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Editorial Team
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Asked: 2026-06-09T01:26:18+00:00

I want to take my users from my table and display them with a

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I want to take my users from my table and display them with a html table.i want their Name Avatar displayed. All of this information is in the table.
I want to display this in a table with 4 cols by as many rows needed.

This is my full code as asked for:

<html><head><title>MySQL Table Viewer</title></head><body>
<?php
$db_host = '';
$db_user = '';
$database = '';
$db_pwd = '';
$table = '';
if (!mysql_connect($db_host, $db_user, $db_pwd))
die("Can't connect to database");
if (!mysql_select_db($database))
die("Can't select database");
// sending query
$result = mysql_query("SELECT * FROM {$table} WHERE livedj=1");
if (!$result) {
die("Query to show fields from table failed");
}
$max_rows = 6;
$max_cols = 4;
echo "<table>";
$rows = 1;
while ($rows <= $max_rows)
{
echo "<tr>";
$cols = 1; // reset columns to 1 for each row
while ($cols <= $max_cols)
{
     echo "<td>". $row['name'] . "<br />
     <img src=\"http://myurl/path/to/pics/" . $row['avatar'] . "\" height=\"150\" width=\"150\" /></td>";
     $cols++; // column counter
}
echo "</tr>";
$rows++;  //row counter
}
echo "</table>";
?>
</body></html>

This only give out a page with the correct table made but it is not displaying any information from mysql…

and yes I have deleted the database and passwords settings here and my copy has the correct settings.

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1 Answer

  1. Editorial Team

    I hope you meant to say your SQL table has 6 data fields (7 with the ‘livedj’ you seem to be using in your WHERE clause), and as many data entries as needed.

    I am also assuming you want your output to look something like this:

    +------+--------+-----------------------------+
| Name | Avatar | Other stuff (email,fb,...)  |
+------+--------+-----------------------------+
| John | :-)    | blah                        |
| ...  |  ...   | ...                         |
+------+--------+-----------------------------+

    Then that code would be:

    <?php
$result = mysql_query("SELECT name, address, twitter, email, facebook, avatar FROM {$table} WHERE livedj=1 ");
if (!$result) {
die("Query to show fields from table failed");
}
echo "<table border='1'>";

while($row = mysql_fetch_array($result)){
  echo "<tr><td>".$result["name"]."</td><td>"
  ."<img alt='avatar' src='path/to/img/folder/".$result["avatar"]." /></td><td>"
  .$result["address"]."</td><td>".$result["email"]."</td><td>"
  .$result["twitter"]."</td><td>".$result["facebook"]"</td></tr>\n";
}
echo "</table>
?>
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