$query = "INSERT INTO clients(ID_Client,Upgrade,Nome,Email,Cod_Postal,Localidade,Contacto,NIF,Factura,Data_Factura,N_Serie,Codigo,Marca,Modelo) VALUES (NULL,'$Upgrade',
'$Nome',
'$Email',
'$Cod_Postal',
'$Localidade',
'$Contacto',
'$NIF',
'$Factura',
'$N_Serie',
'$Data_Factura', ".$N_Serie.",
'$Codigo',
'$Marca',
'$Modelo')";
What it's wrong about this ? I got the same number of fields / values ..
You've got '$N_Serie', twice, remove the first one.
Columns matching values below:
$query = "
INSERT INTO clients(
ID_Client,
Upgrade,
Nome,
Email,
Cod_Postal,
Localidade,
Contacto,
NIF,
Factura,
Data_Factura,
N_Serie,
Codigo,
Marca,
Modelo
) VALUES (
NULL,
'$Upgrade',
'$Nome',
'$Email',
'$Cod_Postal',
'$Localidade',
'$Contacto',
'$NIF',
'$Factura',
'$Data_Factura',
'$N_Serie',
'$Codigo',
'$Marca',
'$Modelo'
)";
Related
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, $conn);
Something like this. I hope this helps
mysql_query( "INSERT INTO users(email, password) VALUES ('$email','$password')" );
$user_id = mysql_insert_id( );
mysql_query( "INSERT INTO business_details ( business_id, name, address, city, state, country, pincode) VALUES ('$category','$business_name', '$business_address', '$city', '$state', '$country', '$pincode')" );
$idB = mysql_insert_id( );
mysql_query( "INSERT INTO user_profiles (name, phone, user_id, business_details_id) VALUES ('$person_name', '$phone_number', '$user_id', '$idB')" );
What will be the best way to insert all queries in one? Is it possible?
No, it isn't supported in MySQL - only Oracle can do it.
MySQL doesn't support multi-table insertion in a single INSERT statement
I want the date to be added to an orders table I have created, once the user proceeds to checkout. The code I currently has just prints:
"Error: Column count doesn't match value count at row 1"
Here's my code:
$sql = "INSERT INTO orders (customer_id, total, order_date) VALUES ('$name', '$total', 'DATETIME: Auto NOW()', NOW() )";
The name and total columns store but the date does not. How can I solve this?
This way:
$sql = "INSERT INTO orders (customer_id, total, order_date) VALUES ('$name', '$total', NOW())";
You want to insert data inside customer_id, total, order_date (3 rows) but you are sendig '$name', '$total', 'DATETIME: Auto NOW()', NOW(), FOUR.
Your error means that you that the number of fields does not match the number of values. That seems correct: your query tries to insert 4 values into 3 fields. You'd probably have to rewrite the query to
$sql = "INSERT INTO orders (customer_id, total, order_date) VALUES ('$name', '$total', NOW() )";
It looks like you need 3 values: customer_id, total, order_date
but you give 4 :'$name', '$total', 'DATETIME: Auto NOW()', NOW()
Maybe it should look like this:
"INSERT INTO orders (customer_id, total, order_date) VALUES ('$name', '$total', NOW() )";
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 am working on building an employee database but seem to have run into an interesting issue.
When I run my query to add a new user/employee, I get the error:
Column count doesn't match value count at row 1
From what I have researched, this seems to be an error with inserting more/less values than what is declared in the first part of an insert statement example:
INSERT INTO table (col1, col2) VALUES (val1, val2, val3)
The thing is though, I have looked over my query and the columns and values match perfectly (count wise). I have even looked for things in my query such as missing quotes, commas, etc.
Here is my code (query):
$db->query("INSERT INTO userdata (
Username,
Email,
Phone,
Password,
FirstName,
LastName,
Address,
City,
State,
Zip,
JobTitle,
UserGroup,
JobStatus,
Points,
Status,
BirthDate,
HireDate,
HourlyRate,
DateAdded,
SSN
) VALUES (
'$Username',
'$Email',
'$Phone',
'$Password',
'$FirstName',
'$LastName',
'$Address',
'$City',
'$State',
'$Zip',
'$JobTitle',
'$Group',
'$JobStatus',
0,
'$Status',
'$BirthDate',
'$HireDate',
'$HourlyRate'
'$TodaysDate',
'$SSN'
)") or die(mysqli_error($db));
Some things to note:
This not all of the columns in the table have data inserted here (I think its possible to do this and things such as auto incrementing ID's will fill themselves in and others will be left blank)
From the variable dumps I have done, all of these variables are valid.
I am really confused about this and any help would be appreciated.
Check the following portion again:
INSERT INTO userdata(...,
JobStatus,
UserGroup,
Points,
UserGroup,
Status,
.....
,)VALUES(...,
$JobStatus',
0,
'$Group',
'$Status',
......
)
In values(, ?, ?, ?), after jobstatus, there should be UserGroup and then Points. And UserGroup appeared twice.