dpquu9206 2016-07-21 08:43 采纳率: 0%
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PHP MySQLi查询没有返回值

I've tried to get this to work in several ways, but I can't get this query to return the Cost value from the database to the $cost variable:

$query2 =  "SELECT Cost FROM 'item' WHERE Item = '$item'";
$cost=  $db->query($query2);

It seems to be empty when I try to echo it.

(The $item variable is selected from a dropdown list generated from the item-table in the mySQL-db. This works fine and if I echo the value from $item, it returns the name of the item as expected.)

Anybody sees what I'm doing wrong?

I could post my complete code if necessary, but I believe this explanation may be sufficient.

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  • dsfds4551 2016-07-21 08:50
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    You have to declared 'fetch_array' and Make reference this URL : http://php.net/manual/en/mysqli-result.fetch-array.php

    For Example:-

    $query = "SELECT Name, CountryCode FROM City ORDER by ID LIMIT 3";
    
    $result = $mysqli->query($query);
    
    $row = $result->fetch_array(MYSQLI_NUM);
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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