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我花了兩個小時試圖解決錯誤所以最後問了這個問題。未定義的變量mysql-php
我的表是這樣的
tablename- schemeOverview
| slno |schemename | amount| date2 | date3 | date4 |collected |
|---------|-----------|-------|-------|-------|-------|----------|
| 1 | Cell | 1000 | DUE | DUE | | 0 |
| 2 | Cell | 1000 | PAID | PAID | | 0 |
| 3 | Cell | 1000 | DUE | DUE | DUE | 0 |
| 4 | Cell | 1000 | PAID | PAID | | 100 |
| 5 | Cell | 1000 | DUE | DUE | | 0 |
the table was created in phpmyadmin and the default value for collected is 0.
我試圖獲取指定schemename收集費爾德的價值。
這裏是代碼
echo "$scheme_name";
$sql12 = " SELECT collected FROM schemeOverview WHERE schemename = '".$scheme_name."'";
$reslut12 = mysqli_query($conn,$sql12);
var_dump($result12);
if(!$result12)
{
echo "Error fetching data: " .mysqli_error($conn);
}
else{
echo "Success";
}
while ($row12 = $result12->fetch_assoc())
{
var_dump($row12);
}
這是錯誤我得到
sapneshad
Notice: Undefined variable: result12 in E:\xampp\htdocs\insertPayment.php on line 195
NULL
Notice: Undefined variable: result12 in E:\xampp\htdocs\insertPayment.php on line 196
Error fetching data :
Notice: Undefined variable: result12 in E:\xampp\htdocs\insertPayment.php on line 203
Fatal error: Uncaught Error: Call to a member function fetch_assoc() on null in E:\xampp\htdocs\insertPayment.php:203 Stack trace: #0 {main} thrown in E:\xampp\htdocs\insertPayment.php on line 203