Question

我是一个MYSQL / PHP新手，我确信这是一个简单的问题。我正在尝试从一个表中计算几个问题和受访者的平均值，并使用该值更新Group表。

例如，表答案由表B中我想要的（name，group_id，TaskClarity1，TaskClarity2，TaskClarity3）组成（group_id，avg（TaskClarity1，TaskClarity2，TaskClarity3））。

这就是我所拥有的......

$avg_task_clarity_1 = mysql_query("SELECT AVG(TaskClarity1) WHERE gruppid = '$group_id'");
$avg_task_clarity_2 = mysql_query("SELECT AVG(TaskClarity2) WHERE gruppid = '$group_id'");
$avg_task_clarity_3 = mysql_query("SELECT AVG(TaskClarity3) WHERE gruppid = '$group_id'");
$avg_task_clarity = ($avg_task_clarity_1+$avg_task_clarity_2+$avg_task_clarity_3)/3;
$print_task_clarity_1" UPDATE results SET results.TaskClarity = '$avg_task_clarity'"; 

if (mysql_query($print_task_clarity_1)) { echo $print_task_clarity_1; } else { echo "Error TaskClarity1: " . mysql_error();

Answer 1

首先，mysql_query（）返回一个资源，然后您需要从中提取信息。您的查询不会指定任何表名（我将其命名为MyTable）。此外，您可以通过一个查询获得所有三个平均值。

以下是我将如何开始：

$table = "MyTable";
$sql = "SELECT AVG(TaskClarity1) AS avgClarity1,
               AVG(TaskClarity2) AS avgClarity2,
               AVG(TaskClarity3) AS avgClarity1
        FROM $table WHERE gruppid = '$group_id'";

$resource = mysql_query($sql);    //execute the query

if (! $resource = mysql_query($sql) ){
    echo "Error reading from table $table";
    die;
}

if (! mysql_num_rows($resource ) ){
    echo "No records found in $table";
}
else {
    $row = mysql_fetch_assoc($resource);   // fetch the first row
    $avg_task_clarity_1 = $row['avgClarity1'];
    $avg_task_clarity_2 = $row['avgClarity2'];
    $avg_task_clarity_3 = $row['avgClarity3'];
    $avg_task_clarity =
        ($avg_task_clarity_1+$avg_task_clarity_2+$avg_task_clarity_3)/3;

    //...
    // other stuff you want to do
}

请评论这是否有用，我会修改我的答案。

MySQL - 将平均值从一个表更新到另一个表

1 个答案: