我的代码应该检查数据库以查看custID是否存在,如果存在,则更新信息。它没有,它需要将客户信息添加到数据库。
目前,当我使用我的代码时,每次在网站上下订单时,都会向数据库添加一个新的custID。
正在发生这些错误:
如果这些信息不充分或不清楚,我很乐意提供更多代码和解释。
//The information is passed through a session object from a previous page.
if (ISSET($_SESSION['fname'])) {
session_start();
$email = $_SESSION['email'];
$fname = $_SESSION['fname'];
$lname = $_SESSION['lname'];
$street = $_SESSION['street'];
$city = $_SESSION['city'];
$state = $_SESSION['state'];
$zip = $_SESSION['zip'];
$safeID = $_SESSION['safeID'];
$custID = $safeID / 507921;
}
include_once("Connection.php");
include_once("header.html");
//check if customer is already in database
$sql = "SELECT *
FROM bookcustomers
where custID = '$custID'";
$result = mysqli_query($link, $sql)
or die('SQL syntax error: ' . mysqli_error($link));
if (mysqli_num_rows($result) > 0 ) {
$sql = "UPDATE bookcustomers
set fname = '$fname',
lname = '$lname',
email = '$email',
street = '$street',
city = '$city',
state = '$state',
zip = '$zip'
WHERE custID = '$custID'";
$result = mysqli_query($link, $sql)
or die('SQL syntax error: ' . mysqli_error($link));
}
else {
$sql = "INSERT into bookcustomers (fname,
lname,
email,
street,
city,
state,
zip)
VALUES ('$fname',
'$lname',
'$email',
'$street',
'$city',
'$state',
'$zip')";
$result = mysqli_query($link, $sql)
or die('SQL syntax error: ' . mysqli_error($link));
$custID = mysqli_insert_id($link);
}
答案 0 :(得分:0)
你可以试试这个吗,移动session_start();
if (ISSET($_SESSION['fname'])) {
的顶部。
<?php
session_start();
if (ISSET($_SESSION['fname'])) {
$email = $_SESSION['email'];
$fname = $_SESSION['fname'];
$lname = $_SESSION['lname'];
$street = $_SESSION['street'];
$city = $_SESSION['city'];
$state = $_SESSION['state'];
$zip = $_SESSION['zip'];
$safeID = $_SESSION['safeID'];
$custID = $safeID / 507921;
}
include_once("Connection.php");
include_once("header.html");
//check if customer is already in database
$sql = "SELECT *
FROM bookcustomers
where custID = '$custID'";
$result = mysqli_query($link, $sql)
or die('SQL syntax error: ' . mysqli_error($link));
if (mysqli_num_rows($result) > 0 ) {
$sql = "UPDATE bookcustomers
set fname = '$fname',
lname = '$lname',
email = '$email',
street = '$street',
city = '$city',
state = '$state',
zip = '$zip'
WHERE custID = '$custID'";
$result = mysqli_query($link, $sql)
or die('SQL syntax error: ' . mysqli_error($link));
}
else {
$sql = "INSERT into bookcustomers (fname,
lname,
email,
street,
city,
state,
zip)
VALUES ('$fname',
'$lname',
'$email',
'$street',
'$city',
'$state',
'$zip')";
$result = mysqli_query($link, $sql)
or die('SQL syntax error: ' . mysqli_error($link));
$custID = mysqli_insert_id($link);
}
?>
答案 1 :(得分:0)
session_start()根据通过GET或POST请求传递的会话标识符创建会话或恢复当前会话,或通过cookie传递。
如果在php文件中更改top if
session_start();
if (ISSET($_SESSION['fname'])) {
$email = $_SESSION['email'];
$fname = $_SESSION['fname'];
$lname = $_SESSION['lname'];
$street = $_SESSION['street'];
$city = $_SESSION['city'];
$state = $_SESSION['state'];
$zip = $_SESSION['zip'];
$safeID = $_SESSION['safeID'];
$custID = $safeID / 507921;
}
include_once("Connection.php");
include_once("header.html");
只要您正确创建会话并在上一页设置fname会话变量,这将恢复您的会话。
如果你正确设置了值并将if子句更改为上面的那个,那么它应该可以工作。