-1
我有一個查詢,我想COUNT行。問題是,當我嘗試計算完全相同的查詢時,它正在打印9,但是當我運行查詢並打印行時,它只有5個(因爲它應該是)在連接表上做一個COUNT
以下是查詢的工作原理和顯示5行:
$results_quiz = $pdo->query("SELECT
sr.statistic_ref_id,
sr.quiz_id,
sr.user_id,
sr.total_time,
qm.name AS q_name,
qm.category_id,
qm.subcategory_id,
sc.sub_category_id,
sc.sub_category_name AS quiz_type,
pl.time,
pl.points,
COUNT(qs.correct_count) AS count_correct
FROM pro_quiz_statistic_ref AS sr
JOIN pro_quiz_master qm ON qm.id = sr.quiz_id
JOIN pro_quiz_subcategory sc ON sc.sub_category_id = qm.subcategory_id
JOIN user_points_log pl ON pl.quiz_id = sr.quiz_id AND pl.user_id = '$get_id'
JOIN pro_quiz_statistic qs ON qs.statistic_ref_id = sr.statistic_ref_id
WHERE
sr.user_id = '$get_id' AND
qs.correct_count = '1'
GROUP BY
sr.statistic_ref_id,
qs.correct_count
ORDER BY qm.name ASC");
上面的代碼工作像它應該,但是當我用這個:
$count_results_quiz = $pdo->query("SELECT COUNT('
sr.statistic_ref_id,
sr.quiz_id,
sr.user_id,
sr.total_time,
qm.name AS q_name,
qm.category_id,
qm.subcategory_id,
sc.sub_category_id,
sc.sub_category_name AS quiz_type,
pl.time,
pl.points,
COUNT(qs.correct_count) AS count_correct')
FROM pro_quiz_statistic_ref AS sr
JOIN pro_quiz_master qm ON qm.id = sr.quiz_id
JOIN pro_quiz_subcategory sc ON sc.sub_category_id = qm.subcategory_id
JOIN user_points_log pl ON pl.quiz_id = sr.quiz_id AND pl.user_id = '$get_id'
JOIN pro_quiz_statistic qs ON qs.statistic_ref_id = sr.statistic_ref_id
WHERE
sr.user_id = '$get_id' AND
qs.correct_count = '1'
GROUP BY
sr.statistic_ref_id,
qs.correct_count
ORDER BY qm.name ASC")->fetchColumn();
它打印 「9」。有人知道爲什麼
請參閱http://meta.stackoverflow.com/questions/333952/why-should-i-provide-an-mcve-for-what-seems-to-me-to-be-a-very-simple- sql-query – Strawberry
,因爲有9行'sr.user_id ='$ get_id'AND qs.correct_count ='1'' for them is true –
但是,當我運行查詢時沒有顯示9行計數?當我刪除「AND qs.correct_count ='1'」時,它仍然在打印9,儘管表格中綁定了10行user_id – Kenneth