php: display each element of an array as a string

Question

hi guys its my first day with php. i guess this is an easy one for you. can someone help me to display each element of my array $result[] as a string in $result_device[]?

<?php
  $mysqli = new mysqli("localhost", "root", "password", "db");      

    $sql = "SELECT erker, brunnen, stehlampe, computer, mediacenter FROM 
    devices";
    if (!$result = $mysqli->query($sql)) {
        echo "Query: " . $sql . "\n";
        echo "Errno: " . $mysqli->errno . "\n";
        echo "Error: " . $mysqli->error . "\n";
        exit;
    }

  $result = $result->fetch_row();
  $i = 0;
  foreach ($result as list ($result_device)) {
    if ($result_device[i]==0){
      $result_device[i] = "ausgeschaltet";
      }
    elseif ($result_device[i]==1){
      $result_device[i] = "eingeschaltet";
      }
      $i++;

  }
  echo $result_device[0];
  echo $result_device[1];
  echo $result_device[2];
  echo $result_device[3];
  echo $result_device[4];


  $result->close();
  $mysqli->close();
  ?>

thank you very much!

First print your result in side the foreach. print_r($result_device) and check your array and add your array here — Priyank
– Priyank, Commented Feb 14, 2018 at 4:32
You used a foreach loop just a few lines above. Have a look at the manual php.net/manual/en/control-structures.foreach.php and i think you will sort it out. — Andreas
– Andreas, Commented Feb 14, 2018 at 4:32
This doesn't seem right $result = $result->fetch_row(); Do you really want to over-write the object? — Tim Morton
– Tim Morton, Commented Feb 14, 2018 at 4:36
i is not going to use the declared $i value. In fact, you don't need to use a counter ($i) if you are just generating an indexed array. You can push values into your result array, by using $result_device[]='whatever'; What coding tutorials did you read to arrive at this code? — mickmackusa
– mickmackusa ♦, Commented Feb 14, 2018 at 4:38
its okay for me to overwrite $result, then i dont need another variable i guess or? — Kai H.
– Kai H., Commented Feb 14, 2018 at 4:41

Death-is-the-real-truth · Accepted Answer · 2018-02-14 04:48:45Z

2

You need to improve your code like below:-

<?php
$mysqli = new mysqli("localhost", "root", "password", "db");

if ($mysqli->connect_errno) {
    printf("Connect failed: %s\n", $mysqli->connect_error);
    exit();
}

$result_device = [];
if ($result = $mysqli->query("SELECT erker, brunnen, stehlampe, computer, mediacenter FROM devices")) {
    while ($row = $result->fetch_row()) {
       print_r($row); 
       // Based on oputput modify condition and assign values to $result_device
    }

    $result->close();
}

$mysqli->close();
?>

answered Feb 14, 2018 at 4:48

Death-is-the-real-truth

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3 Comments

Kai H. Over a year ago

ok thank you very much!. im trying to understand and i go on from there!

Kai H. Over a year ago

i spent some time with it but finally it works. thanks for your help

Death-is-the-real-truth Over a year ago

@KaiH. glad to help you and it's really good that you tried yourself.That's how you will learn

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