loop through database and show in table

Question

I am trying to loop though my users database to show each username in the table in their own row. I can only get it to show one user but loops through this the number of rows there are. Code is below

<?php
require_once ('../login/connection.php');
include ('functions.php');

$query = "SELECT * FROM users";
$results=mysql_query($query);
$row_count=mysql_num_rows($results);
$row_users = mysql_fetch_array($results);

echo "<table>";
    for ($i=0; $i<$row_count; $i++)
    {
    echo "<table><tr><td>".($row_users['email'])."</td></tr>";
    }
    echo "</table>";
?>

Thanks

Paul Dixon · Accepted Answer · 2009-05-11 16:31:09Z

15

mysql_fetch_array fetches a single row - you typically use it in a while loop to eat all the rows in the result set, e.g.

echo "<table>";

while ($row_users = mysql_fetch_array($results)) {
    //output a row here
    echo "<tr><td>".($row_users['email'])."</td></tr>";
}

echo "</table>";

answered May 11, 2009 at 16:31

Paul Dixon

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1 Comment

Syno Over a year ago

Outdated answer, make sure to use MySQLi instead of MySQL php.net/manual/en/mysqli-result.fetch-array.php

Parrots · Accepted Answer · 2009-05-11 16:31:26Z

4

You're only fetching one row:

$row_users = mysql_fetch_array($results);

You're never calling a fetch again so you're not really looping through anything.

I'd suggest changing your loop to the following:

echo "<table>";
while ($row = mysql_fetch_array($results)) {
    echo "<tr><td>".($row['email'])."</td></tr>";
}
echo "</table>";

The while loop will loop through the results and assign a row to $row, until you run out of rows. Plus no need to deal with getting the count of results at that point. This is the "usual" way to loop through results from a DB in php.

answered May 11, 2009 at 16:31

Parrots

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Comments

JerryB · Accepted Answer · 2017-06-22 13:14:44Z

4

In the new MYSQLI, I use this coding

$query = "SELECT * FROM users";
$results=mysqli_query($con,$query);
$row_count=mysqli_num_rows($results);

echo "<table>";
while ($row = mysqli_fetch_array($results)) {
echo "<tr><td>".($row['id'])."</td></tr>";
}
echo "</table>";

mysqli_query($con,$query); 
mysqli_close($con);

answered Jun 22, 2017 at 13:14

JerryB

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Comments

Marc W · Accepted Answer · 2009-05-11 16:30:17Z

0

Your problem is in this line:

echo "<table><tr><td>".($row_users['email'])."</td></tr>";

You're echoing out a <table> tag again. Remove that so your code looks like this:

echo "<table>";
    for ($i=0; $i<$row_count; $i++)
    {
    echo "<tr><td>".($row_users['email'])."</td></tr>";
    }
echo "</table>";

answered May 11, 2009 at 16:30

Marc W

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Comments

Nick Presta · Accepted Answer · 2009-05-11 16:31:49Z

0

You don't need to output a table within a table the way you're doing it.

$result = mysql_query("SELECT `email` FROM `users`");
$num_emails = mysql_num_rows($result);

echo "<table><caption>There are/is $num_emails email(s)</caption>";
while ($row = mysql_fetch_assoc($result)) {
    echo "<tr><td>{$row['email']}</td></tr>";
}
echo '</table>';

Something like that should work.

answered May 11, 2009 at 16:31

Nick Presta

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