Question

我有这段代码存储来自总列table_2013的总和并将其存储到$total2013。现在我想要回显这个总和，我想要回显total2013加上total2014之和的总和，而不是打印我得到的总和

“资源ID＃4”

            $t2013 ="SELECT SUM(total) FROM table_2013";
            $total2013 = (mysql_fetch_assoc($t2013));

            $t2014 ="SELECT SUM(total) FROM table_2014";
            $total2014 = (mysql_fetch_assoc($t2014));

            echo "$total2013";
            echo "$total2014";

            echo "$total2013 + $total2014";

Answer 1

您没有执行查询。执行它们 -

$t2013 ="SELECT SUM(total) FROM table_2013";
$total2013 = (mysql_fetch_assoc(mysql_query($t2013)));

$t2014 ="SELECT SUM(total) FROM table_2014";
$total2014 = (mysql_fetch_assoc(mysql_query($t2014)));

您还需要提供列名称 -

echo $total2013['SUM(total)'] + $total2014['SUM(total)'];

Answer 2

请试试这个：

$con=mysqli_connect("localhost","my_user","my_password","my_db");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }


$t2013 ="SELECT SUM(total) as total2013 FROM table_2013";
$result2013=mysqli_query($con,$t2013);
// Associative array
$row2013=mysqli_fetch_assoc($result2013);
$total2013 = $row['total2013'];

$t2014 ="SELECT SUM(total) as total2014  FROM table_2014";
$result2014=mysqli_query($con,$t2014);
// Associative array
$row2014=mysqli_fetch_assoc($result2014);
$total2014 = $row['total2014'];

                echo "total2013";
                echo "total2014";

                echo $toal = $total2013 +  $total2014;

Answer 3

您需要先运行查询：

$t2013 ="SELECT SUM(total) AS total FROM table_2013";
$q2013 = mysql_query($t2013);
$r2013 = mysql_fetch_assoc($q2013);
$total2013 = $r2013['total'];

$t2014 ="SELECT SUM(total) AS total FROM table_2014";
$q2014 = mysql_query($t2014);
$r2014 = mysql_fetch_assoc($q2014);
$total2014 = $r2014['total'];

echo "$total2013";
echo "$total2014";

echo "$total2013 + $total2014";

Answer 4

您有两个错误。 1.您的查询未执行。 2.你正在回显一个字符串（echo“$ total2013 + $ total2014”;）。它应该回应你放在“......”中的同样的东西，它不会总结这两个变量。你应该回应如下：

echo $total2013 + $total2014;

由于

Mysql为变量赋值并回显它

4 个答案: