Using the given below php script, I connect to database and insert data in it. But the data is not getting inserted in my database table. It is also not throwing any error. Where is my code wrong?
<?php
$host = '127.0.0.1';
$uname='root';
$pswd='';
$myDB='portal';
if($myConn = new mysqli($host,$uname,$pswd))
echo 'Connected to MySQL server successfully.</br>';
else
echo 'Unable to connect to server</br>';
$database = mysqli_select_db($myConn,$myDB);
if($database)
echo 'Connected to database...</br>';
else
echo 'Database not found!</br>';
$var1='string1';
$var2='string2';
$query= "INSERT INTO users VALUES ($var1,$var2)";
$result = mysqli_query($myConn,$query) or die(mysqli_error($myConn));
?>
You have to add single quotes around the values:
$query= "INSERT INTO users VALUES ('$var1','$var2')
Or better use prepared statements. See this for an example.
In your statement, you must define the names of the target tables in your database, that the values should be inserted into, like this:
$query= "INSERT INTO users (Name,Age) VALUES ('$name','$age')";
if users table have only two columns, or two plus an auto-incrementing id the query is:
INSERT INTO users VALUES ('$var1','$var2')
if there are more columns or a non primary id the query is:
INSERT INTO users (col1,col2) VALUES ('$var1','$var2')
You also miss a parameter in the connection instantiation:
$mysqli = new mysqli($host, $uname,$pswd, $myDB);
Related
I've been spending a couple of hours trying to write mysqli queries to insert a new row in a database (with a primary key ID) and then select the ID of the new row. My code as it currently is:
<?php
include('connectionData.php');
$conn = mysqli_connect($server, $user, $pass, $dbname, $port)
or die('Connection error');
if(isset($_POST['submit'])) {
$pnum = $_POST['pnum'];
$phone_insert_text = "INSERT INTO `voterdatabase`.`phone` (`pnum`) VALUES (?)";
$phone_insert_query = $conn->prepare($phone_insert_text);
$phone_insert_query->bind_param('s', $pnum);
$phone_insert_query->execute();
$phone_select_text = "SELECT phone_id FROM voterdatabase.phone WHERE pnum=?";
$phone_select_query = $conn->prepare($phone_select_text);
$phone_select_query->bind_param('s', $pnum);
$phone_select_query->execute();
$phone_select_query->bind_result($phone_id);
echo $phone_id;
?>
$phone_insert_query executes without issue. But $phone_select_query doesn't appear to run at all, as echo $phone_id; has no effect. What might be going on here? I'm able to run the query directly in MySQLWorkbench.
Note that I previously tried doing this in one query using SELECT LAST_INSERT_ID();, but mysqli fails to execute any query containing that.
Please try this
$lastInsertID= mysqli_insert_id($conn);
Use insert_id property:
<?php
include('connectionData.php');
$conn = mysqli_connect($server, $user, $pass, $dbname, $port)
or die('Connection error');
if(isset($_POST['submit'])) {
$pnum = $_POST['pnum'];
$phone_insert_text = "INSERT INTO `voterdatabase`.`phone` (`pnum`) VALUES (?)";
$phone_insert_query = $conn->prepare($phone_insert_text);
$phone_insert_query->bind_param('s', $pnum);
$phone_insert_query->execute();
$phone_id = $conn->insert_id;
echo $phone_id;
?>
If you wish to be able to use the available functions to get the last inserted id, like mysqli_insert_id(), your table must have an AUTO_INCREMENT column. If not you will not get the id.
Also, even if you have the required columns, this will require two calls. To get around this, what you could do is something like create a stored procedure to do your insert for you and return the inserted id from the procedure.
Could someone let me know how to insert a set of values into a table across different databases .
Example : I have a table by name TABLE inside Database1 , Database2 ....
Is there a way where I can insert values into the TABLE present in both the databses.
I tried doing
Insert into Database1,Database2.Table(type)values('1');
But it did not work .
Thanks for your help
Unfortunately Table names, column names, etc. cannot be dynamic in mysql, you can try to create dynamic queries
$databases = list of dbs;
foreach ($databases as $db){
$query = "Insert into".$db.".tableName()....."
}
since you are using php you can use mysqli_select_db(connection,dbname); to select different database using it.
for Eg.
//create a connection
$conn = mysqli_connect("host","user","password","db_name");
//list of databases
$databases = "Your Array of database";
foreach ($databases as $database){
// Change database to $database
mysqli_select_db($conn,$database);
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect :" . mysqli_connect_error();
}
// code for $database eg. - your insert query...
}
Assuming that I have two tables, names and phones,
and I want to insert data from some input to the tables, in one query. How can it be done?
You can't. However, you CAN use a transaction and have both of them be contained within one transaction.
START TRANSACTION;
INSERT INTO table1 VALUES ('1','2','3');
INSERT INTO table2 VALUES ('bob','smith');
COMMIT;
http://dev.mysql.com/doc/refman/5.1/en/commit.html
MySQL doesn't support multi-table insertion in a single INSERT statement. Oracle is the only one I'm aware of that does, oddly...
INSERT INTO NAMES VALUES(...)
INSERT INTO PHONES VALUES(...)
Old question, but in case someone finds it useful... In Posgresql, MariaDB and probably MySQL 8+ you might achieve the same thing without transactions using WITH statement.
WITH names_inserted AS (
INSERT INTO names ('John Doe') RETURNING *
), phones_inserted AS (
INSERT INTO phones (id_name, phone) (
SELECT names_inserted.id, '123-123-123' as phone
) RETURNING *
) SELECT * FROM names_inserted
LEFT JOIN phones_inserted
ON
phones_inserted.id_name=names_inserted.id
This technique doesn't have much advantages in comparison with transactions in this case, but as an option... or if your system doesn't support transactions for some reason...
P.S. I know this is a Postgresql example, but it looks like MariaDB have complete support of this kind of queries. And in MySQL I suppose you may just use LAST_INSERT_ID() instead of RETURNING * and some minor adjustments.
I had the same problem. I solve it with a for loop.
Example:
If I want to write in 2 identical tables, using a loop
for x = 0 to 1
if x = 0 then TableToWrite = "Table1"
if x = 1 then TableToWrite = "Table2"
Sql = "INSERT INTO " & TableToWrite & " VALUES ('1','2','3')"
NEXT
either
ArrTable = ("Table1", "Table2")
for xArrTable = 0 to Ubound(ArrTable)
Sql = "INSERT INTO " & ArrTable(xArrTable) & " VALUES ('1','2','3')"
NEXT
If you have a small query I don't know if this is the best solution, but if you your query is very big and it is inside a dynamical script with if/else/case conditions this is a good solution.
my way is simple...handle one query at time,
procedural programming
works just perfect
//insert data
$insertQuery = "INSERT INTO drivers (fname, sname) VALUES ('$fname','$sname')";
//save using msqli_query
$save = mysqli_query($conn, $insertQuery);
//check if saved successfully
if (isset($save)){
//save second mysqli_query
$insertQuery2 = "INSERT INTO users (username, email, password) VALUES ('$username', '$email','$password')";
$save2 = mysqli_query($conn, $insertQuery2);
//check if second save is successfully
if (isset($save2)){
//save third mysqli_query
$insertQuery3 = "INSERT INTO vehicles (v_reg, v_make, v_capacity) VALUES('$v_reg','$v_make','$v_capacity')";
$save3 = mysqli_query($conn, $insertQuery3);
//redirect if all insert queries are successful.
header("location:login.php");
}
}else{
echo "Oopsy! An Error Occured.";
}
Multiple SQL statements must be executed with the mysqli_multi_query() function.
Example (MySQLi Object-oriented):
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO names (firstname, lastname)
VALUES ('inpute value here', 'inpute value here');";
$sql .= "INSERT INTO phones (landphone, mobile)
VALUES ('inpute value here', 'inpute value here');";
if ($conn->multi_query($sql) === TRUE) {
echo "New records created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
See the code below, there are two databases connections.
First it get the data from first connection and then insert into second database connection but it will not insert - I can an error saying Unknown column 'fullname' in 'field list'
When I tried SQL query manually in phpMyAdmin and it work fine...
$db_new = mysql_connect('localhost', 'root', 'password');
if (!mysql_select_db("menu_new", $db_new)) {
die("Cant connect menu_new DATABASE");
}
$db_old = mysql_connect('localhost', 'root', 'password');
if (!mysql_select_db("old_menu", $db_old)) {
die("Cant connect old_menu DATABASE");
}
$SQL_old = "SELECT * FROM old_table";
$q = mysql_query($SQL_old, $db_old);
while ($row = mysql_fetch_assoc($q)) {
$name = $row['name'];
$SQL = "INSERT INTO tbl_name (fullname) values ('$name')";
//Problem Here - It wont insert into second database
mysql_query($SQL, $db_new) or die(mysql_error($db_new));
}
Nothing is strange in this behavior. Just add the $link parameter on your mysql_connect calls, and set it to true. By default it is False and it means that reusing this function with the same parameters, which is what you're doing as the db name is not on your mysql-connect, will reuse existing connexion with same parameters. So you have two variables but only one connexion.
It means you're simply moving the db used in this connexion. Prefixing with the db name fixed the problem as MySQL allow inter-base manipulations from the same connexion if it's on the same db server.
Thanks #Konerak for suggestion and that does work!
To Insert/Select data from Database connection, you will have to include database name before the table name.
For Example
From:
mysql_query("INSERT into tbl_name (fullname) values ('1')", $db_new) or die(mysql_error($db_new));
To:
mysql_query("INSERT into menu_new.tbl_name (fullname) values ('1')", $db_new) or die(mysql_error($db_new));
That is really odd though.
The 'id' field of my table auto increases when I insert a row. I want to insert a row and then get that ID.
I would do it just as I said it, but is there a way I can do it without worrying about the time between inserting the row and getting the id?
I know I can query the database for the row that matches the information that was entered, but there is a high change there will be duplicates, with the only difference being the id.
$link = mysqli_connect('127.0.0.1', 'my_user', 'my_pass', 'my_db');
mysqli_query($link, "INSERT INTO mytable (1, 2, 3, 'blah')");
$id = mysqli_insert_id($link);
See mysqli_insert_id().
Whatever you do, don't insert and then do a "SELECT MAX(id) FROM mytable". Like you say, it's a race condition and there's no need. mysqli_insert_id() already has this functionality.
Another way would be to run both queries in one go, and using MySQL's LAST_INSERT_ID() method, where both tables get modified at once (and PHP does not need any ID), like:
mysqli_query($link, "INSERT INTO my_user_table ...;
INSERT INTO my_other_table (`user_id`) VALUES (LAST_INSERT_ID())");
Note that Each connection keeps track of ID separately (so, conflicts are prevented already).
The MySQL function LAST_INSERT_ID() does just what you need: it retrieves the id that was inserted during this session. So it is safe to use, even if there are other processes (other people calling the exact same script, for example) inserting values into the same table.
The PHP function mysql_insert_id() does the same as calling SELECT LAST_INSERT_ID() with mysql_query().
As to PHP's website, mysql_insert_id is now deprecated and we must use either PDO or MySQLi (See #Luke's answer for MySQLi). To do this with PDO, proceed as following:
$db = new PDO('mysql:dbname=database;host=localhost', 'user', 'pass');
$statement = $db->prepare('INSERT INTO people(name, city) VALUES(:name, :city)');
$statement->execute([':name' => 'Bob', ':city' => 'Montreal']);
echo $db->lastInsertId();
As #NaturalBornCamper said, mysql_insert_id is now deprecated and should not be used. The options are now to use either PDO or mysqli. NaturalBornCamper explained PDO in his answer, so I'll show how to do it with MySQLi (MySQL Improved) using mysqli_insert_id.
// First, connect to your database with the usual info...
$db = new mysqli($hostname, $username, $password, $databaseName);
// Let's assume we have a table called 'people' which has a column
// called 'people_id' which is the PK and is auto-incremented...
$db->query("INSERT INTO people (people_name) VALUES ('Mr. X')");
// We've now entered in a new row, which has automatically been
// given a new people_id. We can get it simply with:
$lastInsertedPeopleId = $db->insert_id;
// OR
$lastInsertedPeopleId = mysqli_insert_id($db);
Check out the PHP documentation for more examples: http://php.net/manual/en/mysqli.insert-id.php
I just want to add a small detail concerning lastInsertId();
When entering more than one row at the time, it does not return the last Id, but the first Id of the collection of last inserts.
Consider the following example
$sql = 'INSERT INTO my_table (varNumb,userid) VALUES
(1, :userid),
(2, :userid)';
$sql->addNewNames = $db->prepare($sql);
addNewNames->execute(array(':userid' => $userid));
echo $db->lastInsertId();
What happens here is that I push in my_table two new rows. The id of the table is auto-increment. Here, for the same user, I add two rows with a different varNumb.
The echoed value at the end will be equal to the id of the row where varNumb=1, which means not the id of the last row, but the id of the first row that was added in the last request.
An example.
$query_new = "INSERT INTO students(courseid, coursename) VALUES ('', ?)";
$query_new = $databaseConnection->prepare($query_new);
$query_new->bind_param('s', $_POST['coursename']);
$query_new->execute();
$course_id = $query_new->insert_id;
$query_new->close();
The code line $course_id = $query_new->insert_id; will display the ID of the last inserted row.
Hope this helps.
Try like this you can get the answer:
<?php
$con=mysqli_connect("localhost","root","","new");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_query($con,"INSERT INTO new values('nameuser','2015-09-12')");
// Print auto-generated id
echo "New record has id: " . mysqli_insert_id($con);
mysqli_close($con);
?>
Have a look at following links:
http://www.w3schools.com/php/func_mysqli_insert_id.asp
http://php.net/manual/en/function.mysql-insert-id.php
Also please have a note that this extension was deprecated in PHP 5.5 and removed in PHP 7.0
I found an answer in the above link http://php.net/manual/en/function.mysql-insert-id.php
The answer is:
mysql_query("INSERT INTO tablename (columnname) values ('$value')");
echo $Id=mysql_insert_id();
Try this... it worked for me!
$sql = "INSERT INTO tablename (row_name) VALUES('$row_value')";
if (mysqli_query($conn, $sql)) {
$last_id = mysqli_insert_id($conn);
$msg1 = "New record created successfully. Last inserted ID is: " . $last_id;
} else {
$msg_error = "Error: " . $sql . "<br>" . mysqli_error($conn);
}
Another possible answer will be:
When you define the table, with the columns and data it'll have. The column id can have the property AUTO_INCREMENT.
By this method, you don't have to worry about the id, it'll be made automatically.
For example (taken from w3schools )
CREATE TABLE Persons
(
ID int NOT NULL AUTO_INCREMENT,
LastName varchar(255) NOT NULL,
FirstName varchar(255),
Address varchar(255),
City varchar(255),
PRIMARY KEY (ID)
)
Hope this will be helpful for someone.
Edit: This is only the part where you define how to generate an automatic ID, to obtain it after created, the previous answers before are right.