Lambda expression with department name with branch name
0
<Employee>
<Data>
<Details type = "Personal">
<Detail Name ="John" Associate Job="Job">
<Department Name="Law" >
<Branch>New York</Branch>
<Branch>Florida</Branch>
</Department>
<Department Name="Lecture" >
<Branch>London</Branch>
<Branch>Brit</Branch>
</Department>
</Detail>
</Details>
</Data>
</Employee>
Output
Law -- New York,
Florida
Lecture -- London,
Brit
Lambda expression for above XML format :---
var employee = (from r in document.Descendants("Detail").Where(r => (string)r.Attribute("Name") == "John")select new { key = r.Element("Department").Attribute("Name").Value, value = (from type in (r.Element("Department").Elements("Branch")) select type.Value).ToArray() })
.ToDictionary(t => t.key, t => t.value);
Only one records is coming
Law -- New York,
Florida
Missing : -
Lecture -- London,
Brit
c# .net linq c#-4.0 lambda
asked Nov 5 '18 at 12:08
surajsuraj
62
add a comment |
0
<Employee>
<Data>
<Details type = "Personal">
<Detail Name ="John" Associate Job="Job">
<Department Name="Law" >
<Branch>New York</Branch>
<Branch>Florida</Branch>
</Department>
<Department Name="Lecture" >
<Branch>London</Branch>
<Branch>Brit</Branch>
</Department>
</Detail>
</Details>
</Data>
</Employee>
Output
Law -- New York,
Florida
Lecture -- London,
Brit
Lambda expression for above XML format :---
var employee = (from r in document.Descendants("Detail").Where(r => (string)r.Attribute("Name") == "John")select new { key = r.Element("Department").Attribute("Name").Value, value = (from type in (r.Element("Department").Elements("Branch")) select type.Value).ToArray() })
.ToDictionary(t => t.key, t => t.value);
Only one records is coming
Law -- New York,
Florida
Missing : -
Lecture -- London,
Brit
c# .net linq c#-4.0 lambda
asked Nov 5 '18 at 12:08
surajsuraj
62
1
Pure code-writing requests are off-topic on Stack Overflow — we expect questions here to relate to specific programming problems — but we will happily help you write it yourself! Tell us what you've tried, and where you are stuck. This will also help us answer your question better.
– Micha Wiedenmann
Nov 5 '18 at 12:17
add a comment |
0
0
0
<Employee>
<Data>
<Details type = "Personal">
<Detail Name ="John" Associate Job="Job">
<Department Name="Law" >
<Branch>New York</Branch>
<Branch>Florida</Branch>
</Department>
<Department Name="Lecture" >
<Branch>London</Branch>
<Branch>Brit</Branch>
</Department>
</Detail>
</Details>
</Data>
</Employee>
Output
Law -- New York,
Florida
Lecture -- London,
Brit
Lambda expression for above XML format :---
var employee = (from r in document.Descendants("Detail").Where(r => (string)r.Attribute("Name") == "John")select new { key = r.Element("Department").Attribute("Name").Value, value = (from type in (r.Element("Department").Elements("Branch")) select type.Value).ToArray() })
.ToDictionary(t => t.key, t => t.value);
Only one records is coming
Law -- New York,
Florida
Missing : -
Lecture -- London,
Brit
c# .net linq c#-4.0 lambda
asked Nov 5 '18 at 12:08
surajsuraj
62
<Employee>
<Data>
<Details type = "Personal">
<Detail Name ="John" Associate Job="Job">
<Department Name="Law" >
<Branch>New York</Branch>
<Branch>Florida</Branch>
</Department>
<Department Name="Lecture" >
<Branch>London</Branch>
<Branch>Brit</Branch>
</Department>
</Detail>
</Details>
</Data>
</Employee>
Output
Law -- New York,
Florida
Lecture -- London,
Brit
Lambda expression for above XML format :---
var employee = (from r in document.Descendants("Detail").Where(r => (string)r.Attribute("Name") == "John")select new { key = r.Element("Department").Attribute("Name").Value, value = (from type in (r.Element("Department").Elements("Branch")) select type.Value).ToArray() })
.ToDictionary(t => t.key, t => t.value);
Only one records is coming
Law -- New York,
Florida
Missing : -
Lecture -- London,
Brit
c# .net linq c#-4.0 lambda
c# .net linq c#-4.0 lambda
asked Nov 5 '18 at 12:08
surajsuraj
62
asked Nov 5 '18 at 12:08
surajsuraj
62
edited Nov 22 '18 at 4:26
suraj
asked Nov 5 '18 at 12:08
surajsuraj
62
asked Nov 5 '18 at 12:08
surajsuraj
62
asked Nov 5 '18 at 12:08
surajsuraj
62
62
1
Pure code-writing requests are off-topic on Stack Overflow — we expect questions here to relate to specific programming problems — but we will happily help you write it yourself! Tell us what you've tried, and where you are stuck. This will also help us answer your question better.
– Micha Wiedenmann
Nov 5 '18 at 12:17
add a comment |
1
Pure code-writing requests are off-topic on Stack Overflow — we expect questions here to relate to specific programming problems — but we will happily help you write it yourself! Tell us what you've tried, and where you are stuck. This will also help us answer your question better.
– Micha Wiedenmann
Nov 5 '18 at 12:17
1
1
Pure code-writing requests are off-topic on Stack Overflow — we expect questions here to relate to specific programming problems — but we will happily help you write it yourself! Tell us what you've tried, and where you are stuck. This will also help us answer your question better.
– Micha Wiedenmann
Nov 5 '18 at 12:17
Pure code-writing requests are off-topic on Stack Overflow — we expect questions here to relate to specific programming problems — but we will happily help you write it yourself! Tell us what you've tried, and where you are stuck. This will also help us answer your question better.
– Micha Wiedenmann
Nov 5 '18 at 12:17
add a comment |
1 Answer
1
active
oldest
votes
0
You can select all the Department elements first:
var employee = (from r in document.Descendants("Detail").Where(r => (string)r.Attribute("Name") == "John").SelectMany(x => x.Elements("Department"))
select new { key = r.Attribute("Name").Value,
value = (from type in r.Elements("Branch") select type.Value).ToArray() })
.ToDictionary(t => t.key, t => t.value);
answered Nov 22 '18 at 5:13
XiaosuXiaosu
3701619
Thank xiaosu, it is working now
– suraj
Nov 29 '18 at 4:41
@suraj Please mark this as answer if it is accepted by you.
– Xiaosu
Nov 29 '18 at 4:59
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
votes
0
You can select all the Department elements first:
var employee = (from r in document.Descendants("Detail").Where(r => (string)r.Attribute("Name") == "John").SelectMany(x => x.Elements("Department"))
select new { key = r.Attribute("Name").Value,
value = (from type in r.Elements("Branch") select type.Value).ToArray() })
.ToDictionary(t => t.key, t => t.value);
answered Nov 22 '18 at 5:13
XiaosuXiaosu
3701619
Thank xiaosu, it is working now
– suraj
Nov 29 '18 at 4:41
@suraj Please mark this as answer if it is accepted by you.
– Xiaosu
Nov 29 '18 at 4:59
add a comment |
0
You can select all the Department elements first:
var employee = (from r in document.Descendants("Detail").Where(r => (string)r.Attribute("Name") == "John").SelectMany(x => x.Elements("Department"))
select new { key = r.Attribute("Name").Value,
value = (from type in r.Elements("Branch") select type.Value).ToArray() })
.ToDictionary(t => t.key, t => t.value);
answered Nov 22 '18 at 5:13
XiaosuXiaosu
3701619
Thank xiaosu, it is working now
– suraj
Nov 29 '18 at 4:41
@suraj Please mark this as answer if it is accepted by you.
– Xiaosu
Nov 29 '18 at 4:59
add a comment |
0
0
0
You can select all the Department elements first:
var employee = (from r in document.Descendants("Detail").Where(r => (string)r.Attribute("Name") == "John").SelectMany(x => x.Elements("Department"))
select new { key = r.Attribute("Name").Value,
value = (from type in r.Elements("Branch") select type.Value).ToArray() })
.ToDictionary(t => t.key, t => t.value);
answered Nov 22 '18 at 5:13
XiaosuXiaosu
3701619
You can select all the Department elements first:
var employee = (from r in document.Descendants("Detail").Where(r => (string)r.Attribute("Name") == "John").SelectMany(x => x.Elements("Department"))
select new { key = r.Attribute("Name").Value,
value = (from type in r.Elements("Branch") select type.Value).ToArray() })
.ToDictionary(t => t.key, t => t.value);
answered Nov 22 '18 at 5:13
XiaosuXiaosu
3701619
answered Nov 22 '18 at 5:13
XiaosuXiaosu
3701619
answered Nov 22 '18 at 5:13
XiaosuXiaosu
3701619
answered Nov 22 '18 at 5:13
XiaosuXiaosu
3701619
3701619
Thank xiaosu, it is working now
– suraj
Nov 29 '18 at 4:41
@suraj Please mark this as answer if it is accepted by you.
– Xiaosu
Nov 29 '18 at 4:59
add a comment |
Thank xiaosu, it is working now
– suraj
Nov 29 '18 at 4:41
@suraj Please mark this as answer if it is accepted by you.
– Xiaosu
Nov 29 '18 at 4:59
Thank xiaosu, it is working now
– suraj
Nov 29 '18 at 4:41
Thank xiaosu, it is working now
– suraj
Nov 29 '18 at 4:41
@suraj Please mark this as answer if it is accepted by you.
– Xiaosu
Nov 29 '18 at 4:59
@suraj Please mark this as answer if it is accepted by you.
– Xiaosu
Nov 29 '18 at 4:59
add a comment |
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1
Pure code-writing requests are off-topic on Stack Overflow — we expect questions here to relate to specific programming problems — but we will happily help you write it yourself! Tell us what you've tried, and where you are stuck. This will also help us answer your question better.
– Micha Wiedenmann
Nov 5 '18 at 12:17