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hi everyboy;

 
 1. mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, boolean given
 2. Warning: Invalid argument supplied fo   foreach()
 
i want to join quoteText with author name and email. Thanks
 
    here are the tables
 
    author table (id, name, email)
 
    user_quotes table(id, City, quoteText)
    
   
Here are my code
 
    <?php
    if ($_SERVER["REQUEST_METHOD"]=="POST") {
    $id_select = $_POST['id'];
    $action = "DELETE FROM user_quotes WHERE id = '$id_select'";
    $requet= mysqli_query($db_connection, $action);
    }
   $requete = "SELECT joke.id, quoteText, name, email FROM user_quotes INNER
                JOIN author ON authorID = author.id";
   $resultat = mysqli_query($db_connection, $requete);
 
   while ($rows = mysqli_fetch_assoc($resultat)) {
 
   $quotes []= array('id' =>$rows['id'], 'text' =>$rows['quoteText']
                'name'=>$rows['name'], 'email'=>$rows['email']);}
   ?>
  <body>
   <?php foreach ($quotes as $quote):?>
   <form action="?deletequote" method="post">
   
   <?php echo htmlspecialchars($quote['text'], ENT_QUOTES, 'UTF-8'); ?>
 
   <input type="hidden" name="id" value="<?php echo $quote['id'];?>">
  
   <?php echo htmlspecialchars($quote['name'], ENT_QUOTES, 'UTF-8'); ?>   
 
   <?php echo htmlspecialchars($quote['email'], ENT_QUOTES, 'UTF-8'); ?>
 
   <input type="submit" value="Delete">
   </p>
   </form>
   <?php endforeach; ?>

Try changing  joke.id  in the query to just  id

 

If the error still persists then change change  $resultat = mysqli_query($db_connection, $requete);  to be

 $resultat = mysqli_query($db_connection, $requete) or trigger_error('SQL Error: ' . $db_connection->error);
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