PHP: Displaying table data using fetchall() [duplicate]

Question

Quick PHP/MySQL question:

I'm trying to display the data from a MySQL database table as an HTML table, and for some reason my code doubling the output of each piece of data.

This is my code:

$rowarray = $statement->fetchall();
print "<tr>\n";  
foreach ($rowarray as $row) {
    foreach ($row as $col) {
        print "\t<td>$col</td>\n";
    }
    print "</tr>\n";
}

My results look something similar to this:

User ID | User Name | First Name | Last Name

1 1 User Name User Name First Name First Name Last Name Last Name

Etc etc. You get the idea. Why this is happening? By the way, if I manually add the column information by referring to row[] subscripts 0-3, everything is displayed properly; it's only when I use the nested foreach statements that the data is duplicated.

jeroen · Accepted Answer · 2011-12-05 22:56:12Z

3

You are getting both a numeric and a name-indexed value back from PDO fetchall. To get just the numeric index:

$rowarray = $statement->fetchall(PDO::FETCH_NUM);

answered Dec 5, 2011 at 22:56

jeroen

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