如何在php和MySQL中提高查詢速度以便快速顯示

我有兩張表：候選人表存儲候選人信息和表candidate_subjects存儲結果。我想從兩個表查詢和網頁如何在php和MySQL中提高查詢速度以便快速顯示

CREATE TABLE candidate_subjects (
    id INT(10) PRIMARY KEY NOT NULL AUTO_INCREMENT, 
    candidate_id INT(11), 
    exam_type_id INT(10), 
    subject_id INT(10), 
    ca_score INT(11), 
    exam_score INT(6), 
    score_grade VARCHAR(10), 
    date_created VARCHAR(10), 
    date_modified TIMESTAMP 
); 

INSERT INTO `candidate_subjects` (`id`, `candidate_id`, `exam_type_id`, 
`subject_id`, `ca_score`, `exam_score`, `score_grade`, `date_created`, 
`date_modified`) VALUES 
    (1, 2, 1, 32, 22, 61, A, '2017-02-01', '2017-08-28 13:10:33'), 
    (2, 2, 1, 5, 21, 38, B, '2017-02-01', '2017-08-28 13:10:33'), 
    (3, 2, 1, 14, 21, 51, A, '2017-02-01', '2017-08-28 13:10:33'), 
    (4, 2, 1, 1, 19, 34, B, '2017-02-01', '2017-08-28 13:10:33'), 
    (5, 2, 1, 2, 23, 39, B, '2017-02-01', '2017-08-28 13:10:33'), 
    (6, 2, 1, 38, 20, 32, B, '2017-02-01', '2017-08-28 13:10:33'), 
    (7, 2, 1, 53, 24, 47, A, '2017-02-01', '2017-08-28 13:10:33'), 
    (8, 4, 1, 32, 19, 61, A, '2017-02-01', '2017-08-28 13:11:27'), 
    (9, 4, 1, 5, 22, 41, B, '2017-02-01', '2017-08-28 13:11:27'), 
    (10, 4, 1, 14, 20, 46, A, '2017-02-01', '2017-08-28 13:11:27'), 
    (11, 4, 1, 1, 23, 37, B, '2017-02-01', '2017-08-28 13:11:27'), 
    (12, 4, 1, 2, 21, 36, B, '2017-02-01', '2017-08-28 13:11:27'), 
    (13, 4, 1, 38, 22, 34, B, '2017-02-01', '2017-08-28 13:11:27'), 
    (14, 4, 1, 53, 24, 52, A, '2017-02-01', '2017-08-28 13:11:27'), 
    (15, 5, 1, 32, 20, 62, A, '2017-02-01', '2017-08-28 13:11:44'), 
    (16, 5, 1, 5, 22, 38, B, '2017-02-01', '2017-08-28 13:11:44'); 


CREATE TABLE candidates (
    id INT(11) PRIMARY KEY NOT NULL AUTO_INCREMENT, 
    exam_no VARCHAR(15), 
    surname VARCHAR(50), 
    other_names VARCHAR(100), 
    school_id INT(11), 
    registration_completed INT(11), 
    exam_scores_completed INT(5), 
    remark VARCHAR(10) 
); 

INSERT INTO candidates (id, exam_no, surname, other_names, school_id, 
registration_completed, exam_scores_completed, remark) VALUES 
(1, '1171052001', 'ABADO', 'MASENENGEN', 1052, 1, '1', 'RESIT'), 
(2, '1170938001', 'AGBA', 'NGUHER', 938, 1, '1', 'FAIL'), 
(3, '1170071001', 'ABEE', 'SESUUR', 71, 1, '1', 'PASS'), 
(4, '1170938002', 'AHEN', 'REBECCA DOOSUUN', 938, 1, '1', 'PASS');

我有得到考生的等級在candidate_subjects表中的主題功能上顯示的結果。

function getgradebycandsub($candidate_id, $sub){ 

     $s = "SELECT score_grade FROM `candidate_subjects` WHERE candidate_subjects.candidate_id='$candidate_id' AND candidate_subjects.subject_id='$sub'"; 

      $r = mysql_query($s); 
      $c = mysql_fetch_array($r); 
      return $c['score_grade']; 
    }

我有以下循環：

 $s = "SELECT id, exam_no, surname, other_names, remark FROM candidates ORDER BY surname"; 
    $r = mysql_query($s); 

echo "<tr><td align='center'>S/N</td><td align='center'>EXAM NO.</td><td align='center'>NAMES OF CANDIDATES</td><td align='center'>ENG</td><td align='center'>MATH</td><td align='center'>B.STD</td><td align='center'>C.ART</td><td align='center'>FREN</td><td align='center'>HAU</td><td align='center'>BST</td><td align='center'>ARAB</td><td align='center'>P.VOC</td><td align='center'>R.N.V</td><td align='center'>REMARK</td></tr>"; 
while ($candidates = mysql_fetch_array($r)) { 

     echo "<tr>"; 
     echo "<td height='30'>" . ++$sno . "</td>"; 
     echo "<td>" . $candidates['exam_no'] . "</td>"; 
     echo "<td>" . $candidates['surname'] . "&nbsp;&nbsp; " . $candidates['other_names'] . "</td>"; 
     echo "<td align='center'>".getgradebycandsub($candidates['id'], '1')."</td>"; 
     echo "<td align='center'>".getgradebycandsub($candidates['id'], '2')."</td>"; 
     echo "<td align='center'>".getgradebycandsub($candidates['id'], '5')."</td>"; 
     echo "<td align='center'>".getgradebycandsub($candidates['id'], '14')."</td>"; 
     echo "<td align='center'>".getgradebycandsub($candidates['id'], '22')."</td>"; 
     echo "<td align='center'>".getgradebycandsub($candidates['id'], '53')."</td>"; 
     echo "<td align='center'>".getgradebycandsub($candidates['id'], '32')."</td>"; 
     echo "<td align='center'>".getgradebycandsub($candidates['id'], '33')."</td>"; 
     echo "<td align='center'>".getgradebycandsub($candidates['id'], '38')."</td>"; 
     echo "<td align='center'>".getgradebycandsub($candidates['id'], '27')."</td>"; 
     echo "<td align='center'>".$candidates['remark']."</td>"; 
     echo "</tr>"; 
    }

我的挑戰是，它需要5個小時，顯示4000條記錄。 UPDATE
期望的輸出是S/No，考試編號，姓名，ENG，MATH，B.STD，C.ART，FREN，HAU，BST，ARAB，P.VOC，R.N.V，備註。這是候選人的詳細信息，如考試編號，姓名，成績在科目和備註

來源

2017-09-23 AMM

使用'join'來顯示錶格數據的 –

@RameshS這很棒。你能幫忙說明一下嗎？謝謝 – AMM

@AMM我已經發布瞭解釋'JOIN'使用的答案。讓我知道，如果你有任何疑問。 – Ravi

使用JOIN而不是function調用會提高性能。因爲在每個函數調用中，您都再次執行查詢，該查詢將轉到數據庫並獲取導致性能的結果。

SELECT score_grade, id, exam_no, surname, other_names, remark 
FROM candidate_subjects cs 
INNER JOIN candidates ON 
cs.candidate_id=c.id ORDER BY id, subject_id;

它會通過學生ID和主題返回結果順序。

來源

2017-09-23 05:31:56 Ravi

@AMM：這段代碼正確的使用了這個，謝謝Ravi –

如何在php和MySQL中提高查詢速度以便快速顯示

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