Display values from php mysql array separately

Question

What im trying to do is to save vales of "option" from all records in array to be able to echo them out later. I don't know how to display them seperatly. Feel stuck. Any sugestions?

My code:

$result = $mysqli->query("SELECT option FROM `myoptions` WHERE id IN (1, 7, 18, 24);");
$row = array();
while($row = $result->fetch_array())  { 
    echo $row[] = $row["option"];
}

echo $option1 = $row[0];
echo $option2 = $row[1]; 
echo $option3 = $row[2];
echo $option4 = $row[3];

i want to use my 'option' variables in a different part of the page and my problem is that variables are empty. — Nita
– Nita, Commented Aug 15, 2016 at 7:19
Can you show us the code, where you are trying to use it, and getting the empty variables? — Mubashar Abbas
– Mubashar Abbas, Commented Aug 15, 2016 at 7:21

ScaisEdge · Accepted Answer · 2016-08-15 07:19:33Z

1

YOu should store the value in an array

$cnt = 0;
while($row = $result->fetch_array())  { 

  $for_future_use[$cnt] = $row["option"]; 
  $cnt++; 
}

and then use later

foreach($for_future_use as $key=>$value) {
   echo $value;
}

answered Aug 15, 2016 at 7:19

ScaisEdge

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Comments

Dave · Accepted Answer · 2016-08-15 07:29:57Z

1

try this,

$my_array = array("a" => "Cat","b" => "Dog", "c" => "Horse");
$i = 1;
$option = "option";
foreach($my_array as $val)
{
    ${$option.$i} = $val;
    $i++;
}

echo "\$option1  =>  ".$option1."  \$option2 =>  ".$option2."  \$option3  =>  ".$option3;

DEMO : https://3v4l.org/OhQSm

i hope it will be helpful.

answered Aug 15, 2016 at 7:29

Dave

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