douluanzhao6689 2017-09-08 18:47
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注意:类mysqli_result的对象无法转换为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;
?>
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1条回答 默认 最新

  • duanli3277 2017-09-08 18:52
    关注

    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
      }
    ?>
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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