CODEWITHSUNDEEP
phparrays
CuriousMind
CuriousMind

Reputation: 34175

fill array value from database

Consider the following PHP array:

$config= array(
 "Plan" => array( "Type" => "dropdown", "Options" => "value1, value2, value3"),
);

Now, Instead of hardcoding these values(value1, value2 etc); I want these values(value1,value2 etc) to be fetched from a database. 

$rs = mysql_query('SELECT value from tbloptions');
$optvalue = mysql_fetch_array($rs);

I have been thinking about this from past half hours but I can't figure out on How to proceed. Can anyone help?

Upvotes: 1

Views: 2043

Answers (3)

Mike
Mike

Reputation: 21659

mysql_fetch_assoc and mysql_fetch_array return one row at a time, so you have to call the function until there are no more rows. For each row, grab the value and insert it into a numerically indexed array. The join function takes each value of that array, and concatenates them into a string, with the specified delimiter:

while ($row = mysql_fetch_assoc($rs)) {
    $opt[] = $row['value'];
}

$config['Options'] = implode(', ', $opt);

Upvotes: 2

Doug Kavendek
Doug Kavendek

Reputation: 3694

Look into mysql_fetch_assoc.

$result = mysql_query('SELECT value from tbloptions');
$rows = "";
while( $row = mysql_fetch_assoc( $result ) )
{
   $rows[] = $row;
}
mysql_free_result( $rows );

Upvotes: 0

TaylorOtwell
TaylorOtwell

Reputation: 7337

Looks like you just need to build an array of options and implode it...

foreach ($optvalue as $option)
{
    $options[] = $option['value']
}

$config['Options'] = implode(', ', $options);

Upvotes: 1

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