当我使用MySQLi时,当我这样做时,我会遇到一些数学错误:
<?php
include "mysqliconnect.php"; //Some connect details
$gQuery = $db->query("SELECT Waarde FROM Codes WHERE Code='".$db->real_escape_string($_POST['codej'])."'");
while($row1 = $gQuery->fetch_assoc()){
$gQuery2 = $db->query("SELECT balance FROM Users WHERE gnaam='Sombie'");
while($row2 = $gQuery2->fetch_assoc()) {
$money3 = $row1+$row2;
?>变量$money3不可用。
如果我确实回显了$money3,那么它将什么都不会显示。有人能告诉我它有什么问题吗?
谢谢!
发布于 2014-08-14 13:23:49
你所做的是完全错误的。使用下面的代码
<?php
include "mysqliconnect.php"; //Some connect details
$gQuery = $db->query("SELECT Waarde FROM Codes WHERE Code='".$db->real_escape_string($_POST['codej'])."'");
while($row1 = $gQuery->fetch_assoc()){
$gQuery2 = $db->query("SELECT balance FROM Users WHERE gnaam='Sombie'");
while($row2 = $gQuery2->fetch_assoc()) {
$money3 = $row1['Waarde']+$row2['balance'];
?>希望这对你有帮助
https://stackoverflow.com/questions/25308991
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