Question

我有一个连接到数据库的查询，它按设计工作，但我觉得我需要更多的代码。有人可以帮助我吗？

基本上我试图获得多个列的总和，然后为结果分配一个变量。变量必须每次都是唯一的，所以我可以在以后单独回显它们。

是否可以使用数组选择不同的列，然后使用相同的数组来生成变量？

我希望这一切都是我想要实现的目标。

这个例子只是我拥有的12个查询中的3个。

$sql = ("SELECT SUM(Case_closed) AS sum_closed FROM production WHERE Datum = '$actual_time' AND jaar = '$year' ");
    $result = mysqli_query($connect, $sql)  ;
            $Case_closed    = $result->fetch_object()->sum_closed;


$sql = ("SELECT SUM(Case_handled) AS sum_handled FROM production WHERE Datum = '$actual_time'  AND jaar = '$year' ");
    $result = mysqli_query($connect, $sql)  ;
        $Case_handled   = $result->fetch_object()->sum_handled;


$sql = ("SELECT SUM(swap) AS sum_swap FROM production WHERE Datum = '$actual_time'  AND jaar = '$year' ");
    $result = mysqli_query($connect, $sql)  ;
        $swap   = $result->fetch_object()->sum_swap;

Answer 1

您可以使用多个SUM（示例基于预准备语句）：

SELECT SUM(`Case_closed`) AS `sum_closed`, SUM(`Case_handled`) AS `sum_handled`, SUM(`swap`) AS `sum_swap` 
FROM `production` 
WHERE `Datum` = ?  
AND `jaar` = ?

Little Bobby说your script is at risk for SQL Injection Attacks.了解prepared的MySQLi语句。即使escaping the string也不安全！

Answer 2

通过跟随@little Bobby他的建议找到了解决方案来做一个准备好的声明，但是我没有参加专栏。它目前设置为($type)，但我希望(?)

function resultaten($type){

$connection = mysqli_connect("localhost","root","","dbname");

if(mysqli_connect_errno()){
echo mysqli_connect_error();
exit(); 
}   
$query = ("SELECT SUM($type) AS sum_gesloten FROM productie WHERE Datum=? AND jaar=?");

$time = time();
$actual_time = date('d-m-Y', $time);
$year = date("Y");
$stmt = mysqli_prepare($connection,$query);

if($stmt){
mysqli_stmt_bind_param($stmt, 'ss', $actual_time, $year);
mysqli_stmt_bind_result($stmt, $typeresult);
mysqli_stmt_execute($stmt);
mysqli_stmt_fetch($stmt);

if(!empty($type)){
    echo "$type $typeresult";       
}else{
    echo "No results!";
}   }else{

echo "Error creating statement object!";
}
}

$ Case_gesloten = resultaten（＆＃39; Case_gesloten＆＃39;）; echo $ Case_gesloten;

更轻松地访问我的mysqli查询

2 个答案: