how do I combine the mysql insert and select statement into a prepared statement?? Will this part of the code help me to duplicate information into a database when I hit the "copy" button?
Thanks.
INSERT INTO assignment_speeches_copy (id, subject, body, tags, image)
SELECT * FROM assignment_speeches
WHERE id = $id";
$stmt = $mysqli->prepare("INSERT INTO assignment_speeches_copy VALUES (?,?,?,?,?)"); (how do I continue?)
You don't have to combine anything. INSERT ... SELECT is a simple statement....
$stmt = $mysqli->prepare(
"INSERT INTO assignment_speeches_copy (id, subject, body, tags, image)
SELECT (id, subject, body, tags, image)
FROM assignment_speeches
WHERE id = ?"
);
Related
I would like to insert values into one table by selecting fields from another table and also add information such date and login_id (these do not come from the table where I am selecting).
$sql = "INSERT INTO questions_to_answer (login_id, question_id, Question_Category, QuestionType, Question, Meaning) VALUES (?,?,?,?,?,?)";
$stmt= $DB_con->prepare($sql);
$stmt->execute($test_user_id,
"SELECT question_id,
Question_Category,
QuestionType,
Question,
Meaning all_questions
WHERE personality_profile_questions.Question_Category = 1");
You need to have the SELECT in the prepare, you can't have SQL commands in the execute. If in the execute it will be treated as a literal string, quoted, and escaped.
Try:
$sql = "INSERT INTO questions_to_answer (login_id, question_id, Question_Category, QuestionType, Question, Meaning)
SELECT ?, question_id, Question_Category, QuestionType, Question, Meaning
WHERE personality_profile_questions.Question_Category = 1";
$stmt= $DB_con->prepare($sql);
$stmt->execute(array($test_user_id));
Assuming $test_user_id = login_id. I'm not clear what all_questions is.
So I'm trying to insert 4 values into a table. I'm getting 3 values from POST and the other one I want to get it from another table. This is how I thought about implementing it but it doesn't seem to be working. Any suggestions?
$query = "INSERT INTO topics (subject, data, uid, role) VALUES (:user, :pass, :uid, SELECT role FROM users WHERE uid=:uid) ";
In SQL, all subqueries need to be surrounded by their own parentheses. So, you can fix your query by using:
INSERT INTO topics (subject, data, uid, role)
VALUES (:user, :pass, :uid, (SELECT role FROM users WHERE uid = :uid));
Personally, I much prefer the INSERT . . . SELECT version of SELECT:
INSERT INTO topics (subject, data, uid, role)
SELECT :user, :pass, :uid, u.role
FROM users u
WHERE uid = :uid;
So I have 3 tables: donor, blood_type, user_account. I am trying to populate the donor table which contains user_id and blood_id, but there is no join between the blood_group and the user_account table so I tried this, but it didn't work. Can someone please tell what I am doing wrong? I am very new to php and databases.
<?php
if(isset($_POST['submit'])) {
$conn = mysqli_connect("localhost", "root" , "");
if(!$conn) {
die("Cannot connect: ");
}
mysqli_select_db($conn,"blood_bank_project");
$sql = "INSERT INTO user_account(username, password) VALUES ('$_POST[user]', '$_POST[psw]');";
$sql .="INSERT INTO donor(first_name,last_name,email_add,gender, birthday, telephone, city, last_donation,user_id, blood_id)VALUES('$_POST[fname]', '$_POST[lname]', '$_POST[email]', '$_POST[gender]', '$_POST[Birthday]', '$_POST[Telephone]', '$_POST[city]', '$_POST[lastdonation]')";
$sql .="UPDATE donor SET blood_id = (SELECT blood_id from blood_type where blood_group= '$_POST[bloodgroup]');";
$sql .="UPDATE donor SET user_id = (SELECT user_id from user_account where username= '$_POST[user]')";
if(mysqli_multi_query($conn, $sql)){
echo'executed';
}
}
?>
You can use a SELECT clause to produce the values for an INSERT. In this case, you can use that to select the appropriate values from the other tables.
INSERT INTO donor (user_id, blood_id, first_name,last_name,email_add,gender, birthday, telephone, city, last_donation)
SELECT u.user_id, b.blood_id,
'$_POST[fname]', '$_POST[lname]', '$_POST[email]', '$_POST[gender]', '$_POST[Birthday]', '$_POST[Telephone]', '$_POST[city]', '$_POST[lastdonation]'
FROM user_accounts AS u
CROSS JOIN blood_type AS b
WHERE u.username = '$_POST[user]' AND b.blood_group= '$_POST[bloodgroup]'
I also strongly recommend you use prepared queries instead of substituting $_POST variables, as the latter subjects you to SQL-injection. I also recommend against using mysqli_multi_query -- it's rarely needed and only makes checking for success harder. If you insert into user_accounts using a separate query, you can then use mysqli_insert_id($conn) to get the user_id assigned when you inserted into user_accounts, instead of using the above JOIN. You can also use the MySQL built-in function LAST_INSERT_ID() to get it.
$stmt = mysqli_prepare($conn, "INSERT INTO user_account(username, password) VALUES (?, ?);") or die("Can't prepare user_account query: " . mysqli_error($conn));
mysqli_stmt_bind_param($stmt, "ss", $_POST['user'], $_POST['psw']);
mysqli_execute($stmt);
$stmt2 = mysqli_prepare($conn, "
INSERT INTO donor (user_id, blood_id, first_name,last_name,email_add,gender, birthday, telephone, city, last_donation)
SELECT LAST_INSERT_ID(), b.blood_id, ?, ?, ?, ?, ?, ?, ?, ?
FROM blood_type AS b
WHERE b.blood_group= ?") or die ("Can't prepare donor query: " . mysqli_error($conn));
mysqli_stmt_bind_param($stmt2, "sssssssss", $_POST['fname'], $_POST['lname'], $_POST['email'], $_POST['gender'], $_POST['Birthday'], $_POST['Telephone'], $_POST['city'], $_POST['lastdonation'], $_POST['bloodgroup']);
mysqli_execute($stmt2);
theres a few things wrong with that code snippet:
Line 15: You've got a rogue 'w' at the start of the line before your $sql variable
All of your $_POST'ed parameters need to be in the format $_POST['parameter'] (Missing quotes, remember to escape your already quoted ones in places)
The where clause sub-select query in line 14 is selecting from a table that does not exist (blood_type)
I guess what your trying to achieve is a mapping between 'user_account' and 'donor' of which you may be better either storing a foreign key in the user account table of the 'donor_id', or a matrix/mapping table that links the two together.
The matrix/mapping table would hold the primary key date from both user_account and donor to create your matrix.
You can then get to either table information from the other knowing just one side of the information.
I'd also make sure your escaping your inbound variables in your queries to prevent any SQL Injection attacks (see here)
I am trying to insert data into a database after the user clicks on a link from file one.php. So file two.php contains the following code:
$retrieve = "SELECT * FROM catalog WHERE id = '$_GET[id]'";
$results = mysqli_query($cnx, $retrieve);
$row = mysqli_fetch_assoc($results);
$count = mysqli_num_rows($results);
So the query above will get the information from the database using $_GET[id] as a reference.
After this is performed, I want to insert the information retrieved in a different table using this code:
$id = $row['id'];
$title = $row['title'];
$price = $row['price'];
$session = session_id();
if($count > 0) {
$insert = "INSERT INTO table2 (id, title, price, session_id)
VALUES('$id', '$title', '$price', '$session');";
}
The first query $retrieve is working but the second $insert is not. Do you have an idea why this is happening? PS: I know I will need to sanitize and use PDO and prepared statements, but I want to test this first and it's not working and I have no idea why. Thanks for your help
You're not executing the query:
$insert = "INSERT INTO table2 (id, title, price, session_id)
VALUES('$id', '$title', '$price', '$session');";
}
it needs to use mysqli_query() with the db connection just as you did for the SELECT and make sure you started the session using session_start(); seeing you're using sessions.
$insert = "INSERT INTO table2 (id, title, price, session_id)
VALUES('$id', '$title', '$price', '$session');";
}
$results_insert = mysqli_query($cnx, $insert);
basically.
Plus...
Your present code is open to SQL injection. Use mysqli with prepared statements, or PDO with prepared statements.
If that still doesn't work, then MySQL may be complaining about something, so you will need to escape your data and check for errors.
http://php.net/manual/en/mysqli.error.php
Sidenote:
Use mysqli_affected_rows() to check if the INSERT was truly successful.
http://php.net/manual/en/mysqli.affected-rows.php
Here's an example of your query in PDO if you'req planning to use PDO in future.
$sql = $pdo->prepare("INSERT INTO table2 (id, title, price, session_id) VALUES(?, ?, ?, ?");
$sql->bindParam(1, $id);
$sql->bindParam(2, $title);
$sql->bindParam(3, $price);
$sql->bindParam(4, $session_id);
$sql->execute();
That's how we are more safe.
$fname = addslashes($fname);
$lname = addslashes($lname);
$dob = addslashes($dob);
$email = $_POST['email'];
$sql =
"INSERT INTO subscriber
(fname, lname, dob)
VALUES
('".$fname."', '".$lname."', '".$dob."')
WHERE email='".$email."'";
$register = mysql_query($sql) or die("insertion error");
I am getting error in sql query "insertion error". Query is inserting data into DB after removing WHERE statement. What is the error.
You can't use where in an insert statement. You might be thinking of an update instead?
$sql = "update subscriber set fname='".$fname."', lname = '".$lname."', dob = '".$dob."' WHERE email='".$email."'";
If your email is a unique value, you can also combine an insert with an update like this:
insert into
subscriber (fname, lname, dob, email)
values ('".$fname."', '".$lname."', '".$dob."', '".$email."')
on duplicate key update set fname='".$fname."', lname='".$lname."', dob='".$dob."'
This second syntax will insert a row if there isn't one with a matching email (again, this has to be set to a unique constraint on the table) and if there is one there already, it will update the data to the values you passed it.
Basically INSERT statement cannot have where. The only time INSERT statement can have where is when using INSERT INTO...SELECT is used.
The only syntax for select statement are
INSERT INTO TableName VALUES (val1, val2, ..., colN)
and
INSERT INTO TableName (col1, col2) VALUES (val1, val2)
The other one is the
INSERT INTO tableName (col1, col2)
SELECT col1, col2
FROM tableX
WHERE ....
basically what it does is all the records that were selected will be inserted on another table (can be the same table also).
One more thing, Use PDO or MYSQLI
Example of using PDO extension:
<?php
$dbh = new PDO('mysql:host=localhost;dbname=test', $user, $pass);
$stmt = $dbh->prepare("INSERT INTO REGISTRY (name, value) VALUES (?, ?)");
$stmt->bindParam(1, $name);
$stmt->bindParam(2, $value);
// insert one row
$name = 'one';
$value = 1;
$stmt->execute();
?>
this will allow you to insert records with single quotes.
Oops !!!! You cannot use a WHERE clause with INSERT statement ..
If you are targeting a particular row then please use UPDATE
$sql = "Update subscriber set fname = '".$fname."' , lname = '".$lname."' , dob = '".$dob."'
WHERE email='".$email."'";
$register = mysql_query($sql) or die("insertion error");