[英]Display mysql SUM in PHP
我試圖在PHP中輸出MySQL查詢,以前我一直在使用此語句來打印echo("<td>$line[meal_name]</td></tr>");
在下面的代碼中,我嘗試以這種方式進行操作,但是我遇到了錯誤,並且正在努力找出原因
$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>");
您應該為列指定名稱嗎? 例如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>");
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