将mysql结果集php会话传递给其他页面。 [英] pass mysql resultset php session veriable to other page.
问题描述
Hello Friends。
我正在做我的学术项目。
我想在page1.php上的会话变量中获取表值并将其传递给page2.php。但是我收到了一个警告：
警告：main（）：无法在C：\ xampp \中获取mysqli_result第17行的htdocs \ AirTicket \ booking.php
;
我的page1.php代码：
session_start（）;
$ _SESSION ['res'];
？>
< ; html>
< body>
$ con = mysqli_connect（localhost，user，**** ，****）;
$ query =select * from ****;
$ _SESSION ['res'] = mysqli_query（$ con ，$ query）;
？>
Page2.php
< / body>
< / html>
< br $>
page2.php代码：
< html>
<头>
< / head>
< body>
session_start（）;
echo $ _SESSION [' res'];
foreach（$ _SESSION ['res']为$ row）
{
echoEmpId：;
echo $ row [EmpId];
echoSalary;
echo $ row [薪水];
打印（
）;
}
？>
< / body>
< / html>
在page2.php上执行后，它显示：
警告：main（）：无法在C：\ xampp \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\第17行的booking.php
;
Hello Friends.
I am doing my academic project.
I want to fetch table values in a session variable on page1.php and pass it to page2.php. but i got a warning :
"Warning: main(): Couldn't fetch mysqli_result in C:\xampp\htdocs\AirTicket\booking.php on line 17
";
my page1.php code:
session_start();
$_SESSION['res'];
?>
<html>
<body>
$con = mysqli_connect("localhost", "user", "****", "****");
$query = "select * from ****";
$_SESSION['res'] = mysqli_query($con, $query);
?>
Page2.php
</body>
</html>
page2.php Code:
<html>
<head>
</head>
<body>
session_start();
echo $_SESSION['res'];
foreach ($_SESSION['res'] as $row)
{
echo "EmpId: ";
echo $row["EmpId"];
echo "Salary";
echo $row["Salary"];
print("
");
}
?>
</body>
</html>
After execute it on page2.php it shows:
"Warning: main(): Couldn't fetch mysqli_result in C:\xampp\htdocs\AirTicket\booking.php on line 17
";
推荐答案
_SESSION ['res'];
？> ;
< html>
< body>
_SESSION['res'];
?>
<html>
<body>
con = mysqli_connect（localhost，user，****，*** *）;
con = mysqli_connect("localhost", "user", "****", "****");
query =select * from ****;
query = "select * from ****";
这篇关于将mysql结果集php会话传递给其他页面。的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
