意外'=';在代码中,尝试在一个php文件中执行两个SQL查询
我试图从“userpaytoget”表中获取现有的“usercouponinhand”值,并将其作为“get”标记添加到收到的“id”值中。然后将其更新到相同的“userpaytoget”表“usercouponinhand”列 但不幸的是,我在意外'=';在代码中,尝试在一个php文件中执行两个SQL查询,php,mysql,Php,Mysql,我试图从“userpaytoget”表中获取现有的“usercouponinhand”值,并将其作为“get”标记添加到收到的“id”值中。然后将其更新到相同的“userpaytoget”表“usercouponinhand”列 但不幸的是,我在result=$conn->query($sql)行中看到了这个“Unexpected=”错误 守则如下: <?php include('usercoupondelete.php'); ?> <?php $servername = "
result=$conn->query($sql)行中看到了这个“Unexpected=”错误代码>
守则如下:
<?php include('usercoupondelete.php'); ?>
<?php
$servername = "localhost";
$dbusername = "root";
$dbpassword = "";
$dbname = "";
$mobile = $_SESSION['mobile'];
$date = date('M-d,Y H:i:s');
$date2 = date('M-d,Y');
$conn = new mysqli ($servername, $dbusername, $dbpassword, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM userpaytoget WHERE mobile = '$mobile' ";
result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$usercouponinhand = $row["usercouponinhand"];
$couponvalue = $_GET["id"];
$totalvalue = $couponvalue + $usercouponinhand ;
$sql2 = "UPDATE userpaytoget SET usercouponinhand = '$totalvalue', date = '$date', date2 = '$date2'
WHERE mobile = '$mobile'";
if ($conn->query($sql2) === TRUE) {
echo '<a href="usercoupondelete"></a>';
}
else {
echo "ERROR" . $sql2 . "<br>" . $conn->error;
}
}
} else {
echo "None";
}
$conn->close();
?>
非常感谢您的帮助。您缺少一个$
它应该是$result=$conn->query($sql)代码>result=$conn->query($sql)代码>您在结果之前错过了$您可以接受SQL注入。参数化您的查询。您能建议我如何避免这种情况吗使用PDO
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