我的输入是在xml文件下面
<?php
require "Constants.php";
if($_SERVER['REQUEST_METHOD'] == 'POST')
{
$DefaultId = 0;
$image = $_POST['image'];
$mobile = $_POST['mobile'];
// if(isset($_POST['image']))
//{
$ImagePath = "imageUploads/$mobile.jpg";
$ServerURL = "yourPath/$ImagePath";
$InsertSQL = "INSERT INTO info (img) values('$ServerURL') where
mobile=$mobile";
if(mysqli_query($conn, $InsertSQL)){
file_put_contents($ImagePath,base64_decode($ImageData));
echo "Your Image Has Been Uploaded.";
mysqli_close($conn);
}
else{
echo "Please Try Again";
}
//}
}
?>
我已经添加了以下代码,但是它要么明智地返回“位置”,要么明智地返回“否”,但不会同时返回。
<Employees>
<Department Position="9">
<Employee No="7" Status="True" />
<Employee No="6" Status="True" />
<Employee No="8" Status="True" />
</Department>
<Department Position="4">
<Employee No="7" Status="True" />
<Employee No="8" Status="True" />
<Employee No="6" Status="True" />
</Department>
</Employees>
Out put should be sorted by department position and employee "No"
<Employees>
<Department Position="4">
<Employee No="6" Status="True" />
<Employee No="7" Status="True" />
<Employee No="8" Status="True" />
</Department>
<Department Position="9">
<Employee No="6" Status="True" />
<Employee No="7" Status="True" />
<Employee No="8" Status="True" />
</Department>
答案 0 :(得分:4)
sortSignalList.OrderBy(x => x.Position).ThenBy(x=>x.No).ToList();
答案 1 :(得分:0)
尝试。...
sortSignalList.OrderBy(x => x.Position).ThenBy(x=>x.No).ToList()
答案 2 :(得分:0)
列表也可以排序 ThenByDescending
var thenByDescResult =sortSignalList.OrderBy(s => s.Position).ThenByDescending(s => s.No);