I am trying to store database value in PHP variable but id shows undefined index error

Question

I am trying to store the id from the database into an array using PHP but it shows an error.

<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "webmirchi";
$conn = new mysqli($servername, $username, $password, $dbname);

if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT id FROM jobinfo";
$result = $conn->query($sql);
$data = array();
while($row = $result->fetch_row())
{
$data[] = $row['id'];
}                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                         
$arrlength = count($data);
for($x = 0; $x < $arrlength; $x++) {

echo $data[$x];
echo "<br>";
}
$conn->close();
?>

Notice: Undefined index: id in E:\xampp\htdocs\demo1.php on line 19

Why have the `for` and `count`? Just `echo` in the `while`, or use a `foreach`. Try `var_dump($row)` that should show you what you actually have. — user3783243, Jul 15 '19 at 10:53

O.S.Kaya · Answer 1 · 2019-07-15T11:14:11.207

0

You have to target the index not column name 'id'. So this will work:

$sql = "SELECT id FROM jobinfo order by 1 asc";

if ($result = $conn->query($sql)) {
   $data = array();

   while($row = $result->fetch_row())
   {
      $data[] = $row[0];
   }

   $arrlength = count($data);
   for($x = 0; $x < $arrlength; $x++) {
      echo $data[$x];
      echo "<br>";
   }
}

edited Jul 15 '19 at 11:14

answered Jul 15 '19 at 10:53

O.S.Kaya

108
2
8

1

thank you. but it shows the output 6 7 2 3 1 like this it should show 1 2 3 6 7 – Iram Haware Jul 15 '19 at 10:58
@IramHaware That is a different issue. Use an `order by` in your query. – user3783243 Jul 15 '19 at 11:09
Did it solve your problem? @IramHaware – O.S.Kaya Jul 15 '19 at 11:41

I am trying to store database value in PHP variable but id shows undefined index error

1 Answers1