, `first_name`, `last_name`, `username`, `password`, `address1`, `address2`, `postcode`, `access`, `expires`) VALUES (NULL, '$fname', '$lname', '$email', '', '$add1', '$add2', '$postcode', '0', '')";
$result = mysql_query($query);
if($result){
echo"query inserted";
}else{
echo "nope";
}
Instead of echo "nope"; I suggest something like :
echo 'error while inserting : ['.mysql_errno().'] '.mysql_error();
echo 'query : '.$query;
This way you will be able to see the exact error and the query that was executed.
It can be a lot of things :
Constraint error with a foreign key
Data type error
Non-existent field
Wrong database or table name
Instead of...
$query = "INSERT INTO `databasename`.`member_users` ..."
do
$query = "INSERT INTO member_users ..."
Hope it works. :)
If databasename and member_users are variables then,
Instead of
$query = "INSERT INTO databasename.member_users...
do
$query = "INSERT INTO $databasename.$member_users...