douxun4860 2015-05-12 07:57
浏览 120

我们不能在PHP中将$ result称为变量

i have query in my php code

$dbc = mysqli_connect('localhost', 'root','', 'delivery')
            or die('Error connecting to MySQL server.');

        $count = "SELECT MAX(id_pelanggan) FROM pelanggan";

        $result = mysqli_query($dbc, $count)
            or die('Error select query');

        if (empty($result)) {
            $id_pelanggan = 1;
        }
        else {
            $id_pelanggan = $result + 1;
        }

and the result was

Object of class mysqli_result could not be converted to int in C:\xampp\htdocs\delivery\addcustomer.php

whereas in mysql id_pelanggan datatype is int. can anyone help me to make it works?

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2条回答 默认 最新

  • douyong1886 2015-05-12 07:59
    关注
    • You try to assign your query to $id_pelanggan variable instead of the value from your query.
    • Fetch the result using *_fetch array() of your query
    • I've changed your if() condition because your $result won't be empty no matter what happens. The number of result maybe.

    Put this and replace your if else condition:

    if(mysqli_num_rows($result) == 0){ /* IF FOUND 0 RESULT */
      $id_pelanggan = 1;
    }
    
    else {
    
      while($row = mysqli_fetch_array($result)){
        $maxid = $row["id_pelanggan"];
      }
    
      $id_pelanggan = $maxid + 1;
    
    } /* END OF ELSE */
    
    评论

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