douqianke7467
2017-11-23 11:10
浏览 159
已采纳

注意:未定义的索引:第13行的C:\ wamp64 \ www \ quiz emove epeated.php中的Ans

How to resolve this?

Notice: Undefined index: Ans in C:\wamp64\www\quizemoveepeated.php on line 13

My code:

require_once '../class.user.php';
$user_home = new USER();
$lstmtf = $user_home->runQuery("SELECT COUNT(Ans)
FROM answer AS a
LEFT JOIN students_records AS s ON a.Sr = s.Sr
WHERE s.Sr IS NULL");
    $lstmtf->execute();
$reg_rst = $lstmtf->fetch(PDO::FETCH_ASSOC);
$registered= $reg_rst['Ans'];
        echo $registered;

My table answer has column name Ans.

Actually, I want to count the numbers of rows which do not have values in students_records.

For Eg:

students_records

+----+-----+-----+ 
| Sr | SRN | ARN |
+----+-----+-----+
| 1  | ge  | aj  |
| 2  | ge  | bd  |
+----+-----+-----+  

answer

+----+-----+-----+
| Sr | SRN | ARN |
+----+-----+-----+
| 1  | ge  | aj  |
| 2  | ge  | aj  |
| 3  | ge  | ne  |
| 4  | ge  | bd  |
+----+-----+-----+ 

Here count should be 1. As the value "ne" in column ARN of table answer is no where in the rows of column ARN in table students_records.

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2条回答 默认 最新

  • duangou1933 2017-11-23 12:08
    已采纳

    You can get the desire output by following

    SELECT count(a.Ans) as ans
    FROM answer AS a
    LEFT JOIN students_records AS s ON a.ARN = s.ARN
    WHERE   s.ARN IS NULL
    

    Here is working example

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  • douyin2883 2017-11-23 11:12

    Use like this

    $lstmtf = $user_home->runQuery("SELECT COUNT(Ans) as Ans
    FROM answer AS a
    LEFT JOIN students_records AS s ON a.Sr = s.Sr
    WHERE s.Sr IS NULL");
    
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