Make a foreach array from mysql SELECT

Question

I have the following code:

$result = mysqli_query($con, "SELECT * FROM da_questions");

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

}

mysqli_close($con);

which selects everything from the table da_questions. Now this is in a file called config.php which file is required in all my pages of my website. Now I want to do the following on my external pages... let's say index.php

<?php foreach($results as $question): ?>

   <p><?php echo $question['question_title']; ?></p>
   <br />
   <p><?php echo $question['question_text']; ?></p>
   <br />

<?php endforeach; ?>

My file outlay is like shown below:

index.php

config/
  config.php

How can I do this?

not sure I understand. why not just $results = array(); while($row = mysqli_fetch_array($reult)) { $results[] = $row; } — Frank Conry
– Frank Conry, Commented Jan 10, 2014 at 5:24
@FrankConry Or even easier $results = $result->fetch_all(MYSQLI_ASSOC) — Phil
– Phil, Commented Jan 10, 2014 at 5:26
@FrankConry post in answer? So I can abide by stackoverflow policies... — Mitch Evans
– Mitch Evans, Commented Jan 10, 2014 at 5:27

user2936213 · Accepted Answer · 2014-01-10 05:32:12Z

1

Why not just

$results = array(); while($row = mysqli_fetch_array($reult)) { $results[] = $row; }

or as @Phil pointed out:

$results = $result->fetch_all(MYSQLI_ASSOC)

edited Jan 10, 2014 at 5:32

user2936213

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answered Jan 10, 2014 at 5:29

Frank Conry

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Comments

sanjeev · Accepted Answer · 2014-01-10 05:32:12Z

0

With your code

   $result = mysqli_query($con, "SELECT * FROM da_questions");
   $results = array();
   while($row = mysqli_fetch_array($result))
   {
    $results[] = $row;
   }

   mysqli_close($con);

in the index.php write as below

  include('config/config.php');

 <?php foreach($results as $question): ?>

    <p><?php echo $question['question_title']; ?></p>
    <br />
    <p><?php echo $question['question_text']; ?></p>
   <br />

   <?php endforeach; ?>

answered Jan 10, 2014 at 5:32

sanjeev

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