2014-10-27 86 views
-5

我无法回应我的mysql查询的结果。 PHP代码是:php-无法打印mysqli_result

$getc = "SELECT credits FROM members WHERE Phone ='".$_SESSION['Phone']."'"; 
$credits = mysqli_query($dbc,$getc); 
echo "YOU HAVE $credits CREDITS LEFT"; 

但上面的显示以下错误: 开捕致命错误:类mysqli_result的对象不能转换成字符串file.php上线204

有什么不对?

+0

什么'$ cr'的价值?你用'$ credits'做什么? – andrewsi 2014-10-27 19:08:45

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旁注:确保'session_start();'被加载。在打开'<?php'标签后立即在文件顶部添加错误报告 'error_reporting(E_ALL); ini_set('display_errors',1);'和'或死(mysqli_error($ dbc))'到'mysqli_query()' – 2014-10-27 19:08:57

+0

抱歉,问题编辑 – 2014-10-27 19:10:07

回答

1

你只需做一个查询,而不是获取的结果,mysqli_query()返回mysqli_result而不是场

$getc = "SELECT credits FROM members WHERE Phone ='".$_SESSION['Phone']."'"; 
$result = mysqli_query($dbc,$getc); 
$row = mysqli_fetch_assoc($result); 
echo "You have {$row['credits']} left!";