从SQL COUNT()和GROUP BY获取PHP变量及其值

Question

我试图从“思维技能”(如分析、评估等)中提取变量并使之成为变量。并将它们的值设置为每个测试中的测试项数。我被困住了，所以任何帮助都将不胜感激。( SQL语句似乎工作得很好。)

我想要的示例：$analyzing = 7，$applying = 13，等等...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"] ;
}

Answer 1

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']

我不推荐这样做，但如果您真的希望将信息从数组中取出并放入局部变量中，您可以这样做

extract ($summary)