Get mysql results according to an array

Question

I want to get mysql results according to an array.

I have this code but it brings one result and repeat it and i want all results

$connectdb  =   mysql_connect('localhost','root','') or die('nonno');
$selectdb   =   mysql_select_db('test',$connectdb) or die('fofofo');
$se_right   =   mysql_query("select * from ads ") or die(mysql_error());
$row        =   mysql_fetch_object($se_right);
$array = array(
                "id"    =>  $row->id,
                "name"  =>  $row->adsurl
            );
$se_ridght = mysql_query("select * from ads ") or die(mysql_error());

while($roww = mysql_fetch_object($se_ridght))
{
    foreach($array as $key =>$results)
    {
        echo $key.':'.$results.'<br />';
    }
}

what do you mean by accoeding to an array?

Muhammad Raheel
– Muhammad Raheel

2013-01-12 06:38:37 +00:00
Commented Jan 12, 2013 at 6:38 — Muhammad Raheel
– Muhammad Raheel, Commented Jan 12, 2013 at 6:38

웃웃웃웃웃 · Accepted Answer · 2013-01-12 06:36:29Z

3

Try this.

$connectdb=mysql_connect('localhost','root','') or die('nonno');
$selectdb= mysql_select_db('test',$connectdb) or die('fofofo');
$se_right = mysql_query("select * from ads ") or die(mysql_error());
$row = mysql_fetch_object($se_right);
$array = array(
"id" => $row->id,
"name" =>$row->adsurl
);
$index = 0;
while($roww = mysql_fetch_assoc($se_ridght))
{
 $yourArray[$index] = $roww;
 $index++;
}
}

print_r($yourArray);

edited Jan 12, 2013 at 6:36

answered Jan 12, 2013 at 6:19

웃웃웃웃웃

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2 Comments

Flenston F Over a year ago

it returns the first result of mysql table iwant it to returns it all

웃웃웃웃웃 Over a year ago

I have made some edits to the post...I hope you will get the result now @Flenston F...

PassKit · Accepted Answer · 2013-01-12 06:26:32Z

1

Your issue seems to be because you are setting $array only once and outside of the while loop. The following code builds the $array in the while loop. After the loop is finished, the $array variable is available for use later in your project.

$connectdb=mysql_connect('localhost','root','') or die('nonno');
$selectdb= mysql_select_db('test',$connectdb) or die('fofofo');

$array = array();

$se_ridght = mysql_query("select * from ads ") or die(mysql_error());

while($roww = mysql_fetch_object($se_ridght)){
  $array[] = array("id" => $roww->id, "name" => $roww->adsurl);
  echo $roww->id . ':' . $roww->adsurl . '<br />';
}

edited Jan 12, 2013 at 6:26

answered Jan 12, 2013 at 6:20

PassKit

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2 Comments

Flenston F Over a year ago

returns Parse error: syntax error, unexpected '=>' (T_DOUBLE_ARROW) in C:\xampp\htdocs\~34~ads2.php on line 14

PassKit Over a year ago

Try again with $array[] = array("id" => $roww->id, "name" => $roww->adsurl);

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