Remove items contained in an ArrayList from another ArrayList in PowerShell











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0
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I've got two ArrayLists listA and listB. listB is a subset of listA.
Now I want to remove all items contained in listB from listA.



Here is how my lists look like:





Name     ID    Domain

----     --    ------

item1    456   domain1

item2    716   domain2

item3    421   domain2

item4    796   domain1






Name     ID    Domain

----     --    ------

item2    716   domain2

item4    796   domain1


I've already tried using



$listA = $listA | Where-Object {$listB -notcontains $_}


but this did not work on my data.






powershell arraylist compare







share|improve this question




















  • 5




    did you try compare-object?
    – 4c74356b41
    Nov 12 at 8:30










  • @4c74356b41 is right. If you had simple string values in either arraylists your example would work.
    – Mötz
    Nov 12 at 8:48






  • 2




    Your approach would only work if both lists contained the same object. It does not work if you have different objects with the same property values (identity vs. equality). Use Compare-Object as 4c74356b41 suggested.
    – Ansgar Wiechers
    Nov 12 at 9:08










  • Thanks this worked fine for me. I haven't thought about using compare-object in this contaxt.
    – Ramona
    Nov 12 at 11:51















up vote
0
down vote

favorite












I've got two ArrayLists listA and listB. listB is a subset of listA.
Now I want to remove all items contained in listB from listA.



Here is how my lists look like:





Name     ID    Domain

----     --    ------

item1    456   domain1

item2    716   domain2

item3    421   domain2

item4    796   domain1






Name     ID    Domain

----     --    ------

item2    716   domain2

item4    796   domain1


I've already tried using



$listA = $listA | Where-Object {$listB -notcontains $_}


but this did not work on my data.






powershell arraylist compare







share|improve this question




















  • 5




    did you try compare-object?
    – 4c74356b41
    Nov 12 at 8:30










  • @4c74356b41 is right. If you had simple string values in either arraylists your example would work.
    – Mötz
    Nov 12 at 8:48






  • 2




    Your approach would only work if both lists contained the same object. It does not work if you have different objects with the same property values (identity vs. equality). Use Compare-Object as 4c74356b41 suggested.
    – Ansgar Wiechers
    Nov 12 at 9:08










  • Thanks this worked fine for me. I haven't thought about using compare-object in this contaxt.
    – Ramona
    Nov 12 at 11:51













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I've got two ArrayLists listA and listB. listB is a subset of listA.
Now I want to remove all items contained in listB from listA.



Here is how my lists look like:





Name     ID    Domain

----     --    ------

item1    456   domain1

item2    716   domain2

item3    421   domain2

item4    796   domain1






Name     ID    Domain

----     --    ------

item2    716   domain2

item4    796   domain1


I've already tried using



$listA = $listA | Where-Object {$listB -notcontains $_}


but this did not work on my data.






powershell arraylist compare







share|improve this question















I've got two ArrayLists listA and listB. listB is a subset of listA.
Now I want to remove all items contained in listB from listA.



Here is how my lists look like:





Name     ID    Domain

----     --    ------

item1    456   domain1

item2    716   domain2

item3    421   domain2

item4    796   domain1






Name     ID    Domain

----     --    ------

item2    716   domain2

item4    796   domain1


I've already tried using



$listA = $listA | Where-Object {$listB -notcontains $_}


but this did not work on my data.






powershell arraylist compare




powershell arraylist compare






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 12 at 10:41









Ansgar Wiechers

139k12120183




139k12120183










asked Nov 12 at 8:11









Ramona

689




689








  • 5




    did you try compare-object?
    – 4c74356b41
    Nov 12 at 8:30










  • @4c74356b41 is right. If you had simple string values in either arraylists your example would work.
    – Mötz
    Nov 12 at 8:48






  • 2




    Your approach would only work if both lists contained the same object. It does not work if you have different objects with the same property values (identity vs. equality). Use Compare-Object as 4c74356b41 suggested.
    – Ansgar Wiechers
    Nov 12 at 9:08










  • Thanks this worked fine for me. I haven't thought about using compare-object in this contaxt.
    – Ramona
    Nov 12 at 11:51














  • 5




    did you try compare-object?
    – 4c74356b41
    Nov 12 at 8:30










  • @4c74356b41 is right. If you had simple string values in either arraylists your example would work.
    – Mötz
    Nov 12 at 8:48






  • 2




    Your approach would only work if both lists contained the same object. It does not work if you have different objects with the same property values (identity vs. equality). Use Compare-Object as 4c74356b41 suggested.
    – Ansgar Wiechers
    Nov 12 at 9:08










  • Thanks this worked fine for me. I haven't thought about using compare-object in this contaxt.
    – Ramona
    Nov 12 at 11:51








5




5




did you try compare-object?
– 4c74356b41
Nov 12 at 8:30




did you try compare-object?
– 4c74356b41
Nov 12 at 8:30












@4c74356b41 is right. If you had simple string values in either arraylists your example would work.
– Mötz
Nov 12 at 8:48




@4c74356b41 is right. If you had simple string values in either arraylists your example would work.
– Mötz
Nov 12 at 8:48




2




2




Your approach would only work if both lists contained the same object. It does not work if you have different objects with the same property values (identity vs. equality). Use Compare-Object as 4c74356b41 suggested.
– Ansgar Wiechers
Nov 12 at 9:08




Your approach would only work if both lists contained the same object. It does not work if you have different objects with the same property values (identity vs. equality). Use Compare-Object as 4c74356b41 suggested.
– Ansgar Wiechers
Nov 12 at 9:08












Thanks this worked fine for me. I haven't thought about using compare-object in this contaxt.
– Ramona
Nov 12 at 11:51




Thanks this worked fine for me. I haven't thought about using compare-object in this contaxt.
– Ramona
Nov 12 at 11:51












1 Answer
1






active

oldest

votes

















up vote
1
down vote













You can do this using the Compare-Object cmdlet.

If your lists are like this:



$listA = @()

$listA += [PSCustomObject]@{Name = 'item1' ; ID = '456'; Domain = 'domain1'}

$listA += [PSCustomObject]@{Name = 'item2' ; ID = '716'; Domain = 'domain2'}

$listA += [PSCustomObject]@{Name = 'item3' ; ID = '421'; Domain = 'domain2'}

$listA += [PSCustomObject]@{Name = 'item4' ; ID = '796'; Domain = 'domain1'}



$listB = @()

$listB += [PSCustomObject]@{Name = 'item2' ; ID = '716'; Domain = 'domain2'}

$listB += [PSCustomObject]@{Name = 'item4' ; ID = '796'; Domain = 'domain1'}


then to remove all objects in $listA that are also in $listB taking all properties into account, do this:



$listA = $listA | Where-Object {(Compare-Object -ReferenceObject $_ -DifferenceObject $listB -Property Name,ID,Domain).SideIndicator -eq '<=' }


After this, $listA will have only these two members left:




Name  ID  Domain 

----  --  ------ 

item1 456 domain1

item3 421 domain2



Edit



Instead of actually naming the properties to compare like in the above, you can also collect them in a variable. For PS versions 3 and up you do this:



$props = $listA[0].psobject.properties.name


PowerShell versions below 3.0 use:



$props = $listA[0].psobject.properties | ForEach-Object { $_.name }


Then you can change the line to



$listA = $listA | Where-Object {(Compare-Object -ReferenceObject $_ -DifferenceObject $listB -Property $props).SideIndicator -eq '<=' }


This of course only if both lists have the same property names to compare..






share|improve this answer























  • Good point, didn't notice that - I'll remove the comment.
    – LotPings
    Nov 12 at 14:03











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote













You can do this using the Compare-Object cmdlet.

If your lists are like this:



$listA = @()

$listA += [PSCustomObject]@{Name = 'item1' ; ID = '456'; Domain = 'domain1'}

$listA += [PSCustomObject]@{Name = 'item2' ; ID = '716'; Domain = 'domain2'}

$listA += [PSCustomObject]@{Name = 'item3' ; ID = '421'; Domain = 'domain2'}

$listA += [PSCustomObject]@{Name = 'item4' ; ID = '796'; Domain = 'domain1'}



$listB = @()

$listB += [PSCustomObject]@{Name = 'item2' ; ID = '716'; Domain = 'domain2'}

$listB += [PSCustomObject]@{Name = 'item4' ; ID = '796'; Domain = 'domain1'}


then to remove all objects in $listA that are also in $listB taking all properties into account, do this:



$listA = $listA | Where-Object {(Compare-Object -ReferenceObject $_ -DifferenceObject $listB -Property Name,ID,Domain).SideIndicator -eq '<=' }


After this, $listA will have only these two members left:




Name  ID  Domain 

----  --  ------ 

item1 456 domain1

item3 421 domain2



Edit



Instead of actually naming the properties to compare like in the above, you can also collect them in a variable. For PS versions 3 and up you do this:



$props = $listA[0].psobject.properties.name


PowerShell versions below 3.0 use:



$props = $listA[0].psobject.properties | ForEach-Object { $_.name }


Then you can change the line to



$listA = $listA | Where-Object {(Compare-Object -ReferenceObject $_ -DifferenceObject $listB -Property $props).SideIndicator -eq '<=' }


This of course only if both lists have the same property names to compare..






share|improve this answer























  • Good point, didn't notice that - I'll remove the comment.
    – LotPings
    Nov 12 at 14:03















up vote
1
down vote













You can do this using the Compare-Object cmdlet.

If your lists are like this:



$listA = @()

$listA += [PSCustomObject]@{Name = 'item1' ; ID = '456'; Domain = 'domain1'}

$listA += [PSCustomObject]@{Name = 'item2' ; ID = '716'; Domain = 'domain2'}

$listA += [PSCustomObject]@{Name = 'item3' ; ID = '421'; Domain = 'domain2'}

$listA += [PSCustomObject]@{Name = 'item4' ; ID = '796'; Domain = 'domain1'}



$listB = @()

$listB += [PSCustomObject]@{Name = 'item2' ; ID = '716'; Domain = 'domain2'}

$listB += [PSCustomObject]@{Name = 'item4' ; ID = '796'; Domain = 'domain1'}


then to remove all objects in $listA that are also in $listB taking all properties into account, do this:



$listA = $listA | Where-Object {(Compare-Object -ReferenceObject $_ -DifferenceObject $listB -Property Name,ID,Domain).SideIndicator -eq '<=' }


After this, $listA will have only these two members left:




Name  ID  Domain 

----  --  ------ 

item1 456 domain1

item3 421 domain2



Edit



Instead of actually naming the properties to compare like in the above, you can also collect them in a variable. For PS versions 3 and up you do this:



$props = $listA[0].psobject.properties.name


PowerShell versions below 3.0 use:



$props = $listA[0].psobject.properties | ForEach-Object { $_.name }


Then you can change the line to



$listA = $listA | Where-Object {(Compare-Object -ReferenceObject $_ -DifferenceObject $listB -Property $props).SideIndicator -eq '<=' }


This of course only if both lists have the same property names to compare..






share|improve this answer























  • Good point, didn't notice that - I'll remove the comment.
    – LotPings
    Nov 12 at 14:03













up vote
1
down vote










up vote
1
down vote









You can do this using the Compare-Object cmdlet.

If your lists are like this:



$listA = @()

$listA += [PSCustomObject]@{Name = 'item1' ; ID = '456'; Domain = 'domain1'}

$listA += [PSCustomObject]@{Name = 'item2' ; ID = '716'; Domain = 'domain2'}

$listA += [PSCustomObject]@{Name = 'item3' ; ID = '421'; Domain = 'domain2'}

$listA += [PSCustomObject]@{Name = 'item4' ; ID = '796'; Domain = 'domain1'}



$listB = @()

$listB += [PSCustomObject]@{Name = 'item2' ; ID = '716'; Domain = 'domain2'}

$listB += [PSCustomObject]@{Name = 'item4' ; ID = '796'; Domain = 'domain1'}


then to remove all objects in $listA that are also in $listB taking all properties into account, do this:



$listA = $listA | Where-Object {(Compare-Object -ReferenceObject $_ -DifferenceObject $listB -Property Name,ID,Domain).SideIndicator -eq '<=' }


After this, $listA will have only these two members left:




Name  ID  Domain 

----  --  ------ 

item1 456 domain1

item3 421 domain2



Edit



Instead of actually naming the properties to compare like in the above, you can also collect them in a variable. For PS versions 3 and up you do this:



$props = $listA[0].psobject.properties.name


PowerShell versions below 3.0 use:



$props = $listA[0].psobject.properties | ForEach-Object { $_.name }


Then you can change the line to



$listA = $listA | Where-Object {(Compare-Object -ReferenceObject $_ -DifferenceObject $listB -Property $props).SideIndicator -eq '<=' }


This of course only if both lists have the same property names to compare..






share|improve this answer














You can do this using the Compare-Object cmdlet.

If your lists are like this:



$listA = @()

$listA += [PSCustomObject]@{Name = 'item1' ; ID = '456'; Domain = 'domain1'}

$listA += [PSCustomObject]@{Name = 'item2' ; ID = '716'; Domain = 'domain2'}

$listA += [PSCustomObject]@{Name = 'item3' ; ID = '421'; Domain = 'domain2'}

$listA += [PSCustomObject]@{Name = 'item4' ; ID = '796'; Domain = 'domain1'}



$listB = @()

$listB += [PSCustomObject]@{Name = 'item2' ; ID = '716'; Domain = 'domain2'}

$listB += [PSCustomObject]@{Name = 'item4' ; ID = '796'; Domain = 'domain1'}


then to remove all objects in $listA that are also in $listB taking all properties into account, do this:



$listA = $listA | Where-Object {(Compare-Object -ReferenceObject $_ -DifferenceObject $listB -Property Name,ID,Domain).SideIndicator -eq '<=' }


After this, $listA will have only these two members left:




Name  ID  Domain 

----  --  ------ 

item1 456 domain1

item3 421 domain2



Edit



Instead of actually naming the properties to compare like in the above, you can also collect them in a variable. For PS versions 3 and up you do this:



$props = $listA[0].psobject.properties.name


PowerShell versions below 3.0 use:



$props = $listA[0].psobject.properties | ForEach-Object { $_.name }


Then you can change the line to



$listA = $listA | Where-Object {(Compare-Object -ReferenceObject $_ -DifferenceObject $listB -Property $props).SideIndicator -eq '<=' }


This of course only if both lists have the same property names to compare..







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 12 at 13:48

























answered Nov 12 at 13:35









Theo

3,3111518




3,3111518












  • Good point, didn't notice that - I'll remove the comment.
    – LotPings
    Nov 12 at 14:03


















  • Good point, didn't notice that - I'll remove the comment.
    – LotPings
    Nov 12 at 14:03
















Good point, didn't notice that - I'll remove the comment.
– LotPings
Nov 12 at 14:03




Good point, didn't notice that - I'll remove the comment.
– LotPings
Nov 12 at 14:03


















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body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} 鏡平學校 地址 小學部：俾利喇街128號 中學部：馬場大馬路389號 详细信息 其它名称 葡萄牙語： Escola Keang Peng 英语： Keang Peng School 学校类别 私立 校训 誠、毅、精、博 宗教背景 鏡湖醫院慈善會 创校日期 1892年 创始人 余倩嫻 校长 黎世祺 語言 中文 電話號碼 +853 28423118 学校网址 http://www.keangpeng.edu.mo 鏡平學校小學部 鏡平學校中學部 鏡平學校 (葡萄牙語： Escola Keang Peng ; 英语： Keang Peng School )是澳門鏡湖醫院屬下的一所學校。現任校長為黎世祺。幼稚園和小學在黑沙環俾利喇街；中學部在黑沙環馬場大馬路。 目录 1 歷史 2 辦學理念 2.1 辦學宗旨 2.2 辦學目標 2.3 育人目標 3 歷任校長 4 教學相關 4.1 幼稚園(學前教育) 5 参考资料 6 外部連結 歷史 鏡平學校是鏡湖慈善會屬下的教育機構, 創辦於1892年, 為澳門辦校歷史最悠久的學校。 鏡湖平民聯合小學合校至今已六十五周年了，學校的前身是鏡湖小學與平民學校。1948年間，兩校的校董會董事決定將兩校合併。當年，鏡湖醫院慈善會董事何善衡先生捐出望廈地段作校址之用，董事會合力籌募建校經費。 合校初期 1949年中，新校舍落成，命名為鏡湖平民聯合小學。此外，還在下環區和新橋區附設分校，正校也附設有夜校，學生人數增至一千四百多人。但由於學額有限，每年招收的學生遠遠未能滿足家長的要求。1956年，正校又在操場靠近俾利喇街地段增建一座兩層高的，擁有八個課室的校舍，學生人數增至二千人。 隨著社會的發展需要，為了進一步辦好鏡平學校，1990年鏡湖慈善會、校董會和董事們，本著關心教育造福清貧學生，培養社會人才的精神，開展了籌募鉅款改建學
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ꓛꓣだゔៀៅຸ໢ທຮ໕໒ ,ໂ'໥໓າ໼ឨឲ៵៭ៈゎゔit''䖳𥁄卿' ☨₤₨こゎもょの;ꜹꟚꞖꞵꟅꞛေၦေɯ,ɨɡ𛃵𛁹ޝ޳ޠ޾,ޤޒޯ޾𫝒𫠁သ𛅤チョ'サノބޘދ𛁐ᶿᶇᶀᶋᶠ㨑㽹⻮ꧬ꧹؍۩وَؠ㇕㇃㇪ ㇦㇋㇋ṜẰᵡᴠ 軌ᵕ搜۳ٰޗޮ޷ސޯ𫖾𫅀ल, ꙭ꙰ꚅꙁꚊꞻꝔ꟠Ꝭㄤﺟޱސꧨꧼ꧴ꧯꧽ꧲ꧯ'⽹⽭⾁⿞⼳⽋២៩ញណើꩯꩤ꩸ꩮᶻᶺᶧᶂ𫳲𫪭𬸄𫵰𬖩𬫣𬊉ၲ𛅬㕦䬺𫝌𫝼,,𫟖𫞽ហៅ஫㆔ాఆఅꙒꚞꙍ,Ꙟ꙱エ ,ポテ,フࢰࢯ𫟠𫞶 𫝤𫟠ﺕﹱﻜﻣ𪵕𪭸𪻆𪾩𫔷ġ,ŧآꞪ꟥,ꞔꝻ♚☹⛵𛀌ꬷꭞȄƁƪƬșƦǙǗdžƝǯǧⱦⱰꓕꓢႋ神 ဴ၀க௭எ௫ឫោ ' េㇷㇴㇼ神ㇸㇲㇽㇴㇼㇻㇸ'ㇸㇿㇸㇹㇰㆣꓚꓤ₡₧ ㄨㄟ㄂ㄖㄎ໗ﾂڒذ₶।ऩछएोञयूटक़कयँृी,冬'𛅢𛅥ㇱㇵㇶ𥄥𦒽𠣧𠊓𧢖𥞘𩔋цѰㄠſtʯʭɿʆʗʍʩɷɛ,əʏダヵㄐㄘＲ｛ｇỚṖḺờṠṫảḙḭᴮᵏᴘᵀᵷᵕᴜᴏᵾq﮲ﲿﴽﭙ軌ﰬﶚﶧ﫲Ҝжюїкӈㇴffצּ﬘﭅﬈軌'ffistfflſtffतभफɳɰʊɲʎ𛁱𛁖𛁮𛀉 𛂯𛀞నఋŀŲ 𫟲𫠖𫞺ຆຆ ໹້໕໗ๆทԊꧢꧠ꧰ꓱ⿝⼑ŎḬẃẖỐẅ ,ờỰỈỗﮊDžȩꭏꭎꬻ꭮ꬿꭖꭥꭅ㇭神 ⾈ꓵꓑ⺄㄄ㄪㄙㄅㄇstA۵䞽ॶ𫞑𫝄㇉㇇゜軌𩜛𩳠Jﻺ‚Üမ႕ႌႊၐၸဓၞၞၡ៸wyvtᶎᶪᶹစဎ꣡꣰꣢꣤ٗ؋لㇳㇾㇻㇱ㆐㆔,,㆟Ⱶﾔﾏފ޼ޝަݿݞݠݷݐ',ݘ,ݪݙݵ𬝉𬜁𫝨𫞘くせぉて¼óû×ó£…𛅑הㄙくԗԀ5606神45,神796'𪤻𫞧ꓐ㄁ㄘɥɺꓵꓲ3''7034׉ⱦⱠˆ“𫝋ȍ,ꩲ軌꩷ꩶꩧꩫఞ۔فڱێظペサ神ナᴦᵑ47 9238їﻂ䐊䔉㠸﬎ffiﬣ,לּᴷᴦᵛᵽ,ᴨᵤ ᵸᵥᴗᵈꚏꚉꚟ⻆rtǟƴ𬎎

Оҁ菊湮ऍ؞ږᶣᶯꝮ꟟ꜧꜨꞤ្軌,វ៫ឌ֜׸֕㇮,,㇚㇖ٞٚڭ'ػ ث ګۮك౵౟ఁ𫠒𫝵𫞲𫟸ォス・ゥヰ'ゲִֿ׿꙼꙼⾎⾌⿐ݶݿݓݩݢ⺤㆜㆓㆞㆞ⱶⱸⱮ꧸ꧻH౜య఍ఔఀꙄꙀꙇﺂﻅﻔﻜﹾ⽜⾈⾔⿗⿕嫉礶𛀜𛀙𛂭𛂝ⱱ௮ж,ҡф 'ӇҠѺْ؅ؘ𛅝𛄻𛅯𛅃𛅭 '𛅩おｶｻﾉｭᶀᶗ𛅦𛄻𛅨𛅐׵׽ט֣ץ׶׻׫׬ꓳꓷꓶꓛꓢ, ㆼㆺㆪㆦ ௭XIUㄛㄔㄫ⃁₠⃏⃏₣₡धू޹ޞޫ޿އޛޟޖވ'޺ލިぶꬱꭦぢぜ㆛㆞㆟㇮㇫㇫ꓻꓶ㇑㇗㇈㇝㆕㆗㆓໫໲꣱𫤠ﬢ𢡊調留醙難ꓪꓟލޏݼݿݥݟⱬⱦⱶࢢ8𣫽𠻙𠣼s D,,B$h០ល៵៱ឬꓟ㆜㆕㆜㆖㆐㆕㆖㆗㆟எஓ꣥꣤ឱអɖʠɒɰʗɧʘ𬳉𬹲𫡸𬄯ﾍﾅﾝ･ㆮㆲﬗﬔ𫅡𛅘𛅥ꭉꬴꙥꚊꚌꙈꙦꚁꚚꙠꚝꬾ,꭯ꭩ꭭ᴼᴥᴄᵲᵱᵽЄеޛޞ꙳Ƞ𪡶తయఠఞభ౲ఁꭏṲṙḖɻʝꩻꩼｗ＆ｓmgU"D'ꭢꭥɜﶒﳒ꣼ ZR,㆑𛀔𛃦𛃟ݡݐݦ࣮ ࣥࣽࣟ神ࣹ ࣬࣍ࢹࣧࣻޯޔްꩼꩳꩬꩵ꩷⻽⺽⻮⺾ᶀⱣⱮⱤⱬⱡ⽊ggzy𬂑𫲇𬍪𫳵'𬎡𬛞𫢆𬗕𫵎𬰙㆛ꩫꩺ神ꩢꩰꩽາ໨,‘ó×⿈⼿purkz,qg yjລᴷᵢᵋ𥲀ﯜ'﵈ﳞﶎ﷉ﱈ﷊炲郃絒钰湨轢躽龠សㄅㄐㄋ神⺼⺽⻅⺺⻪⻺,û軌ǽ֧ǾNJԭԢԚ'ԙyuj'plrh3 1055ｺゴﻦﺒﹲwD₪صڑٔ42ﵢﮝေ၀်Ｘ'ｒＦ,＂＊！Ｈㆵヿイ, ヮベ𫝑𫠝𫞝𫞹 邏꣨E,神YGYBNT޿'ޤ޵囹أٚډ៎៦៥៲៌ îㆨㆤ ꭗꬼ𛄰₮ ₰₸₺䳬ឋ៛ꓦỲởḴṵ15,',ꝵǢƖȗ៯៑ពㇳㇲฤ๫สัฬษั,ꟾ꟣ꜽꞈꞾꟛ﬜שׁ𬃘𫧶𬀏𫱞𬌀𬯒𫬸𫨋𬯤𬅶𫪎ﳪ﷘ĶŲ៷ឯၝ၌ဩိေ𛂚𛁳𛁑𛀰ʁʧʞダfifflfifiꓱꓠꓣꓻǗDzǧ'ƥชฌ,๭ฃป๛ꧾԂࣀꧡ ㇨㇙⻑⻨ᶤﹿ𣤭𦃈𨂿𣤚ឡខ៓𫞬𫞚𫝼𫗜𫎥֠כֿ﬊ႋၵဨ𛂳٫ۋ٭１㇅㇏㇚㇤㇆㇞Ȼ𡮉𠷯֒ꩤꩠ₹C'BOGZAO𫿫𫸈𫩄𫾳𬡌𬟊ះŖĦŪčŸꙓVFマアユㄑㄈފ𛂜׌׳ו𫫇𫢯𫩥C𫝠𫞪𫝹ㄔ㄁げゕ焺萣躍ẌốỬṀด๱ฦธ㇦㇯,㇃ӫ ﰒ ﴌﳜﶅ﷨ﭽﯮݺݻݞ䊉䂖䥯𫞣𫝩𫝘𫠒𫝝㮛Б㇉ҞӶДӧяоЉ꣱ꓫ,,ꓕꓞꓣ꧹، ؄ۇ۶ظꓣʁɴ𬨉,偺𤜵帨㆘㆚㆒'㆐,神𬳰𬣧𬔏ឥ⻥𛂝𛁌𛀀ݘݑ
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Why https connections are so slow when debugging (stepping over) in Java?

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up vote 0 down vote favorite I'm trying to debug some problems with my Java code, which establishes a http/https connection. I'm wondering why the connection gets ridiculously slow (more than one minute for downloading a small web page) when debugging, even when stepping-over the method which does the network work. And if there is some remedy. Below I post an example (you can change https: to http: , and try debugging this from Eclipse pressing F11 - for stepping, place a breakpoint in the first main() statement, and press F6 when it pauses there. My results (time in milliseconds) : conn time total time http (not stepping) 60 350 http (stepping over) 1100 1500 https (not stepping) 570 1300 https (stepping over) 21000 83000
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