在每个 id php 的单独行上显示一个 id 的所有记录

Question

i have this table resulted from some inner joins

id_outfit   | name | article
----------------------------------------
   56        | one  | shoe.png
   56        | one  | dress.png
   56        | one  | bag.png
   54        | two  | shoe1.png
   54        | two  | dress2.png

Column 'article' contains images.

I try to display in my app all the articles from an id_outfit on a single row, but I don't know how to do that.

For example I try to display all articles that have id 56 on a row,all articles that have id 54 on the next row and so on.

I tried to do this with GROUP BY but for each id it displays only the first image(article).

Any suggestion is very helpful for me, thank you !

Answer 1

Here is a way do it using the PDO syntax:

<?php
try{
    $db = new PDO("mysql:host=localhost;dbname=yourdb", $user, $pass);
}
catch (Exception $e){
    //Handle your connection errors
}

//Request the things you need
$req = $db->query("SELECT id_outfit, article FROM your_table ORDER BY id_outfit DESC");

$tempId = null;

//start to loop through your data
while ($data = $req->fetch()){

  //if the outfit id is different than the last time we went through the loop:
  if($data["id_outfit"] !== $tempId){
    //close the previous row unless this is the first time through the loop
    echo $tempId === null ? "" : "</div>";
    //start a new row
    echo "<div class='row'>";
  }

  //now we display the image
  echo "<img src='" . $data["article"] . "' alt='youralt' >";
  //store the id we had for the next loop to check
  $tempId = $data["id_outfit"];
}

//close the last row
echo "</div>";

?>

This will output a div of class "row" for each different ID in your DB, contain all the images corresponding to that ID.

Then you can style those div using css to display it as a row, most likely using a flexbox:

.row{
  display: flex;
}

(you will need more css than that to make it actually pretty obviously.)