CODEWITHSUNDEEP
phpmysqli
Irfan Fathoni
Irfan Fathoni

Reputation: 13

Notice: Object of class mysqli_result could not be converted to int

I've searched all the questions that are almost identical to my case. but I am still confused .I just learned php programming and got an problem like this: Notice: Object of class mysqli_result could not be converted to int in ... please help me to solve the above problem.

    <?php
 $per_hal=10;
$jumlah_record="SELECT COUNT(*) from user";
$d=mysqli_query($link, $jumlah_record);
if($d == FALSE) { die(mysql_error()); } 
$halaman=ceil($d / $per_hal);  //error here
$page = (isset($_GET['page'])) ? (int)$_GET['page'] : 1;
$start = ($page - 1) * $per_hal;
?>

Upvotes: 1

Views: 2318

Answers (1)

Death-is-the-real-truth
Death-is-the-real-truth

Reputation: 72299

1.$d is the mysqli_result object. first get data from it and then use it.

2.don't mix mysql_* with mysqli_*.

<?php
  $per_hal=10;
  $jumlah_record="SELECT COUNT(*) as total_count from user";
  $d=mysqli_query($link, $jumlah_record);
  if($d) {
    $result = mysqli_fetch_assoc($d); //fetch record
    $halaman=ceil($result['total_count'] / $per_hal);  //error here
    $page = (isset($_GET['page'])) ? (int)$_GET['page'] : 1;
    $start = ($page - 1) * $per_hal;
  }else{
    die(mysqli_error($link));  // you used mysql_error() which is incorrect
  }
?>

Upvotes: 1

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