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php - 在 PHP 中显示 mysql SUM

转载 作者:行者123 更新时间:2023-11-29 22:41:21 24 4
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我正在尝试在 PHP 中输​​出 MySQL 查询,之前我一直使用此语句来打印 echo("<td>$line[meal_name]</td></tr>");在下面的代码中,我尝试这样做,但出现错误,并且我正在努力找出原因

$query = 'SELECT SUM(meal.calorific_output), SUM(activity.calorific_output)
FROM meal
INNER JOIN activity
ON meal.customer_id=activity.customer_id AND meal.day=activity.day
WHERE meal.customer_id =
"'. $_POST['customer_id'] .'"AND meal.day = "'. $day.'"';
$result = mysql_query($query) or die('Query failed: ' . mysql_error());

// Printing results
while ($line = mysql_fetch_array($result)) {

echo"<tr><td>$day</td>";
echo("<td>$line[SUM(meal.calorific_output)]</td>");
echo("<td>$line[SUM(meal.calorific_output)]</td></tr>");

最佳答案

您应该为您的列指定一个名称吗?就像SELECT SUM(meal.calorific_output) as sumMeal,

    $query = 'SELECT SUM(meal.calorific_output) as sumMeal, SUM(activity.calorific_output) as sumActivity
FROM meal
INNER JOIN activity
ON meal.customer_id=activity.customer_id AND meal.day=activity.day
WHERE meal.customer_id =
"'. $_POST['customer_id'] .'"AND meal.day = "'. $day.'"';
$result = mysql_query($query) or die('Query failed: ' . mysql_error());

// Printing results
while ($line = mysql_fetch_array($result)) {

echo"<tr><td>$day</td>";
echo("<td>$line["sumMeal"]</td>");
echo("<td>$line["sumActivity"]</td></tr>");

关于php - 在 PHP 中显示 mysql SUM,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/29329843/

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