注意：未定義的索引：在162行的C：\ wamp \ www \ lamp \ main.php中的enrollment_number

-1

這是我的代碼。它已連接到數據庫，表已成功創建併成功輸入值。在問題中提到的錯誤顯示，當我想查看錶。錯誤是爲登記號碼。和sub1。注意：未定義的索引：在162行的C： wamp www lamp main.php中的enrollment_number

$student_academic_details = "CREATE TABLE student_academic_details 
    (
     enrollment_number varchar(30), 
     sub1 int(2), 
     sub2 int(2), 
     sub3 int(2) 
    )"; 

echo"<br><br>"; 

if (mysqli_query($conn, $student_academic_details)) 
    { 
     echo "Table student_academic_details created successfully"; 
    } 
else 
    { 
     echo "Error creating table: " . mysqli_error($conn); 
    } 

echo"<br><br>"; 

$row2 = "INSERT INTO student_academic_details VALUES ('009/015', '87', '78', '85')"; 

if (mysqli_query($conn, $row2)) 
    { 
     echo "New record created successfully"; 
    } 
else 
    { 
     echo "Error: " . $row2 . "<br>" . mysqli_error($conn); 
    } 



$show = "select * FROM student_personal_details"; 
$result = mysqli_query($conn, $show); 
if (mysqli_num_rows($result) > 0) 
    { 

    while($row = mysqli_fetch_assoc($result)) 
     { 
     echo "id: " . $row["enrollment_number"]. " - sub1: " . $row["sub1"]. "<br>"; 
     } 
    } 
else 
    { 
    echo "0 results"; 
    }

來源

2017-05-28 Vandita Srivastava

enrollment_number列不存在於您的表中只是檢查拼寫或字 –

在這個表student_personal_details你有那個領域？或者錯誤地將表名從student_academic_details打到student_personal_details – Exprator

非常感謝......我不敢相信我自己的愚蠢 –

您創建名爲student_academic_details的表，但從student_personal_details中進行選擇。這些表中可能沒有字段。

來源

2017-05-28 16:28:51

注意：未定義的索引：在162行的C：\ wamp \ www \ lamp \ main.php中的enrollment_number

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