How to transfer several rows from one table to other table (SQL) - php

I have two tables TableA and TableB with colums id, login, pass. Also I have a php array: $array = [1,3,5]. How can I transfer rows from TableA to TableB where id equal each value of my array? id is autoincremented and must be unique.
In my head it looks like this :
INSERT INTO TableB (`login`, `pass`)
SELECT `login`, `pass` FROM TableA WHERE `id` = $array[0]
AND `id` = $array[1] AND `id` = $array[2];
But it does not work
Is there any chances to do it in cycle using WHILE?

You can use a statement such as this:
$ids = [1,3,5];
$query = "INSERT INTO `TableB` (`login`, `pass`) SELECT `login`,`pass` FROM `TableA` WHERE `id` IN (".implode(",", $ids).");";
Although: You should be using prepared statements and parameters for this, this answer is just meant to show the syntax.

it is easy let's say you are using mysqli:
$query= "SELECT * FROM TableA limit 3";
//takes first 3 rows from TableA
//$con being the connection string in config file
$result = mysqli_query($con,$query);
while($row = mysqli_fetch_array($result)){
//takes each row by row
$username = $row['username'];
$password = $row['password'];
//now that we initiated variables the sql query
$query1 = "INSERT INTO TableB (username, password) VALUES ('$username', '$password')";
mysqli_query($con,$query1);
}
Hope this helps
EDIT:
I just read that it should have an array so you should change your query to this:
for($i=0; $i<count(array), $i++){
$query= "SELECT * FROM TableA WHERE ID = '$array[i]'";
//$con being the connection string in config file
$result = mysqli_query($con,$query);
while($row = mysqli_fetch_array($result){
//takes each row by row
$username = $row['username'];
$password = $row['password'];
//now that we initiated variables the sql query
$query1 = "INSERT INTO TableB (username, password) VALUES ('$username', '$password')";
mysqli_query($con,$query1);
}
}

Related

Insert INTO with 3 tables

$sql3 = "INSERT INTO users_addresses (ua_user_id,ua_address_id) VALUES ('','')";
I am new in php and my hint is to link 2 tables id's in in another one called users_addresses.When a user is registered in my database i want the user_id and address_id to clone in users_addresses(ua_user_id,ua_address_id)
My tables
$sql = "INSERT INTO users (user_fname,user_mname,user_lname,user_login,user_email,user_phone)
VALUES ('{$_SESSION['userinfo']['fname']}', '{$_SESSION['userinfo']['mname']}', '{$_SESSION['userinfo']['lname']}', '{$_SESSION['userinfo']['login']}', '{$_SESSION['userinfo']['email']}', '{$_SESSION['userinfo']['phone']}')";
$sql1 = "INSERT INTO addresses (address_line_1,address_line_2,address_zip,address_city,address_province,address_country)
VALUES ('{$_SESSION['addressinfo']['adr1']}', '{$_SESSION['addressinfo']['adr2']}', '{$_SESSION['addressinfo']['zip']}', '{$_SESSION['addressinfo']['city']}', '{$_SESSION['addressinfo']['provinciq']}', '{$_SESSION['addressinfo']['durjava']}')";
$sql2 = "INSERT INTO notes (note_text)
VALUES ('{$_SESSION['noteinfo']['note']}')";
These are my others SQL codes for adding session's data in DB.
Just need get user_id from first sql. If you are using mysqli function, do this
// run your first sql: insert user
mysqli_query($con, $sql);
$user_id = mysqli_insert_id($con); // or mysqli::$insert_id
Next, you have $user_id variable with user id.
$sql1 = "INSERT INTO addresses (address_line_1,address_line_2,address_zip,address_city,address_province,address_country)
VALUES ($'{$_SESSION['addressinfo']['adr1']}', '{$_SESSION['addressinfo']['adr2']}', '{$_SESSION['addressinfo']['zip']}', '{$_SESSION['addressinfo']['city']}', '{$_SESSION['addressinfo']['provinciq']}', '{$_SESSION['addressinfo']['durjava']}')";
mysqli_query($con, $sql);
$address_id = mysqli_insert_id($con); // or mysqli::$insert_id
$sql3 = "INSERT INTO users_addresses (ua_user_id, ua_address_id) VALUES ($user_id, $address_id)";
mysqli_query($con, $sql);
Use mysqli_insert_id() to get the unique ID of the insert table, this example uses Procedural style:
<?php
include 'connection.php';
......
$InsertSQL = "INSERT INTO users (user_fname,user_mname,user_lname,user_login,user_email,user_phone)
VALUES ('{$_SESSION['userinfo']['fname']}',
'{$_SESSION['userinfo']['mname']}',
'{$_SESSION['userinfo']['lname']}',
'{$_SESSION['userinfo']['login']}',
'{$_SESSION['userinfo']['email']}',
'{$_SESSION['userinfo']['phone']}')";
$ResultSQL = mysqli_query($conn, $InsertSQL) or die(mysqli_error($conn)); // <-- execute your query
$UserID = mysqli_insert_id($conn); // <-- get the UserID
$InsertSQL = "INSERT INTO addresses (address_line_1,address_line_2,address_zip,address_city,address_province,address_country)
VALUES ('{$_SESSION['addressinfo']['adr1']}',
'{$_SESSION['addressinfo']['adr2']}',
'{$_SESSION['addressinfo']['zip']}',
'{$_SESSION['addressinfo']['city']}',
'{$_SESSION['addressinfo']['provinciq']}',
'{$_SESSION['addressinfo']['durjava']}')";
$ResultSQL = mysqli_query($conn, $InsertSQL) or die(mysqli_error($conn)); // <-- execute your query
$AddressID = mysqli_insert_id($conn); // <-- get the AddressID
$InsertSQL = "INSERT INTO user_addresses (ua_user_id,ua_address_id)
VALUES ($UserID,$AddressID)"; // <-- INSERT INTO user_address
$ResultSQL = mysqli_query($conn, $InsertSQL) or die(mysqli_error($conn)); // <-- execute your query
?>
You should also look into SQL Injection vulnerability, check out prepared statements.
Hope that helps.

Use returned ID from SCOPE_IDENTITY in new Query

Right now, this is what I have:
$query = "INSERT INTO COMMENTS VALUES ('$user', '$comment', '$star')";
mssql_query($query, $connection);
$commentIDQuery = "SELECT SCOPE_IDENTITY() AS ins_id";
$CI = mssql_query ($commentIDQuery, $connection);
$commentID = mssql_fetch_row($CI);
$idQuery = "SELECT recipeid FROM t_recipe WHERE recipename = '$recipeName'";
$RID = mssql_query($idQuery, $connection);
$recipeID = mssql_fetch_row($RID);
$rcQuery = "INSERT INTO COMMENT_RECIPE VALUES ('$commentID[0]', '$recipeID[0]')";
mssql_query($rcQuery, $connection);
So how would I get that ins_id?
It adds it to the first table, which is comments, but not the relation table.
Using sql server 2008
What about this......
$query = "DECLARE #NewID INT
INSERT INTO COMMENTS VALUES ('$user', '$comment', '$star');
SELECT #NewID = SCOPE_IDENTITY();
INSERT INTO COMMENTS_RECIPE VALUES (#NewID, '$recipeid')";
$stmt = sqlsrv_query($conn,$query);

add mysql row if userID do not exist in table

i'm sending a post request to this code:
$name = (string)$_POST['name'];
$id = (int)$_POST['id'];
$query = "INSERT INTO Users (name, userID) VALUES ('$name', '$id')";
$result = mysqli_query($link,$query);
Which works fine and it adds a row to the table. How do i check wether the userID all ready exist in one of the following rows?
Do it like this
$query = "SELECT COUNT(*) FROM Users WHERE userID = '$id'";
$result = mysqli_query($link,$query);
if ( mysqli_fetch_assoc($result) ) {
$message = "Already exists";
} else {
$query = "INSERT INTO Users (name, userID) VALUES ('$name', '$id')";
$result = mysqli_query($link,$query);
}
Try this
$name = (string)$_POST['name'];
$id = (int)$_POST['id'];
$res = mysqli_query($link, "SELECT * FROM Users WHERE userID = '$id' LIMIT 1 ");
if($row = mysqli_fetch_assoc($res))
{
echo "this user id is already exists";
}
else
{
$query = "INSERT INTO Users (name, userID) VALUES ('$name', '$id')";
$result = mysqli_query($link,$query);
echo "record inserted successfully ";
}
REMEMBER : always use LIMIT 1 when you trying to get exactly one result.
IF you have properly set 'id' as primary key or unique key in your table, you can use the modifier IGNORE in your query you don't get an error when you try to isert a duplicate.
Doing this will result in the row only being inserted if the value of the primary key wasn't already in the table.
$query = "INSERT IGNORE INTO Users (name, userID) VALUES ('$name', '$id')";
IF you haven't set a primary key in your table you will have to do a SELECT query to find out if a row with that id is already in your table.
Make the UserID an Unique Key.
If it already exists, your code will throw an error and the row will not be insterted.
alter table Users
add unique index Unique_user (userID (8000))
Before inserting the values to the table check whether the following user id exists in the table or not
You can do it in this way
$name = $_POST['name'];
$id = $_POST['id'];
$sql = "SELECT * FROM Users WHERE userID = '$id'";
$res= mysqli_query($sql);
$num = mysqli_num_rows($res);
if($num == 0)
{
$query2 = "INSERT INTO Users (name, userID) VALUES ('$name', '$id')";
$result2 = mysqli_query($query2);
echo "record inserted successfully ";
}
else
{
echo "Record Failed !!";
}

How to insert the same UUID in two rows of two different tables

I try next script:
// Insert data into mysql
$qry="INSERT INTO $tbl_name1 (ID, REFERENCE, CODE, NAME) VALUES (UUID(), '$REFERENCE', '$CODE', '$NAME')";
$result=mysql_query($qry);
$qry2="INSERT INTO $tbl_name2 (PRODUCT) VALUES ('$ID')"; <--- Here is a problem
$result=mysql_query($qry2)
I do not know how two insert the same UUID in two tables simultanoiusly. Please help me!
I will appreciate much your support!
DONE!!!
THE WORKING SCRIPT:
$q = "SELECT UUID() AS uid";
$res = mysql_query($q) or die('q error: '.mysql_error());
$row = mysql_fetch_assoc($res);
// Insert data into mysql
$qry="INSERT INTO $tbl_name1 (ID, REFERENCE, CODE, NAME) VALUES ('".$row['uid']."', '$REFERENCE', '$CODE', '$NAME')";
$result=mysql_query($qry) or die('err 034r '.mysql_error());
$qry2="INSERT INTO $tbl_name2 (PRODUCT) VALUES ('".$row['uid']."')";
$result=mysql_query($qry2) or die('gg2345 '.mysql_error());
Just do SELECT UUID() before you send the INSERTs and put the values into the statements in PHP. Something like this (untested):
$result = mysql_query("SELECT UUID() AS UUID") or die('SQL error: ' . mysql_error());
$row = mysql_fetch_assoc($result);
$UUID = $row["UUID"];
$qry="INSERT INTO $tbl_name1 (ID, REFERENCE, CODE, NAME) VALUES ('$UUID', '$REFERENCE', '$CODE', '$NAME')";
$result=mysql_query($qry);
$qry2="INSERT INTO $tbl_name2 (PRODUCT) VALUES ('$UUID ')"; <--- Here is a problem
$result=mysql_query($qry2)
Another way would be the use of a user-defined variable (see SQL Fiddle):
SET #UUID = (SELECT UUID() AS UUID);
INSERT INTO test1 VALUES(#UUID, "foo");
INSERT INTO test1 VALUES(#UUID, "bar");
Assuming the ID is the table Unique Index you could add before $qry2:
$ID = mysql_insert_id();

How to insert same data into two tables in mysql

Is this possible if I want to insert some data into two tables simultaneously?
But at table2 I'm just insert selected item, not like table1 which insert all data.
This the separate query:
$sql = "INSERT INTO table1(model, serial, date, time, qty) VALUES ('star', '0001', '2010-08-23', '13:49:02', '10')";
$sql2 = "INSERT INTO table2(model, date, qty) VALUES ('star', '2010-008-23', '10')";
Can I insert COUNT(model) at table2?
I have found some script, could I use this?
$sql = "INSERT INTO table1(model, serial, date, time, qty) VALUES ('star', '0001', '2010-08-23', '13:49:02', '10')";
$result = mysql_query($sql,$conn);
if(isset($model))
{
$model = mysql_insert_id($conn);
$sql2 = "INSERT INTO table2(model, date, qty) VALUES ('star', '2010-008-23', '10')";
$result = mysql_query($sql,$conn);
}
mysql_free_result($result);
The simple answer is no - there is no way to insert data into two tables in one command. Pretty sure your second chuck of script is not what you are looking for.
Generally problems like this are solved by ONE of these methods depending on your exact need:
Creating a view to represent the second table
Creating a trigger to do the insert into table2
Using transactions to ensure that either both inserts are successful or both are rolled back.
Create a stored procedure that does both inserts.
Hope this helps
//if you want to insert the same as first table
$qry = "INSERT INTO table (one, two, three) VALUES('$one','$two','$three')";
$result = #mysql_query($qry);
$qry2 = "INSERT INTO table2 (one,two, three) VVALUES('$one','$two','$three')";
$result = #mysql_query($qry2);
//or if you want to insert certain parts of table one
$qry = "INSERT INTO table (one, two, three) VALUES('$one','$two','$three')";
$result = #mysql_query($qry);
$qry2 = "INSERT INTO table2 (two) VALUES('$two')";
$result = #mysql_query($qry2);
//i know it looks too good to be right, but it works and you can keep adding query's just change the
"$qry"-number and number in #mysql_query($qry"")
its cant be done in one statment,
if the tables is create by innodb engine , you can use transaction to sure that the data insert to 2 tables
<?php
if(isset($_POST['register'])) {
$username = $_POST['username'];
$password = $_POST['password'];
$email = $_POST['email'];
$website = $_POST['website'];
if($username == NULL OR $password == NULL OR $email == NULL OR $website == NULL) {
$final_report2.= "ALERT - Please complete all fields!";
} else {
$create_chat_user = mysql_query("INSERT INTO `chat_members` (`id` , `name` , `pass`) VALUES('' , '$username' , '$password')");
$create_member = mysql_query("INSERT INTO `members` (`id`,`username`, `password`, `email`, `website`) VALUES ('','$username','$password','$email','$website')");
$final_report2.="<meta http-equiv='Refresh' content='0; URL=login.php'>";
}
}
?>
you can use something like this. it works.
In general, here's how you post data from one form into two tables:
<?php
$dbhost="server_name";
$dbuser="database_user_name";
$dbpass="database_password";
$dbname="database_name";
$con=mysql_connect($dbhost, $dbuser, $dbpass) or die('Error connecting to the database:' . mysql_error());
$mysql_select_db($dbname, $con);
$sql="INSERT INTO table1 (table1id, columnA, columnB)
VALUES (' ', '$_POST[columnA value]','$_POST[columnB value]')";
mysql_query($sql);
$lastid=mysql_insert_id();
$sql2=INSERT INTO table2 (table1id, table2id, columnA, columnB)
VALUES ($lastid, ' ', '$_POST[columnA value]','$_POST[columnB value]')";
//tableid1 & tableid2 are auto-incrementing primary keys
mysql_query($sql2);
mysql_close($con);
?>
//this example shows how to insert data from a form into multiples tables, I have not shown any security measures

Categories