如何使用where和like語句從表中選擇數據並回顯它

Question

我有一個關於如何使用are and like語句從數據庫表中選擇數據的問題。

$hy=mysql_query("select  (Total) AS firstterm 
                 FROM studentmark, subject 
                 where studentmark.student_id='$name' 
                   AND studentmark.YEAR='$ya' 
                   AND subject.code=studentmark.code    
                   AND studentmark.TERM='$term' LIKE 'F%'");

$hm=mysql_num_rows($hy);
$fetch=mysql_fetch_array($hy);

echo $fetch['firstterm'];

問題是未在表中選擇總分為89的術語中的LIKE'F％'（FIRST），而是在總分為73的術語中選擇了LIKE'S％'（SECOND）。 我有什么想念的嗎？

下表

TERM   | CODE |student_id|contAss20Asg|ClassWk10 |Test2nd10|YEAR |EXAM| TOTAL
FIRST  | AGR  | John     |  18        |5         |   7     |2011 | 59 |   89
SECOND |AGR2  |John      |  13        |6         |   4     |2011 | 40 |   73
THIRD  |AGR3  |John      |  18        |6         |   8     |2011 | 34 |   64
FIRST  |BIO   |John      |  12        |3         |   3     |2011 | 55 |   73
SECOND |BIO2  |John      |  14        |8         |   7     |2011 | 56 |   85
THIRD  |BIO3  |John      |  12        |8         |   8     |2011 | 42 |   70

我的代碼如下

<?php echo '</td><td>'?>
  <?php 
    if ($fetch['Total']==NULL){
echo 'missed';
}else 
    $hy=mysql_query("select  (Total) AS secondterm FROM studentmark, subject where studentmark.student_id='$name' AND studentmark.YEAR='$ya' AND subject.code=studentmark.code    AND studentmark.TERM='$term' LIKE 'S%'");
$hm=mysql_num_rows($hy);
$fetch=mysql_fetch_array($hy);
echo $fetch['secondterm'];
?>
<?php echo '</td><td>'?>
  <?php 
    if ($fetch['Total']==NULL){

}else 
    $hy=mysql_query("select  (Total) AS firstterm FROM studentmark, subject where studentmark.student_id='$name' AND studentmark.YEAR='$ya' AND subject.code=studentmark.code    AND studentmark.TERM='$term' LIKE 'F%'");
$hm=mysql_num_rows($hx);

$hm=mysql_num_rows($hy);
$row=mysql_fetch_array($hy);
echo $row['secondterm'];


?>

<?php echo '</td><td>'?>
  <?php 
    if ($fetch['Total']==NULL){
//echo 'missed';
}else 
    $hy=mysql_query("select  (Total) AS thirdterm FROM studentmark, subject where studentmark.student_id='$name' AND studentmark.YEAR='$ya' AND subject.code=studentmark.code    AND studentmark.TERM='$term'");
$hm=mysql_num_rows($hy);

$hm=mysql_num_rows($hy);
$fetch=mysql_fetch_array($hy);
$row=mysql_fetch_array($hy);

echo $fetch['firstterm']+ $row['secondterm'] + $fetch['thirdterm'];

?>

Answer 1

LIKE是一個比較運算符，您需要像使用'='，'>'等一樣將其用作運算符。

studentmark.TERM LIKE '%someval%';

希望這能回答你的問題。

Answer 2

like語法錯誤，因此請嘗試

SELECT TOTAL AS firstterm 
FROM studentmark, subject 
WHERE studentmark.student_id='$name' 
  AND subject.code=studentmark.code   
  AND studentmark.YEAR='$ya' 
  AND studentmark.TERM LIKE 'F%'");

最好也使用JOIN語法進行編碼

SELECT TOTAL AS firstterm 
FROM studentmark
   JOIN subject ON subject.code=studentmark.code
WHERE studentmark.student_id='$name' 
  AND studentmark.YEAR='$ya' 
  AND studentmark.TERM LIKE 'F%'");

同樣，如果他們發明了FOURTH術語，將會發生什么，這將不會產生正確的響應。 由於您的TERM專欄似乎已正式定義為FIRST ， SECOND ， THIRD因此最好完全放松LIKE並執行

SELECT TOTAL AS firstterm 
FROM studentmark
   JOIN subject ON subject.code=studentmark.code
WHERE studentmark.student_id='$name' 
  AND studentmark.YEAR='$ya' 
  AND studentmark.TERM = 'FIRST'");