当它在while循环之外时未定义的变量 - PHP

Question

我想知道为什么我收到错误提示 注意：未定义变量：第 xx 行的 C:\xampp\htdocs\op\ajax\ajaxLanddevNTPPayroll.php 中的 query_add_landdev_temp_payroll 。在过去的 2 周里，它运行良好。当我今天尝试执行代码时，我遇到了这种错误。这是我的代码：

$query_get_payroll = mysqli_query($new_conn, "SELECT ntp_with_ob_payroll_transaction.ntp_id, ntp_with_ob_payroll_transaction.allotment_code, ntp_with_ob_payroll_transaction.category_name, ntp_with_ob_payroll_transaction.block_number, ntp_with_ob_payroll_transaction.activity, ntp_with_ob_payroll_transaction.lot_number, ntp_with_ob_payroll_transaction.labor_cost, other_budget.quantity FROM ntp_with_ob_payroll_transaction JOIN other_budget ON ntp_with_ob_payroll_transaction.ob_id = other_budget.ob_id WHERE other_budget.transaction_id = $transaction_id AND other_budget.ob_number = '$ntp_number' AND is_payroll = 0");

while($row = mysqli_fetch_assoc($query_get_payroll)) {

    $ntp_id = $row['ntp_id'];
    $allotment_code = $row['allotment_code'];
    $category_name = $row['category_name'];
    $block_number = $row['block_number'];
    $activity = $row['activity'];
    $lot_number = $row['lot_number'];
    $labor_cost = $row['labor_cost']; //this is equivalent to unit cost
    $quantity = $row['quantity']; //this is equivalent to total percentage


    $query_add_landdev_temp_payroll = mysqli_query($new_conn, "INSERT INTO temp_payroll_landdev(ntp_id, userid, transaction_id, ntp_number, contractor_name, allotment_code, category_name, block_number, activity, lot_numbers, regular_labor, quantity) VALUES($ntp_id, $_SESSION[userid], $transaction_id, '$ntp_number', '$fullname', '$allotment_code', '$category_name', $block_number, '$activity', '$lot_number', '$labor_cost', '$quantity')");
}

if($query_add_landdev_temp_payroll) {

    echo 1;

    //database file name
    $database_file = $database.'.sql';
    $new_database_file = $new_database.'.sql';

    if(file_exists('backup/'.$new_database_file)) {

        unlink('backup/'.$new_database_file);

        //backup project database
        $command = "C:/xampp/mysql/bin/mysqldump --host=$new_host --user=$new_user --password=$new_pass $new_database > backup/$new_database_file";
        system($command);

    } else {

        //backup project database
        $command = "C:/xampp/mysql/bin/mysqldump --host=$new_host --user=$new_user --password=$new_pass $new_database > backup/$new_database_file";
        system($command);
    }
} else {

    echo 0;
}

Answer 1

如果查询没有返回任何行，mysqli_fetch_assoc() 将第一次返回false，因此您永远不会进入循环，也永远不会分配给变量。如果您随后尝试使用该变量，则会收到该警告。

你可以在循环之前给它一个默认的初始值，或者你可以把你的测试改成：

if (!empty($query_add_landdev_temp_payroll))

顺便说一句，该变量仅包含循环最后一次迭代中INSERT 的结果。也许您应该将if 块放在循环中，紧跟在INSERT 之后？

另外，没有必要在循环中执行INSERT。你可以用一个查询来完成整个事情：

INSERT INTO temp_payroll_landdev(ntp_id, userid, transaction_id, ntp_number, contractor_name, allotment_code, category_name, block_number, activity, lot_numbers, regular_labor, quantity)
SELECT ntp_with_ob_payroll_transaction.ntp_id, ntp_with_ob_payroll_transaction.allotment_code, ntp_with_ob_payroll_transaction.category_name, ntp_with_ob_payroll_transaction.block_number, ntp_with_ob_payroll_transaction.activity, ntp_with_ob_payroll_transaction.lot_number, ntp_with_ob_payroll_transaction.labor_cost, other_budget.quantity 
FROM ntp_with_ob_payroll_transaction 
JOIN other_budget ON ntp_with_ob_payroll_transaction.ob_id = other_budget.ob_id 
WHERE other_budget.transaction_id = $transaction_id AND other_budget.ob_number = '$ntp_number' AND is_payroll = 0

Answer 2

只需像这样更新if 条件，

if(isset($query_add_landdev_temp_payroll) && $query_add_landdev_temp_payroll) 

Answer 3

看起来您在while 命令中使用了一个变量。因此，在该命令之外，该变量将不会定义。尝试在while 命令之前使用$query_add_landdev_temp_payroll = null;。希望对你有帮助！