从 SQL COUNT() 和 GROUP BY 获取 PHP 变量及其值

发布于 2024-09-08 09:54:17 字数 513 浏览 4 评论 0原文

我试图从测试中的“思维技能”（如分析、评估等）中提取并生成变量，并将其值设置为每个测试项目的数量。我被困住了，所以任何帮助将不胜感激。（SQL 语句似乎工作正常。）

我想要的示例：$analyzing = 7、$applying = 13 等...谢谢！

$sql = "SELECT  thinkskill        AS tskill
              , COUNT(thinkskill) AS counttskill
          FROM $c_keytable
      GROUP BY thinkskill
      ORDER BY thinkskill" ;
$result = mysql_query ( $sql ) ;

while ( $row = mysql_fetch_assoc ( $result ) )
{
   // Example: $analyzing = 7  --> 
   ${$row["tskill"]} = $row["counttskill"] ;
}

原文

I'm trying to extract and make variables out of the "thinking skills" (like analyzing, evaluating, etc.) from a test and set their value to the number of test items within each. I'm stuck, so any help would be appreciated. (The SQL statement seems to work fine.)

Example of what I want: $analyzing = 7, $applying = 13, etc.... Thanks!

$sql = "SELECT  thinkskill        AS tskill
              , COUNT(thinkskill) AS counttskill
          FROM $c_keytable
      GROUP BY thinkskill
      ORDER BY thinkskill" ;
$result = mysql_query ( $sql ) ;

while ( $row = mysql_fetch_assoc ( $result ) )
{
   // Example: $analyzing = 7  --> 
   ${$row["tskill"]} = $row["counttskill"] ;
}

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明天过后 2024-09-15 09:54:17

试试这个：

$summary = array ();
while ( $row = mysql_fetch_assoc ( $result ) )
{ 
   $summary[$row["tskill"]] = $row["counttskill"] ;
}
// now you can get the count for 'analyzing' with $summary['analyzing']

我不推荐它，但如果你真的想要将信息从数组中取出并放入局部变量中，你可以这样做

extract ($summary)

Try this:

$summary = array ();
while ( $row = mysql_fetch_assoc ( $result ) )
{ 
   $summary[$row["tskill"]] = $row["counttskill"] ;
}
// now you can get the count for 'analyzing' with $summary['analyzing']

I don't recommend it, but if you really want the info out of the array and into local variables, you can do