Php script to fetch database value in array to see in Drop down list

Question

I want to display a drop-down list which pull out database values. My actual code working fine, but here I have to provide the list in the array itself. What i required is, without editing the code, i needed to pull out values from database to this array("state1","state2"....); which is equal to something like this array("$states");

$sql = "SELECT * FROM `States`";
$result = $conn->query($sql);
$i=0;
while($row = $result->fetch_assoc()) {  
  $t=$row['statename'];
  $t[$i]=$row['statenname'];
  $i++;
}
$countryArr = array( "INDIA" => array("$t"));

I tried the above code but its listing only one value from database.

codisfy · Accepted Answer · 2016-02-21 06:30:58Z

1

Remove this line

$t=$row['statename'];

just keep it to be

$t[$i]=$row['statenname'];

answered Feb 21, 2016 at 6:30

codisfy

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3 Comments

Jackson Jacob Over a year ago

i did that, but here also one value shows in option list, i think the while loop end value is shown in array. that is the reason it shows the last value aas option. is there any way to get values inside array like this : array("$t[1]","$t[2]","$t[3]"));

codisfy Over a year ago

Why are you doing $countryArr = array( "INDIA" => array("$t")); and not $countryArr = array( "INDIA" => $t));

codisfy Over a year ago

Add the expected output array to your question, that will help

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