php - 如何使用变量从php mysql中的两个表中减去

Question

我从我的第一个表 (income) 中得到这个总数，我想从另一个表 (exp) 查询中减去它。所有这些代码都在同一个页面上。那么我如何使用这些查询变量来减去和回显同一页面上但在另一个位置的结果呢？查询 1

<?php 
include("db.php");
$query = "SELECT source, SUM(income.inamount) FROM income"; 
$result = mysql_query($query) or die(mysql_error());

// Print out result
while($row = mysql_fetch_array($result)){
echo "Total Sales  = N". $row['SUM(inamount)'];
echo "<br />";
}
?>


<?php
include("db.php");
$query = "SELECT amount, SUM(exp.amount) FROM exp"; 

$result = mysql_query($query) or die(mysql_error());

// Print out result
while($row = mysql_fetch_array($result)){
echo "Total Expenditure  = N". $row['SUM(amount)'];
echo "<br />";
}
?>

Answer 1

检查数据类型是否设置为 int。

$myVar = $row['SUM(inamount)'] - $row['SUM(amount)'];
echo $myVar;

另一种方法是将这两个值分配给它们自己的变量，然后执行计算:

$totalSales = $row['SUM(inamount)'];
$expense = $row['SUM(amount)'];

$actual = $totalSales - $expense;

echo $actual;

使用 var_dump() 会告诉您 $row 变量中实际保存了哪些值。这将为您提供有关数据是否正确地从数据库提取到变量中的线索。

var_dump($row['SUM(inamount)']);
var_dump($row['SUM(amount)']);

要检查的另一件事是是否设置了实际变量，因此整体解决方案可能如下所示:

<?php 
include("db.php");

//query 1
$query_income = "SELECT source, SUM(income.inamount) FROM income"; 
$result_income = mysql_query($query_income) or die(mysql_error());
while($row = mysql_fetch_array($result_income)){
$inamount = $row['inamount'];
}

//query 2
$query_exp = "SELECT amount, SUM(exp.amount) FROM exp"; 
$result_exp = mysql_query($query_exp) or die(mysql_error());
while($row = mysql_fetch_array($result_exp)){
$amount = $row['amount'];
}

if(isset($inamount, $amount))  {
 $actual = $amount - $inamount;
 echo $actual // obviously you format $actual to your preferred     output
 }
 else {
   echo "Variable data not set";
 }
?>