从一个表中选择多个列并将数据插入到 PHP-MySQL 中不同数据库中的另一个表中

Question

I have to select data from a table in a database and insert them into another table in a different database. The below code returns the $select_query OK, but the $insert_query is not OK. Could you please correct the code and let me know your response?

$host1 = "localhost";
$user1 = "jackpot";
$pass1 = "jackpot";
$db1 = "pinkapple";
$host2 = "localhost";
$user2 = "jackpot";
$pass2 = "jackpot";
$db2 = "blueberry";

$mysql_connection1 = mysql_connect($host1, $user1, $pass1);
mysql_select_db($db1, $mysql_connection1) or die(mysql_error());
$select_query = mysql_query("SELECT field1, field2, field3 FROM tree WHERE date_entered > '2014-01-01 16:22:00'", $mysql_connection1);
$number = mysql_num_rows($select_query);

    if ($select_query) {
        echo "Select Query is OK <br>";
        echo $number ."<br>";
    } else {
        echo "Select Query is  not OK <br>";
    }

$mysql_connection2 = mysql_connect($host2, $user2, $pass2, true);

mysql_select_db($db2, $mysql_connection2) or die(mysql_error());

    while ($row = mysql_fetch_array($select_query)) {
    $field1 = $row['field1'];
    $field2 = $row['field2'];
    $field3 = $row['field3'];
    $insert_query = mysql_query("INSERT INTO jungle (desk1, chair1, table1) VALUES ('$field1', '$field2', '$field3')", $mysql_connection2);
        if ($insert_query) {
            echo "Insert Query is OK <br>";
        } else {
            echo "Insert Query is not OK <br>";
        }
}
mysql_close($mysql_connection1);
mysql_close($mysql_connection2);

below you can see the schematic image of the tables.

Answer 1

You need to establish the values?

"INSERT INTO jungle (desk1, chair1, table1) VALUES (value1, value2, value3)"

You should collect the data from the first select and then you can set the pertinent values in your insert statement

You can do that using the mysql_fetch statement

$mysql_connection2 = mysql_connect($host2, $user2, $pass2, true);
mysql_select_db($db2, $mysql_connection2) or die(mysql_error());

while($row = mysql_fetch_array($select_query))
{
    $field1 = $row['field1'];
    $field2 = $row['field2'];
    $field3 = $row['field3'];
    $insert_query = mysql_query("INSERT INTO jungle (desk1, chair1, table1) VALUES ('$field1', '$field2', '$field3')",$mysql_connection2);

    if ($insert_query) {
      echo "Insert Query is OK <br>";
    } else {
      echo "Insert Query is not OK <br>";
    }

}

This statement loops through your select statement and inserts a row for every returned result using your INSERT statement

But really you should also look into mysqli because mysql is depreciated. This is however the basic logic behind what you are doing

Answer 2

why not trying using mysql Trigger if you on the same mysql server, i though that more efective than using php, you can see this question for more information

http://stackoverflow.com/questions/7128173/mysql-trigger-to-insert-data-into-different-db

AND here

http://stackoverflow.com/questions/20441783/mysql-create-trigger-insert-content-into-another-database

Answer 3

I get the post is old, but I had to do the same thing and used mysqli as is suggested. Only thing I would caution is this example uses just a integer field and a short text field. If you have additional fields you should use prepared statements, however as I know the data here is fixed and contains no special characters so this will be fine for my needs.

<?php
include "database1.php"; // Using database connection file here conn variable is $connj
include "database2.php"; // Using database connection file here conn variable is $conn
                    

// select data and put values in array from first database
                $selectsql = mysqli_query($connj,"SELECT id,name FROM table1 
                ORDER BY id ASC");
                $result = mysqli_query($connj, $selectsql);
                
                while($row = mysqli_fetch_array($selectsql))
                {
                    $id = $row['id'];
                    $name = $row['name'];
                    
//replace query into second table in second database    
                    $replacesql = mysqli_query($conn, "REPLACE INTO table2 (name,id) VALUES ('$name','$username','$id');");
                
                    if($replacesql) 
                        {                   
                        echo "Replace Successful<br>";
                        }
                    else 
                            {
                            echo "replace error<br>";
                            }
                }
?>