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  • CSS Child selectors in IE7 tables

    - by John
    I'm trying to use the CSS child selector in IE7, and it doesn't seem to work. I have nested tables. My outer table has a class name "mytable", and I want the td's of the outer table to show borders. I don't want the inner table td's to have borders. I think I should be able to have CSS that looks like this: .mytable { border-style: solid } .mytable>tr>td { border-style: solid } But the second line seems to have no effect. If I change the second line to make it less specific, it applies to all the td's - I see too many borders. td { border-style: solid } So I think it really is just an issue with the selectors. Pages like this suggest that IE7 should be able to do what I want. Am I doing something silly? Here's the whole HTML file: <html> <head> <style type="text/css"> .mytable { border-style: solid; border-collapse: collapse;} td { border-style: solid; } </style> </head> <body> <table class="mytable"> <tr> <td>Outer top-left</td> <td>Outer top-right</td> </tr> <tr> <td>Outer bottom-left</td> <td> <table> <tr> <td>Inner top-left</td> <td>Inner top-right</td> </tr> <tr> <td>Inner bottom-left</td> <td>Inner bottom-right</td> </tr> <table> </td> </tr> <table> </body> </html>

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  • Oracle's Vision for the Social-Enabled Enterprise

    - by Peggy Chen
    Register Now Join us for the Webcast. Mon., Sept. 10, 2012 10 a.m. PT / 1 p.m. ET Join the conversation: #oracle and #socbiz Mark Hurd President, Oracle Thomas Kurian Executive Vice President, Product Development, Oracle Reggie Bradford Senior Vice President, Product Development, Oracle Dear Colleague, Smart companies are developing social media strategies to engage customers, gain brand insights, and transform employee collaboration and recruitment. Oracle is powering this transformation with the most comprehensive enterprise social platform that lets you: Monitor and engage in social conversations Collect and analyze social data Build and grow brands through social media Integrate enterprisewide social functionality into a single system Create rich social applications Join Oracle President Mark Hurd and senior Oracle executives to learn more about Oracle’s vision for the social-enabled enterprise. Register now for this Webcast. Copyright © 2012, Oracle and/or its affiliates. All rights reserved. Contact Us | Legal Notices and Terms of Use | Privacy Statement

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  • The Road to New Orleans: IT Grand Prix

    - by Enrique Lima
    Four teams race for charity. They need your help. Four teams of MCPs are racing to TechEd in New Orleans on a quest to win $10,000 for the charity of their choice. But they can't win without your help--pick a team, join their pit crew, and earn them points toward victory! While they're on the ground, they need your help in the cloud--pick a team, join their virtual pit crew, and earn them points by meeting online challenges. Join us, be part of this amazing drive to raise awareness and help out by becoming part of the virtual pit crew. I am a pit crew member for the Gold Team.

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  • Unexpected output from Bubblesort program with MSVC vs TCC

    - by Sujith S Pillai
    One of my friends sent this code to me, saying it doesn't work as expected: #include<stdio.h> void main() { int a [10] ={23, 100, 20, 30, 25, 45, 40, 55, 43, 42}; int sizeOfInput = sizeof(a)/sizeof(int); int b, outer, inner, c; printf("Size is : %d \n", sizeOfInput); printf("Values before bubble sort are : \n"); for ( b = 0; b &lt; sizeOfInput; b++) printf("%d\n", a[b]); printf("End of values before bubble sort... \n"); for ( outer = sizeOfInput; outer &gt; 0; outer-- ) { for ( inner = 0 ; inner &lt; outer ; inner++) { printf ( "Comparing positions: %d and %d\n",inner,inner+1); if ( a[inner] &gt; a[inner + 1] ) { int tmp = a[inner]; a[inner] = a [inner+1]; a[inner+1] = tmp; } } printf ( "Bubble sort total array size after inner loop is %d :\n",sizeOfInput); printf ( "Bubble sort sizeOfInput after inner loop is %d :\n",sizeOfInput); } printf ( "Bubble sort total array size at the end is %d :\n",sizeOfInput); for ( c = 0 ; c &lt; sizeOfInput; c++) printf("Element: %d\n", a[c]); } I am using Micosoft Visual Studio Command Line Tool for compiling this on a Windows XP machine. cl /EHsc bubblesort01.c My friend gets the correct output on a dinosaur machine (code is compiled using TCC there). My output is unexpected. The array mysteriously grows in size, in between. If you change the code so that the variable sizeOfInput is changed to sizeOfInputt, it gives the expected results! A search done at Microsoft Visual C++ Developer Center doesn't give any results for "sizeOfInput". I am not a C/C++ expert, and am curious to find out why this happens - any C/C++ experts who can "shed some light" on this? Unrelated note: I seriously thought of rewriting the whole code to use quicksort or merge sort before posting it here. But, after all, it is not Stooge sort... Edit: I know the code is not correct (it reads beyond the last element), but I am curious why the variable name makes a difference.

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  • Coherence Webcast for Developers July 11

    - by jeckels
    Coming on July 11th, we look forward to having you join us for a special Coherence webcast - just for developers! Want to learn how you, the developer, can make applications Big Data and Fast data ready? Want to be able to customize and manage your applications and services to provide real-time data and processing with ease? Then this webcast is for you. Coherence Live Webcast Developers: Deploy Highly-Available Custom Services on Your Data Grid Products July 11, 10am Pacific Time >> Register now! <<  (of course, it's free)Join Brian Oliver of the Coherence team to see how you can create and deploy customized, highly-available services for your data grid, and how real-time data processing will allow you to provide unmatched end-user experiences. We look forward to having you join us.

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  • Empathy auto accept group chat invite

    - by Sivaji
    I'm using empathy 2.34.0 as chat client for account hosted on Google app (server talk.google.com). I'm happy with the features that empathy provides and integration with Google chat, however for group chat when the request is received I need to click on "join" button showed in popup to get started. This makes sense but I would to know if there is any way to automatically join the chat room without clicking the "join" button as I use it only with trusted uses. Besides the messages shared after the invite request and before my entry to chat room is not accessible to me. I looked around the empathy settings but couldn't find anything useful, wondering if I can get some help from here. Thanks.

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  • Non-trivial functions that operate on any monad

    - by Strilanc
    I'm looking for examples of interesting methods that take an arbitrary monad and do something useful with it. Monads are extremely general, so methods that operate on monads are widely applicable. On the other hand, methods I know of that can apply to any monad tend to be... really, really trivial. Barely worth extracting into a function. Here's a really boring example: joinTwice. It just flattens an m m m t into an m t: join n = n >>= id joinTwice n = (join . join) n main = print (joinTwice [[[1],[2, 3]], [[4]]]) -- prints [1,2,3,4] The only non-trivial method for monads that I know of is bindFold (see my answer below). Are there more?

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  • Public JCP EC Meeting on 10 June

    - by Heather VanCura
    The next JCP EC Meeting is open to the public!  We hope you will join us on Tuesday, 10 June at 08:00 AM PDT.  Agenda includes a discussion on the latest JCP.Next news--JSR 364, Broadening JCP Membership. We hope you will join us, but if you cannot attend, the recording and materials will also be public on the JCP.org multimedia page. Meeting details below. ------------------------------------------------------- Topic: Public EC Meeting Date: Tuesday, June 10, 2014 Time: 8:00 am, Pacific Daylight Time (San Francisco, GMT-07:00) Meeting Number: 807 111 580 Meeting Password: 6893 ------------------------------------------------------- To start or join the online meeting ------------------------------------------------------- Go to https://jcp.webex.com/ ------------------------------------------------------- Audio conference information ------------------------------------------------------- +1 (866) 682-4770 (US) Conference code: 5731908 Security code: 6893 Global access numbers

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  • Stairway to T-SQL DML Level 4: The Mathematics of SQL: Part 1

    A relational database contains tables that relate to each other by key values. When querying data from these related tables you may choose to select data from a single table or many tables. If you select data from many tables, you normally join those tables together using specified join criteria. The concepts of selecting data from tables and joining tables together is all about managing and manipulating sets of data. In Level 4 of this Stairway we will explore the concepts of set theory and mathematical operators to join, merge, and return data from multiple SQL Server tables. Get Smart with SQL Backup Pro Powerful centralised management, encryption and more.SQL Backup Pro was the smartest kid at school Discover why.

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  • Oracle Partner Specialists – Sell & Deliver High Value Products to Customers

    - by Richard Lefebvre
    Normal 0 false false false EN-US X-NONE X-NONE MicrosoftInternetExplorer4 Do you want to know where to find useful information about partner training and other activities to complete Oracle Specialization available in the country you are personally based? Go to the EMEA partner enablement blog and read latest information regarding training opportunities ready to join for Cloud Services, Applications, Business Intelligence, Middleware, Database 12c, Engineered System as well as Server & Storage. Recently, we announced new TestFest events in France, which you can join to pass your own Implementation Assessment within the Specialization category you have already chosen. To find out where and when the next TestFest close to your location will take place, please contact [email protected] or watch out for further announcements of TestFest events in your home country. Turnback to the EMEA Partner Enablement Blog from time to time to update your own Specialization and join the latest training for Sales, Presales or Implementation Specialists:  https://blogs.oracle.com/opnenablement/

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  • Unexpected output while using Microsoft Visual Studio 2008 (Express Edition) C++ Command Line Tool

    - by Sujith S Pillai
    One of my friends sent this code to me, saying it doesn't work as expected: #include<stdio.h> void main() { int a [10] ={23, 100, 20, 30, 25, 45, 40, 55, 43, 42}; int sizeOfInput = sizeof(a)/sizeof(int); int b, outer, inner, c; printf("Size is : %d \n", sizeOfInput); printf("Values before bubble sort are : \n"); for ( b = 0; b < sizeOfInput; b++) printf("%d\n", a[b]); printf("End of values before bubble sort... \n"); for ( outer = sizeOfInput; outer > 0; outer-- ) { for ( inner = 0 ; inner < outer ; inner++) { printf ( "Comparing positions: %d and %d\n",inner,inner+1); if ( a[inner] > a[inner + 1] ) { int tmp = a[inner]; a[inner] = a [inner+1]; a[inner+1] = tmp; } } printf ( "Bubble sort total array size after inner loop is %d :\n",sizeOfInput); printf ( "Bubble sort sizeOfInput after inner loop is %d :\n",sizeOfInput); } printf ( "Bubble sort total array size at the end is %d :\n",sizeOfInput); for ( c = 0 ; c < sizeOfInput; c++) printf("Element: %d\n", a[c]); } I am using Micosoft Visual Studio Command Line Tool for compiling this on a Windows XP machine. cl /EHsc bubblesort01.c My friend gets the correct output on a dinosaur machine (code is compiled using TCC there). My output is unexpected. The array mysteriously grows in size, in between. If you change the code so that the variable sizeOfInput is changed to sizeOfInputt, it gives the expected results! A search done at Microsoft Visual C++ Developer Center doesn't give any results for "sizeOfInput". I am not a C/C++ expert, and am curious to find out why this happens - any C/C++ experts who can "shed some light" on this? Unrelated note: I seriously thought of rewriting the whole code to use quicksort or merge sort before posting it here. But, after all, it is not Stooge sort...

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  • Tuesday at OpenWorld: Identity Management

    - by Tanu Sood
    At Oracle OpenWorld? From keynotes, general sessions to product deep dives and executive events, this Tuesday is full of informational, educational and networking opportunities for you. Here’s a quick run-down of what’s happening today: Tuesday, October 2, 2012 KEYNOTE: The Oracle Cloud: Oracle’s Cloud Platform and Applications Strategy 8:00 a.m. – 9:45 a.m., Moscone North, Hall D Leading customers will join Oracle Executive Vice President Thomas Kurian to discuss how Oracle’s innovative cloud solutions are transforming how they manage their business, excite and retain their employees, and deliver great customer experiences through Oracle Cloud. GENERAL SESSION: Oracle Fusion Middleware Strategies Driving Business Innovation 10:15 a.m. – 11:15 a.m., Moscone North - Hall D Join Hasan Rizvi, Executive Vice President of Product in this strategy and roadmap session to hear how developers leverage new innovations in their applications and customers achieve their business innovation goals with Oracle Fusion Middleware. CON9437: Mobile Access Management 10:15 a.m. – 11:15 a.m., Moscone West 3022 The session will feature Identity Management evangelists from companies like Intuit, NetApp and Toyota to discuss how to extend your existing identity management infrastructure and policies to securely and seamlessly enable mobile user access. CON9162: Oracle Fusion Middleware: Meet This Year's Most Impressive Customer Projects 11:45 a.m. – 12:45 a.m., Moscone West, 3001 Hear from the winners of the 2012 Oracle Fusion Middleware Innovation Awards and see which customers are taking home a trophy for the 2012 Oracle Fusion Middleware Innovation Award.  Read more about the Innovation Awards here. CON9491: Enhancing the End-User Experience with Oracle Identity Governance applications 11:45 a.m. – 12:45 p.m., Moscone West 3008 Join experts from Visa and Oracle as they explore how Oracle Identity Governance solutions deliver complete identity administration and governance solutions with support for emerging requirements like cloud identities and mobile devices. CON9447: Enabling Access for Hundreds of Millions of Users 1:15 p.m. – 2:15 p.m., Moscone West 3008 Dealing with scale problems? Looking to address identity management requirements with million or so users in mind? Then take note of Cisco’s implementation. Join this session to hear first-hand how Cisco tackled identity management and scaled their implementation to bolster security and enforce compliance. CON9465: Next Generation Directory – Oracle Unified Directory 5:00 p.m. – 6:00 p.m., Moscone West 3008 Get the 360 degrees perspective from a solution provider, implementation services partner and the customer in this session to learn how the latest Oracle Unified Directory solutions can help you build a directory infrastructure that is optimized to support cloud, mobile and social networking and yet deliver on scale and performance. EVENTS: Executive Edge @ OpenWorld: Chief Security Officer (CSO) Summit 10:00 a.m. – 3:00 p.m. If you are attending the Executive Edge at Open World, be sure to check out the sessions at the Chief Security Officer Summit. Former Sr. Counsel for the National Security Agency, Joel Brenner, will be speaking about his new book "America the Vulnerable". In addition, PWC will present a panel discussion on "Crisis Management to Business Advantage: Security Leadership". See below for the complete agenda. PRODUCT DEMOS: And don’t forget to see Oracle identity Management solutions in action at Oracle OpenWorld DEMOgrounds. DEMOS LOCATION EXHIBITION HALL HOURS Access Management: Complete and Scalable Access Management Moscone South, Right - S-218 Monday, October 1 9:30 a.m.–6:00 p.m. 9:30 a.m.–10:45 a.m. (Dedicated Hours) Tuesday, October 2 9:45 a.m.–6:00 p.m. 2:15 p.m.–2:45 p.m. (Dedicated Hours) Wednesday, October 3 9:45 a.m.–4:00 p.m. 2:15 p.m.–3:30 p.m. (Dedicated Hours) Access Management: Federating and Leveraging Social Identities Moscone South, Right - S-220 Access Management: Mobile Access Management Moscone South, Right - S-219 Access Management: Real-Time Authorizations Moscone South, Right - S-217 Access Management: Secure SOA and Web Services Security Moscone South, Right - S-223 Identity Governance: Modern Administration and Tooling Moscone South, Right - S-210 Identity Management Monitoring with Oracle Enterprise Manager Moscone South, Right - S-212 Oracle Directory Services Plus: Performant, Cloud-Ready Moscone South, Right - S-222 Oracle Identity Management: Closed-Loop Access Certification Moscone South, Right - S-221 For a complete listing, keep the Focus on Identity Management document handy. And don’t forget to converse with us while at OpenWorld @oracleidm. We look forward to hearing from you.

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  • Rotating text using CSS

    - by Renso
    Normal 0 false false false EN-US X-NONE X-NONE /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin-top:0in; mso-para-margin-right:0in; mso-para-margin-bottom:10.0pt; mso-para-margin-left:0in; line-height:115%; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;} Goal: Rotating text using css only. How: Surprisingly IE supports this feature rather well. You could use property filters in IE, but since this is only supported on IE browsers, I would not recommend it. CSS3, still in proposal state, has a "writing-mode" property for doing this. It has been part of IE's browser engine since IE5.5. Now that it is part of the CSS3 draft specification, would be the best way to implement this going forward. Webkit based browsers; Firefox 3.5+, Opera 11 and IE9 implement this feature differently by utilizing the transform property. Without using third-party JavaScript or CSS properties, we can use the CSS3 "writing-mode" property, supported from IE5.5 up to IE8, the latter adding addition formatting options through -ms extensions. <style type="text/css"> .rightToLeft{ writing-mode: tb-rl; } </style> <p class="rightToLeft">This is my text</p> This will rotate the text 90 degrees, starting from the right to the left. Here are all the options: ·         lr-tb – Default value, left to right, top to bottom ·         rl-tb – Right to left, top to bottom ·         tb-rl – Vertically; top to bottom, right to left ·         bt-rl – Vertically; bottom to top, right to left ·         tb-lr – Available in IE8+: -ms-writing-mode; top to bottom, left to right ·         bt-lr – Bottom to top, left to right ·         lr-bt – Left to right, bottom to top What about Firefox, Safari, etc.? The following techniques need to be used on Webkit browsers like Firefox, Opera 11, Google Chrome and IE9. These browsers require their proprietary vendor extensions: -moz-, -webkit-, -o- and -ms-. -webkit-transform: rotate(90deg);    -moz-transform: rotate(90deg); -ms-transform: rotate(90deg); -o-transform: rotate(90deg); transform: rotate(90deg);

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  • How to control the text inside DIV tag by not allowing the text to be increased?

    - by SzamDev
    Hi I have this CSS code : #d_image { float: left; width: 320px; margin-top: 20px; margin-left: 25px; } #d_coll { width: 320px; float: left; } #n_div { width: 240px; text-align: right; padding-top: 10px; float: left; padding-right: 10px; padding-left: 0px; } #n_text { text-align: right; float: left; clear: both; } #im { height: 50px; width: 50px; float: right; padding-right: 15px; } #n_col { clear: both; width: 310px; float: right; border-bottom-width: 2px; border-bottom-style: solid; border-bottom-color: #CCC; padding-bottom: 10px; } #n_tittle { text-decoration: none; text-align: right; padding-bottom: 10px; padding-right: 15px; } #n_tittle a { text-decoration: none; text-align: right; color: #1C60B3; } #n_tittle a:hover { text-decoration: underline; text-align: right; color: #FF0000; } I have this HTML code : <div id="d_coll"> <div id="d_image">sample text</div> <div id="n_text">Here will be photo</div> </div> <div id="n_col"> <div id="n_tittle">sample text</div> <div id="im">small photo will be here</div> <div id="n_div">long text</div> </div> I have proplem in n_div, if I put in it a very long text it will be under im div and I want the text inside it to go to new line when it reach the end because the image should be to the right of the text in im div . How I can do that? Thanks in Advance.

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  • How to align Buttons in a TableLayout to different directions?

    - by Bevor
    Hello, probably I don't understand the layout properties of TableLayout yet. It doesn't seem to be possible to achieve such a flexible table like in HTML, because there are no cells. My target is it to achieve such a layout: Link to draft How can I do that? I thought about using a GridView but this doesn't seem to be useful in XML. My efforts look like this: <TableLayout android:id="@+id/tableLayout" android:layout_width="320sp" android:layout_height="fill_parent" android:layout_gravity="center_horizontal" android:gravity="bottom" android:layout_alignParentBottom="true"> <TableRow android:background="#333333" android:gravity="bottom" android:layout_width="fill_parent"> <Button android:id="@+id/btnUp" android:layout_width="60sp" android:layout_height="50sp" android:gravity="left" android:text="Lift U" /> <Button android:id="@+id/btnScreenUp" android:gravity="right" android:layout_gravity="right" android:layout_width="60sp" android:layout_height="50sp" android:text="Scrn U" /> </TableRow> <TableRow android:background="#444444" android:gravity="bottom" android:layout_gravity="right"> <Button android:id="@+id/btnDown" android:layout_width="60sp" android:layout_height="50sp" android:text="Lift D" /> <Button android:id="@+id/btnScreenLeft" android:layout_width="60sp" android:layout_height="50sp" android:gravity="right" android:layout_gravity="right" android:text="Scrn L" /> <Button android:id="@+id/btnScreenDown" android:layout_width="60sp" android:layout_height="50sp" android:gravity="right" android:layout_gravity="right" android:text="Scrn D" /> <Button android:id="@+id/btnScreenRight" android:layout_width="60sp" android:layout_height="50sp" android:gravity="right" android:layout_gravity="right" android:text="Scrn R" /> </TableRow> </TableLayout>

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  • Why am I getting a " instance has no attribute '__getitem__' " error?

    - by Kevin Yusko
    Here's the code: class BinaryTree: def __init__(self,rootObj): self.key = rootObj self.left = None self.right = None root = [self.key, self.left, self.right] def getRootVal(root): return root[0] def setRootVal(newVal): root[0] = newVal def getLeftChild(root): return root[1] def getRightChild(root): return root[2] def insertLeft(self,newNode): if self.left == None: self.left = BinaryTree(newNode) else: t = BinaryTree(newNode) t.left = self.left self.left = t def insertRight(self,newNode): if self.right == None: self.right = BinaryTree(newNode) else: t = BinaryTree(newNode) t.right = self.right self.right = t def buildParseTree(fpexp): fplist = fpexp.split() pStack = Stack() eTree = BinaryTree('') pStack.push(eTree) currentTree = eTree for i in fplist: if i == '(': currentTree.insertLeft('') pStack.push(currentTree) currentTree = currentTree.getLeftChild() elif i not in '+-*/)': currentTree.setRootVal(eval(i)) parent = pStack.pop() currentTree = parent elif i in '+-*/': currentTree.setRootVal(i) currentTree.insertRight('') pStack.push(currentTree) currentTree = currentTree.getRightChild() elif i == ')': currentTree = pStack.pop() else: print "error: I don't recognize " + i return eTree def postorder(tree): if tree != None: postorder(tree.getLeftChild()) postorder(tree.getRightChild()) print tree.getRootVal() def preorder(self): print self.key if self.left: self.left.preorder() if self.right: self.right.preorder() def inorder(tree): if tree != None: inorder(tree.getLeftChild()) print tree.getRootVal() inorder(tree.getRightChild()) class Stack: def __init__(self): self.items = [] def isEmpty(self): return self.items == [] def push(self, item): self.items.append(item) def pop(self): return self.items.pop() def peek(self): return self.items[len(self.items)-1] def size(self): return len(self.items) def main(): parseData = raw_input( "Please enter the problem you wished parsed.(NOTE: problem must have parenthesis to seperate each binary grouping and must be spaced out.) " ) tree = buildParseTree(parseData) print( "The post order is: ", + postorder(tree)) print( "The post order is: ", + postorder(tree)) print( "The post order is: ", + preorder(tree)) print( "The post order is: ", + inorder(tree)) main() And here is the error: Please enter the problem you wished parsed.(NOTE: problem must have parenthesis to seperate each binary grouping and must be spaced out.) ( 1 + 2 ) Traceback (most recent call last): File "C:\Users\Kevin\Desktop\Python Stuff\Assignment 11\parseTree.py", line 108, in main() File "C:\Users\Kevin\Desktop\Python Stuff\Assignment 11\parseTree.py", line 102, in main tree = buildParseTree(parseData) File "C:\Users\Kevin\Desktop\Python Stuff\Assignment 11\parseTree.py", line 46, in buildParseTree currentTree = currentTree.getLeftChild() File "C:\Users\Kevin\Desktop\Python Stuff\Assignment 11\parseTree.py", line 15, in getLeftChild return root[1] AttributeError: BinaryTree instance has no attribute '__getitem__'

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  • C# - foreach showing strange behavior / for working with no problem

    - by Marks
    Hi there. Today I coded a function that uses two nested foreach loops. After seeing, that it did not work like expected, i debugged it. But I dont see an error, and dont think a simple error can cause the behavior i have noticed. The part looks like this: foreach(MyClass cItem in checkedListBoxItemList.Items) { foreach(MyClass cActiveItem in ActiveItemList) { if (cActiveItem.ID == cItem.ID) /*...check checkbox for item...*/; } } Lets say, checkedListBoxItemList.items holds 4 items of type MyClass, and ActiveItemList is a List< MyClass with 2 Items. The debugger jumps into the outer foreach, reaches inner foreach, executes the if 2 times (once per cActiveItem) and reaches the end of the outer foreach.Now, the debugger jumps back to the head of the outer foreach as it should. But instead of starting the second round of the outer foreach, the debugger suddenly jumps into the MyClass.ToString() method. I can step through this method 4 times (number of items in checkedListBoxItemList.Items) and then ... nothing. Visual Studio shows me my windows form, and the foreach is not continued. When changing the code to int ListCount = checkedListBoxItemList.Items.Count; for(int i=0; i<ListCount; i++) { MyClass cItem = checkedListBoxItemList.Items[i] as MyClass; foreach(MyClass cActiveItem in ActiveItemList) { if (cActiveItem.ID == cItem.ID) /*...check checkbox for item...*/; } } everything works fine and as supposed. I showed the problem to a collegue, but he also didnt understand, what happened. I dont understand why the debugger jumps into the MyClass.ToString() method. I used F10 to step through, so no need to leave the function. And even, if there is a reason, why isnt the foreach loop continued? Im using Visual Studio 2010, if this is of any matter. Please tell me what happened. Thanks.

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  • how to dynamically replace css?

    - by photographer
    I've got an image on the page (class='image') within the div (class='galleria_wrapper') and want to place a caption (class='caption') at the lower right corner under that image. The image could be vertical or horizontal, and making a fixed padding-right value let's say .caption<{float:right;padding-top:5px;padding-right:90px;} is not working for one or another. I need to switch on the fly padding-right value depending on horizontal or vertical image is currently on the page. I can theoretically access image's width through document.getElementByName('image').width although don't understand where to put that code. So, I probably need something like that: document.getElementByClass('caption').padding-right = (document.getElementByName('galleria_wrapper').width - document.getElementByName('image').width)/2 Where do I put this code? I do have that in my css file: .caption{float:right;padding-top:5px;} which places the caption below the image, but to far to the right (div 'galleria_wrapper' is wider than most of the images supposed to be displayed within that area). I have an img tag in the html: <img src=image title='this is caption' /></li> ...and some JavaScript which makes title displayed styled by the "caption" css definition. How do I assign variable value for the padding-right without in-advance knowledge of particular image's width?

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  • Preventing iframe caching in browser

    - by Zarjay
    How do you prevent Firefox and Safari from caching iframe content? I have a simple webpage with an iframe to a page on a different site. Both the outer page and the inner page have HTTP response headers to prevent caching. When I click the "back" button in the browser, the outer page works properly, but no matter what, the browser always retrieves a cache of the iframed page. IE works just fine, but Firefox and Safari are giving me trouble. My webpage looks something like this: <html> <head><!-- stuff --></head> <body> <!-- stuff --> <iframe src="webpage2.html?var=xxx" /> <!-- stuff --> </body> </html> The var variable always changes. Despite the fact that the URL of the iframe has changed (and thus, the browser should be making a new request to that page), the browser just fetches the cached content. I've examined the HTTP requests and responses going back and forth, and I noticed that even if the outer page contains <iframe src="webpage2.html?var=222" />, the browser will still fetch webpage2.html?var=111. Here's what I've tried so far: Changing iframe URL with random var value Adding Expires, Cache-Control, and Pragma headers to outer webpage Adding Expires, Cache-Control, and Pragma headers to inner webpage I'm unable to do any JavaScript tricks because I'm blocked by the same-origin policy. I'm running out of ideas. Does anyone know how to stop the browser from caching the iframed content?

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  • Counting the number of objects that meets a certain criteria

    - by Candy Chiu
    The title doesn't tell the complete story. Please read the message. I have two objects: Adult and Child. Child has a boolean field isMale, and a reference to Adult. Adult doesn't reference Child. public class Adult { long id; } public class Child { long id; boolean isMale; Adult parent; } I want to create a query to list the number of sons each adult has including adults who don't have any sons. I tried: Query 1 SELECT adult, COUNT(child) FROM Child child RIGHT OUTER JOIN child.parent as adult WHERE child.isMale='true' GROUP BY adult which translates to sql select adult.id as col_0_0_, count(child.id) as col_1_0_, ... {omit properties} from Child child right outer join Adult adult on child.parentId=adult.id where child.isMale = 'true' group by adult.id Query 1 doesn't pick up adults that don't have any sons. Query 2: SELECT adult, COUNT(child.isMale) FROM Child child RIGHT OUTER JOIN child.parent as adult GROUP BY adult translates to sql: select adult.id as col_0_0_, count(child.id) as col_1_0_, ... {omit properties} from Child child right outer join Adult adult on child.parentId=adult.id group by adult.id Query 2 doesn't have the right count of sons. Basically COUNT doesn't evaluate isMale. The where clause in Query 1 filtered out Adults with no sons. How do I build a HQL or a Criteria query for this use case? Thanks.

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  • awk + sorting file according to values in the file and write two diffrent files

    - by yael
    hi I have in file file_test values of right eye and left eye How to separate the file_test to file1 and file2 by awk in order to write the equal values on file1 and different values on file2 as the following example down THX file_test: NAME: jim LAST NAME: bakker right eye: |5|< left eye VALUE: |5|< NAME: Jorg LAST NAME: mitchel right eye: |3|< left eye VALUE: |5|< NAME: jimmy LAST NAME: kartter right eye: |6|< left eye VALUE: |5|< NAME: david LAST NAME: kann right eye: |9|< left eye VALUE: |9|< file1: NAME: jim LAST NAME: bakker right eye: |5|< left eye VALUE: |5|< NAME: david LAST NAME: kann right eye: |9|< left eye VALUE: |9|< file2: NAME: Jorg LAST NAME: mitchel right eye: |3|< left eye VALUE: |5|< NAME: jimmy LAST NAME: kartter right eye: |6|< left eye VALUE: |5|<

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  • How can I make this method more Scalalicious

    - by Neil Chambers
    I have a function that calculates the left and right node values for some collection of treeNodes given a simple node.id, node.parentId association. It's very simple and works well enough...but, well, I am wondering if there is a more idiomatic approach. Specifically is there a way to track the left/right values without using some externally tracked value but still keep the tasty recursion. /* * A tree node */ case class TreeNode(val id:String, val parentId: String){ var left: Int = 0 var right: Int = 0 } /* * a method to compute the left/right node values */ def walktree(node: TreeNode) = { /* * increment state for the inner function */ var c = 0 /* * A method to set the increment state */ def increment = { c+=1; c } // poo /* * the tasty inner method * treeNodes is a List[TreeNode] */ def walk(node: TreeNode): Unit = { node.left = increment /* * recurse on all direct descendants */ treeNodes filter( _.parentId == node.id) foreach (walk(_)) node.right = increment } walk(node) } walktree(someRootNode) Edit - The list of nodes is taken from a database. Pulling the nodes into a proper tree would take too much time. I am pulling a flat list into memory and all I have is an association via node id's as pertains to parents and children. Adding left/right node values allows me to get a snapshop of all children (and childrens children) with a single SQL query. The calculation needs to run very quickly in order to maintain data integrity should parent-child associations change (which they do very frequently). In addition to using the awesome Scala collections I've also boosted speed by using parallel processing for some pre/post filtering on the tree nodes. I wanted to find a more idiomatic way of tracking the left/right node values. After looking at the answers listed I have settled on this synthesised version: def walktree(node: TreeNode) = { def walk(node: TreeNode, counter: Int): Int = { node.left = counter node.right = treeNodes .filter( _.parentId == node.id) .foldLeft(counter+1) { (counter, curnode) => walk(curnode, counter) + 1 } node.right } walk(node,1) }

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  • How to exit an if clause

    - by Roman Stolper
    What sorts of methods exist for prematurely exiting an if clause? There are times when I'm writing code and want to put a break statement inside of an if clause, only to remember that those can only be used for loops. Lets take the following code as an example: if some_condition: ... if condition_a: # do something # and then exit the outer if block ... if condition_b: # do something # and then exit the outer if block # more code here I can think of one way to do this: assuming the exit cases happen within nested if statements, wrap the remaining code in a big else block. Example: if some_condition: ... if condition_a: # do something # and then exit the outer if block else: ... if condition_b: # do something # and then exit the outer if block else: # more code here The problem with this is that more exit locations mean more nesting/indented code. Alternatively, I could write my code to have the if clauses be as small as possible and not require any exits. Does anyone know of a good/better way to exit an if clause? If there are any associated else-if and else clauses, I figure that exiting would skip over them.

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  • How to properly translate the "var" result of a lambda expression to a concrete type?

    - by CrimsonX
    So I'm trying to learn more about lambda expressions. I read this question on stackoverflow, concurred with the chosen answer, and have attempted to implement the algorithm using a console app in C# using a simple LINQ expression. My question is: how do I translate the "var result" of the lambda expression into a usable object that I can then print the result? I would also appreciate an in-depth explanation of what is happening when I declare the outer => outer.Value.Frequency (I've read numerous explanations of lambda expressions but additional clarification would help) C# //Input : {5, 13, 6, 5, 13, 7, 8, 6, 5} //Output : {5, 5, 5, 13, 13, 6, 6, 7, 8} //The question is to arrange the numbers in the array in decreasing order of their frequency, preserving the order of their occurrence. //If there is a tie, like in this example between 13 and 6, then the number occurring first in the input array would come first in the output array. List<int> input = new List<int>(); input.Add(5); input.Add(13); input.Add(6); input.Add(5); input.Add(13); input.Add(7); input.Add(8); input.Add(6); input.Add(5); Dictionary<int, FrequencyAndValue> dictionary = new Dictionary<int, FrequencyAndValue>(); foreach (int number in input) { if (!dictionary.ContainsKey(number)) { dictionary.Add(number, new FrequencyAndValue(1, number) ); } else { dictionary[number].Frequency++; } } var result = dictionary.OrderByDescending(outer => outer.Value.Frequency); // How to translate the result into something I can print??

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