dqls67891
dqls67891
2017-09-28 03:29

在循环外面的未定义变量 - PHP

已采纳

I was wondering why I got an error saying Notice: Undefined variable: query_add_landdev_temp_payroll in C:\xampp\htdocs\op\ajax\ajaxLanddevNTPPayroll.php on line xx. for the past 2 weeks, it works perfect. When I tried to execute the code today, I got that kind of error. Here is my code:

$query_get_payroll = mysqli_query($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;
}
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3条回答

  • dongraobei6719 dongraobei6719 4年前

    If the query doesn't return any rows, mysqli_fetch_assoc() will return false the first time, so you'll never go into the loop, and never assign to the variable. If you then try to use the variable you get that warning.

    You can give it a default initial value before the loop, or you can change your test to:

    if (!empty($query_add_landdev_temp_payroll))
    

    BTW, the variable only contains the result of the INSERT in the last iteration of the loop. Maybe you should put the if block inside the loop, right after the INSERT?

    Also, there's no need to do INSERT in a loop. You can do the whole thing with a single query:

    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
    
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  • drzdu44226 drzdu44226 4年前

    Looks like you are using a variable inside a while command. So outside that command this variable will not define. Try use $query_add_landdev_temp_payroll = null; before while command. Hope it help you!

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  • dongshang6790 dongshang6790 4年前

    just update the if condition like this,

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