sqlite3查询多个条件的PHP

Question

I have a database that looks like this:

Owner   Vehicle   Color

Peter   Car       Black
Peter   Bike      Black
Peter   Bike      Red
Peter   Bike      Black
Marc    Car       Black
Marc    Car       Black
Marc    Bike      Red
Marc    Bike      Red

What I need is a query that goes through the whole database and eventually provide me an overview, such as:

• Peter has 1 black car
• Peter has 2 black bikes
• Peter has 1 red bike
• Marc has 2 black cars
• Marc has 2 red bikes

So for each owner I would want to have the count (and name) of the instances of vehicle+color.

What is the best way with PHP to approach this for a database with 900,000 rows. I have a list with the names of the owners, but how do I count the instances of vehicle+color in an effective way?

Answer 1

You'll need to use COUNT with a GROUP BY to aggregate per row. Since you want to know how many of each vehicle and color an owner has, you need to group by all three columns:

SELECT owner, vehicle, color, COUNT(*) as vehicle_count
FROM mytable
GROUP BY owner, vehicle, color

Result:

+-------+---------+-------+---------------+
| owner | vehicle | color | vehicle_count |
+-------+---------+-------+---------------+
| Marc  | Bike    | Red   |             2 |
| Marc  | Car     | Black |             2 |
| Peter | Bike    | Black |             2 |
| Peter | Bike    | Red   |             1 |
| Peter | Car     | Black |             1 |
+-------+---------+-------+---------------+

After executing the query, these rows can then be read out by your PHP code like

echo $row["owner"]." has ".$["vehicle_count"]." ".$vehicles;

or processed however you want to present the information.

Answer 2

you can use the following mysqli code.. don't forget to change the "username", "password", "database "in connection and tablename in $get_owner and $get_num.

search for the owner and the result will be shown.

code:-

<form action="" method="post">
  <input type="text" placeholder="Owner name" name="owner" />
  <input type="submit" value="search" name="search"/>

</form>

<?php

$conn = mysqli_connect ("localhost", "username", "password", "database");

if (mysqli_connect_errno()) {
  echo "Error ".mysqli_connect_errno();
}

if(isset($_POST['search'])) {
  $owner = $_POST['owner'];
    $get_owner = "SELECT * FROM `tablename` WHERE Owner='$owner'";
  $run = mysqli_query($conn, $get_owner);

while ($row =  mysqli_fetch_array($run)) {

    $vehical = $row['Vehicle'];
    $color = $row['Color'];

    $get_num = "SELECT `tablename` FROM `efgh` WHERE Vehicle='$vehical' and Color='$color'";
    $run_num = mysqli_query($conn ,$get_num);
    $chekc = mysqli_num_rows($run_num);

    echo $owner.' has '.$chekc.' '.$color.' '.$vehical.'<br>';
  }
}

 ?>