Search Results

Search found 23206 results on 929 pages for 'non root'.

Page 179/929 | < Previous Page | 175 176 177 178 179 180 181 182 183 184 185 186  | Next Page >

  • Mvc relative path using virtual directory..help!

    - by kevin
    When i drag and drop my image/script/css file into my view, relative path will automatically use to refer on the files. example: <link href="../../Content/style.css" rel="stylesheet" type="text/css" /> <script src="../../Scripts/jquery-min.js" type="text/javascript"></script> <img src="../../Images/logo.jpg" /> It is working fine when i host it on my root directory, but if i'm using virtual directory then only my css file able to refer correctly, the rest will return 404...as it will refer to http://{root}/Images/logo.jpg rather than http://{root}/{virtual directory}/Images/logo.jpg But why css file is working? and how to specify the relative path correctly for both root & virtual directory cases?

    Read the article

  • My linux server takes more than an hour to boot. Suggestions?

    - by jamieb
    I am building a CentOS 5.4 system that boots off a compact flash card using a card reader that emulates an IDE drive. It literally takes about an hour to boot. The ultra-slow part occurs when Grub is loading the kernel. Once that's done, the rest of the boot process only takes about a minute to get to a login prompt. Does anyone have any suggestions? I suspect that it may have to do with UDMA. Everything IDE-related in my BIOS seems to checkout. The read performance hdparm is telling me 1.77 MB/s. Ouch! (But even at that rate, it still shouldn't take an hour to decompress and load the kernel) [root@server ~]# hdparm -tT /dev/hdc /dev/hdc: Timing cached reads: 2444 MB in 2.00 seconds = 1222.04 MB/sec Timing buffered disk reads: 6 MB in 3.39 seconds = 1.77 MB/sec Trying to enable DMA is a no-go though: [root@server ~]# hdparm -d1 /dev/hdc /dev/hdc: setting using_dma to 1 (on) HDIO_SET_DMA failed: Operation not permitted using_dma = 0 (off) Here's some command outputs that might help: System [root@server ~]# uname -a Linux server.localdomain 2.6.18-164.el5xen #1 SMP Thu Sep 3 04:47:32 EDT 2009 i686 i686 i386 GNU/Linux PCI info: [root@server ~]# lspci -v 00:00.0 Host bridge: Intel Corporation 82945G/GZ/P/PL Memory Controller Hub (rev 02) Subsystem: Intel Corporation 82945G/GZ/P/PL Memory Controller Hub Flags: bus master, fast devsel, latency 0 Capabilities: [e0] Vendor Specific Information 00:02.0 VGA compatible controller: Intel Corporation 82945G/GZ Integrated Graphics Controller (rev 02) (prog-if 00 [VGA controller]) Subsystem: Intel Corporation 82945G/GZ Integrated Graphics Controller Flags: bus master, fast devsel, latency 0, IRQ 10 Memory at fdf00000 (32-bit, non-prefetchable) [size=512K] I/O ports at ff00 [size=8] Memory at d0000000 (32-bit, prefetchable) [size=256M] Memory at fdf80000 (32-bit, non-prefetchable) [size=256K] Capabilities: [90] Message Signalled Interrupts: 64bit- Queue=0/0 Enable- Capabilities: [d0] Power Management version 2 00:1d.0 USB Controller: Intel Corporation 82801G (ICH7 Family) USB UHCI Controller #1 (rev 01) (prog-if 00 [UHCI]) Subsystem: Intel Corporation 82801G (ICH7 Family) USB UHCI Controller #1 Flags: bus master, medium devsel, latency 0, IRQ 16 I/O ports at fe00 [size=32] 00:1d.1 USB Controller: Intel Corporation 82801G (ICH7 Family) USB UHCI Controller #2 (rev 01) (prog-if 00 [UHCI]) Subsystem: Intel Corporation 82801G (ICH7 Family) USB UHCI Controller #2 Flags: bus master, medium devsel, latency 0, IRQ 17 I/O ports at fd00 [size=32] 00:1d.2 USB Controller: Intel Corporation 82801G (ICH7 Family) USB UHCI Controller #3 (rev 01) (prog-if 00 [UHCI]) Subsystem: Intel Corporation 82801G (ICH7 Family) USB UHCI Controller #3 Flags: bus master, medium devsel, latency 0, IRQ 18 I/O ports at fc00 [size=32] 00:1d.3 USB Controller: Intel Corporation 82801G (ICH7 Family) USB UHCI Controller #4 (rev 01) (prog-if 00 [UHCI]) Subsystem: Intel Corporation 82801G (ICH7 Family) USB UHCI Controller #4 Flags: bus master, medium devsel, latency 0, IRQ 19 I/O ports at fb00 [size=32] 00:1d.7 USB Controller: Intel Corporation 82801G (ICH7 Family) USB2 EHCI Controller (rev 01) (prog-if 20 [EHCI]) Subsystem: Intel Corporation 82801G (ICH7 Family) USB2 EHCI Controller Flags: bus master, medium devsel, latency 0, IRQ 16 Memory at fdfff000 (32-bit, non-prefetchable) [size=1K] Capabilities: [50] Power Management version 2 Capabilities: [58] Debug port 00:1e.0 PCI bridge: Intel Corporation 82801 PCI Bridge (rev e1) (prog-if 01 [Subtractive decode]) Flags: bus master, fast devsel, latency 0 Bus: primary=00, secondary=01, subordinate=01, sec-latency=32 I/O behind bridge: 0000d000-0000dfff Memory behind bridge: fde00000-fdefffff Prefetchable memory behind bridge: 00000000fdd00000-00000000fdd00000 Capabilities: [50] #0d [0000] 00:1f.0 ISA bridge: Intel Corporation 82801GB/GR (ICH7 Family) LPC Interface Bridge (rev 01) Subsystem: Intel Corporation 82801GB/GR (ICH7 Family) LPC Interface Bridge Flags: bus master, medium devsel, latency 0 Capabilities: [e0] Vendor Specific Information 00:1f.2 IDE interface: Intel Corporation 82801GB/GR/GH (ICH7 Family) SATA IDE Controller (rev 01) (prog-if 80 [Master]) Subsystem: Intel Corporation 82801GB/GR/GH (ICH7 Family) SATA IDE Controller Flags: bus master, 66MHz, medium devsel, latency 0, IRQ 17 I/O ports at <unassigned> I/O ports at <unassigned> I/O ports at <unassigned> I/O ports at <unassigned> I/O ports at f800 [size=16] Capabilities: [70] Power Management version 2 00:1f.3 SMBus: Intel Corporation 82801G (ICH7 Family) SMBus Controller (rev 01) Subsystem: Intel Corporation 82801G (ICH7 Family) SMBus Controller Flags: medium devsel, IRQ 17 I/O ports at 0500 [size=32] 01:04.0 Ethernet controller: Realtek Semiconductor Co., Ltd. RTL-8139/8139C/8139C+ (rev 10) Subsystem: Realtek Semiconductor Co., Ltd. RTL-8139/8139C/8139C+ Flags: bus master, medium devsel, latency 32, IRQ 18 I/O ports at de00 [size=256] Memory at fdeff000 (32-bit, non-prefetchable) [size=256] Capabilities: [50] Power Management version 2 01:06.0 Ethernet controller: Realtek Semiconductor Co., Ltd. RTL-8139/8139C/8139C+ (rev 10) Subsystem: Realtek Semiconductor Co., Ltd. RTL-8139/8139C/8139C+ Flags: bus master, medium devsel, latency 32, IRQ 17 I/O ports at dc00 [size=256] Memory at fdefe000 (32-bit, non-prefetchable) [size=256] Capabilities: [50] Power Management version 2 01:07.0 Ethernet controller: Realtek Semiconductor Co., Ltd. RTL-8139/8139C/8139C+ (rev 10) Subsystem: Realtek Semiconductor Co., Ltd. RTL-8139/8139C/8139C+ Flags: bus master, medium devsel, latency 32, IRQ 19 I/O ports at da00 [size=256] Memory at fdefd000 (32-bit, non-prefetchable) [size=256] Capabilities: [50] Power Management version 2 hdparm ouput: [root@server ~]# hdparm /dev/hdc /dev/hdc: multcount = 0 (off) IO_support = 0 (default 16-bit) unmaskirq = 0 (off) using_dma = 0 (off) keepsettings = 0 (off) readonly = 0 (off) readahead = 256 (on) geometry = 8146/16/63, sectors = 8211168, start = 0 [root@server ~]# hdparm -I /dev/hdc /dev/hdc: ATA device, with non-removable media Model Number: InnoDisk Corp. - iCF4000 4GB Serial Number: 20091023AACA70000753 Firmware Revision: 081107 Standards: Supported: 5 Likely used: 6 Configuration: Logical max current cylinders 8146 8146 heads 16 16 sectors/track 63 63 -- CHS current addressable sectors: 8211168 LBA user addressable sectors: 8211168 device size with M = 1024*1024: 4009 MBytes device size with M = 1000*1000: 4204 MBytes (4 GB) Capabilities: LBA, IORDY(can be disabled) Standby timer values: spec'd by Vendor R/W multiple sector transfer: Max = 2 Current = 2 DMA: mdma0 mdma1 mdma2 udma0 udma1 *udma2 udma3 udma4 Cycle time: min=120ns recommended=120ns PIO: pio0 pio1 pio2 pio3 pio4 Cycle time: no flow control=120ns IORDY flow control=120ns Commands/features: Enabled Supported: * Power Management feature set * WRITE_BUFFER command * READ_BUFFER command * NOP cmd * CFA feature set * Mandatory FLUSH_CACHE HW reset results: CBLID- above Vih Device num = 0 CFA power mode 1: enabled and required by some commands Maximum current = 100ma Checksum: correct

    Read the article

  • how to effectively modify index

    - by daedlus
    Hej everyone, problem : I am looking for right way to convert an index from clustered to non-clustered Description : I have a table as below in sybase db: dbo.UserLog Id | UserId |time | .... This is hash partitioned using UserId. Currently it has 2 indexes UserId : non-clustered time: clustered This table has about 20 million records. I now want to make UserId as clustered index and time as non-clustered index. is it correct to user alter index to change from clustered to non-clustered or do i drop index and recreate. does the fact that userId is used in hash partitioning have any implications to this? To me alter seems way to go but I have not yet tried this.

    Read the article

  • List of common ways those could shut down server unexpectedly ?

    - by SpawnCxy
    After running a bash fork bomb which made my webserver down, I think I should be more careful even not under root.I thought it would be totally fine while I'm not under root.So I ignored the warning and ran the bash fork bomb which is :() { :|:& }; : .(Please don't run it if u don't understand this code cuz it will make you system down).And I think I need a list of common ways those could cause a sever shutting down unexpectly even not under root. Any suggestion would be appreciated. Regards `

    Read the article

  • XSL unique values per node per position

    - by Nathan Colin
    this get ever more complicated :) now i face another issue in last question we managed to take unique values from only one parent node now with: <?xml version="1.0" encoding="ISO-8859-1"?> <roots> <root> <name>first</name> <item> <something>A</something> <something>A</something> </item> <item> <something>B</something> <something>A</something> </item> <item> <something>C</something> <something>P</something> </item> <item> <something>A</something> <something>L</something> </item> <item> <something>A</something> <something>A</something> </item> <item> <something>B</something> <something>A</something> </item> <item> <something>D</something> <something>A</something> </item> </root> <root> <name>second</name> <item> <something>E</something> <something>A</something> </item> <item> <something>B</something> <something>A</something> </item> <item> <something>F</something> <something>A</something> </item> <item> <something>A</something> <something>A</something> </item> <item> <something>A</something> <something>A</something> </item> <item> <something>B</something> <something>H</something> </item> <item> <something>D</something> <something>G</something> </item> </root> </roots> now i need to get the unique values depending only from one node before but just from the elements on the second position <?xml version="1.0" encoding="ISO-8859-1"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output indent="yes" method="text"/> <xsl:key name="item-by-value" match="something" use="concat(normalize-space(.), ' ', generate-id(./ancestor::root))"/> <xsl:key name="rootkey" match="root" use="name"/> <xsl:template match="/"> <xsl:for-each select="key('rootkey','first')"> <xsl:for-each select="item/something[1]"> <xsl:sort /> <xsl:if test="generate-id() = generate-id(key('item-by-value', concat(normalize-space(.), ' ', generate-id(./ancestor::root))))"> <xsl:value-of select="."/> </xsl:if> </xsl:for-each> <xsl:text>_________</xsl:text> <xsl:for-each select="item/something[2]"> <xsl:sort /> <xsl:if test="generate-id() = generate-id(key('item-by-value', concat(normalize-space(.), ' ', generate-id(./ancestor::root))))"> <xsl:value-of select="."/> </xsl:if> </xsl:for-each> </xsl:for-each> </xsl:template> </xsl:stylesheet> with this XSL i get ABCD__LP where the result i need is ABCD__ALP any ideas?

    Read the article

  • How can I run a command on a remote machine with Perl?

    - by Bharath Kumar
    I'm using following code to connect to a remote machine and try to execute one simple command on remote machine. cat tt.pl #!/usr/bin/perl #use strict; use warnings; use Net::Telnet; $telnet = new Net::Telnet ( Timeout=>2, Errmode=>'die'); $telnet->open('172.168.12.58'); $telnet->waitfor('/login:\s*/'); $telnet->print('admin'); $telnet->waitfor('/password:\s*/'); $telnet->print('Blue'); #$telnet->cmd('ver > C:\\log.txt'); $telnet->cmd('mkdir gy'); You have new mail in /var/spool/mail/root [root@localhost]# But when I'm executing this script it is throwing error messages [root@localhost]# perl tt.pl command timed-out at tt.pl line 12 [root@localhost]# Please help me in this

    Read the article

  • how to solve this Problem in asp.net crystal report.

    - by Ayyappan.Anbalagan
    Problem in crystal report After excecuting the bellow code,In my report page it ask like "The report you requested requires further information" server= user= password= databse= protected void Page_Load(object sender, EventArgs e) { MySqlConnection sqlcom = new MySqlConnection("server=localhost;userid=root;password=root;database=hemaepdb;"); MySqlCommand cmd = new MySqlCommand("SP_ViewBillDetails", sqlcom); cmd.CommandType = CommandType.StoredProcedure; cmd.Parameters.Add("p_Invoice_Id", MySqlDbType.Int16).Value = 1; cmd.Parameters.Add("p_Org_id", MySqlDbType.Int16).Value = 1; MySqlDataAdapter adapter = new MySqlDataAdapter(cmd); DataSet dsTest = new DataSet(); sqlcom.Open(); adapter.Fill(dsTest, "Table"); sqlcom.Close(); CrystalReportViewer1.Visible = true; ReportDocument myRpt = new ReportDocument(); myRpt.Load(Server.MapPath("CrystalReport.rpt")); myRpt.SetDatabaseLogon("root", "root", "localhost", "hemaepdb"); myRpt.SetDataSource(dsTest); CrystalReportViewer1.ReportSource = myRpt; }

    Read the article

  • Python MD5 Hash Faster Calculation

    - by balgan
    Hi everyone. I will try my best to explain my problem and my line of thought on how I think I can solve it. I use this code for root, dirs, files in os.walk(downloaddir): for infile in files: f = open(os.path.join(root,infile),'rb') filehash = hashlib.md5() while True: data = f.read(10240) if len(data) == 0: break filehash.update(data) print "FILENAME: " , infile print "FILE HASH: " , filehash.hexdigest() and using start = time.time() elapsed = time.time() - start I measure how long it takes to calculate an hash. Pointing my code to a file with 653megs this is the result: root@Mars:/home/tiago# python algorithm-timer.py FILENAME: freebsd.iso FILE HASH: ace0afedfa7c6e0ad12c77b6652b02ab 12.624 root@Mars:/home/tiago# python algorithm-timer.py FILENAME: freebsd.iso FILE HASH: ace0afedfa7c6e0ad12c77b6652b02ab 12.373 root@Mars:/home/tiago# python algorithm-timer.py FILENAME: freebsd.iso FILE HASH: ace0afedfa7c6e0ad12c77b6652b02ab 12.540 Ok now 12 seconds +- on a 653mb file, my problem is I intend to use this code on a program that will run through multiple files, some of them might be 4/5/6Gb and it will take wayy longer to calculate. What am wondering is if there is a faster way for me to calculate the hash of the file? Maybe by doing some multithreading? I used a another script to check the use of the CPU second by second and I see that my code is only using 1 out of my 2 CPUs and only at 25% max, any way I can change this? Thank you all in advance for the given help.

    Read the article

  • C++ print out a binary search tree

    - by starcorn
    Hello, Got nothing better to do this Christmas holiday, so I decided to try out making a binary search tree. I'm stuck with the print function. How should the logic behind it work? Since the tree is already inserting it in a somewhat sorted order, and I want to print the tree from smallest values to the biggest. So I need to travel to the furthest left branch of the tree to print the first value. Right, so after that how do I remember the way back up, do I need to save the previous node? A search in wikipedia gave me an solution which they used stack. And other solutions I couldn't quite understand how they've made it, so I'm asking here instead hoping someone can enlight me. I also wonder my insert function is OK. I've seen other's solution being smaller. void treenode::insert(int i) { if(root == 0) { cout << "root" << endl; root = new node(i,root); } else { node* travel = root; node* prev; while(travel) { if(travel->value > i) { cout << "travel left" << endl; prev = travel; travel = travel->left; } else { cout << "travel right" << endl; prev = travel; travel = travel->right; } } //insert if(prev->value > i) { cout << "left" << endl; prev->left = new node(i); } else { cout << "right" << endl; prev->right = new node(i); } } } void treenode::print() { node* travel = root; while(travel) { cout << travel->value << endl; travel = travel->left; } }

    Read the article

  • Divide a large amount of text on an arbitrary number of equal parts.

    - by kalininew
    I probably already fed up with their stupid questions, but I have one more question. I have a large piece of text <p> Sed ut perspiciatis, unde omnis iste natus error sit voluptatem accusantium doloremque laudantium, totam rem aperiam eaque ipsa, quae ab illo inventore veritatis et quasi architecto beatae vitae dicta sunt, explicabo. Nemo enim ipsam voluptatem, quia voluptas sit, </p> <p> aspernatur aut odit aut fugit, sed quia consequuntur magni dolores eos, qui ratione voluptatem sequi nesciunt, neque porro quisquam est, qui dolorem ipsum, quia dolor sit, amet, consectetur, adipisci velit, sed quia non numquam eius modi tempora incidunt, ut labore et dolore magnam aliquam quaerat voluptatem. Ut enim ad minima veniam, quis nostrum exercitationem ullam corporis suscipit laboriosam, nisi ut aliquid ex ea commodi consequatur? </p> <p> Quis autem vel eum iure reprehenderit, qui in ea voluptate velit esse, quam nihil molestiae consequatur, vel illum, qui dolorem eum fugiat, quo voluptas nulla pariatur? At vero eos et accusamus et iusto odio dignissimos ducimus, qui blanditiis praesentium voluptatum deleniti atque corrupti, quos dolores et quas molestias excepturi sint, obcaecati cupiditate non provident, similique sunt in culpa, qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum </p> <p> soluta nobis est eligendi optio, cumque nihil impedit, quo minus id, quod maxime placeat, facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Temporibus autem quibusdam et aut officiis debitis aut rerum necessitatibus saepe eveniet, ut et voluptates repudiandae sint et molestiae non recusandae. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat. </p> <p> Sed ut perspiciatis, unde omnis iste natus error sit voluptatem accusantium doloremque laudantium, totam rem aperiam eaque ipsa, quae ab illo inventore veritatis et quasi architecto beatae vitae dicta sunt, explicabo. Nemo enim ipsam voluptatem, quia voluptas sit, </p> <p> Sed ut perspiciatis, unde omnis iste natus error sit voluptatem accusantium doloremque laudantium, totam rem aperiam eaque ipsa, quae ab illo inventore veritatis et quasi architecto beatae vitae dicta sunt, explicabo. Nemo enim ipsam voluptatem, quia voluptas sit, </p> <p> aspernatur aut odit aut fugit, sed quia consequuntur magni dolores eos, qui ratione voluptatem sequi nesciunt, neque porro quisquam est, qui dolorem ipsum, quia dolor sit, amet, consectetur, adipisci velit, sed quia non numquam eius modi tempora incidunt, ut labore et dolore magnam aliquam quaerat voluptatem. Ut enim ad minima veniam, quis nostrum exercitationem ullam corporis suscipit laboriosam, nisi ut aliquid ex ea commodi consequatur? </p> <p> Quis autem vel eum iure reprehenderit, qui in ea voluptate velit esse, quam nihil molestiae consequatur, vel illum, qui dolorem eum fugiat, quo voluptas nulla pariatur? At vero eos et accusamus et iusto odio dignissimos ducimus, qui blanditiis praesentium voluptatum deleniti atque corrupti, quos dolores et quas molestias excepturi sint, obcaecati cupiditate non provident, similique sunt in culpa, qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum </p> <p> soluta nobis est eligendi optio, cumque nihil impedit, quo minus id, quod maxime placeat, facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Temporibus autem quibusdam et aut officiis debitis aut rerum necessitatibus saepe eveniet, ut et voluptates repudiandae sint et molestiae non recusandae. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat. </p> <p> Quis autem vel eum iure reprehenderit, qui in ea voluptate velit esse, quam nihil molestiae consequatur, vel illum, qui dolorem eum fugiat, quo voluptas nulla pariatur? At vero eos et accusamus et iusto odio dignissimos ducimus, qui blanditiis praesentium voluptatum deleniti atque corrupti, quos dolores et quas molestias excepturi sint, obcaecati cupiditate non provident, similique sunt in culpa, qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum </p> <p> Sed ut perspiciatis, unde omnis iste natus error sit voluptatem accusantium doloremque laudantium, totam rem aperiam eaque ipsa, quae ab illo inventore veritatis et quasi architecto beatae vitae dicta sunt, explicabo. Nemo enim ipsam voluptatem, quia voluptas sit, </p> <p> soluta nobis est eligendi optio, cumque nihil impedit, quo minus id, quod maxime placeat, facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Temporibus autem quibusdam et aut officiis debitis aut rerum necessitatibus saepe eveniet, ut et voluptates repudiandae sint et molestiae non recusandae. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat. </p> <p> Sed ut perspiciatis, unde omnis iste natus error sit voluptatem accusantium doloremque laudantium, totam rem aperiam eaque ipsa, quae ab illo inventore veritatis et quasi architecto beatae vitae dicta sunt, explicabo. Nemo enim ipsam voluptatem, quia voluptas sit, </p> At the exit I need to divide the text on the "n" equal parts, so that in these parts was about the same amount of text. Then I these part are arranged in columns and the need for these columns look about the same height. Another condition: Tags you can break (I mean that if the tag "p" contains a lot of text, it can be divided into two parts, to bring in another column). I think this is a monumental task, I shall be grateful for any help.

    Read the article

  • Syntax error when using "with open" in Python (python newbie)

    - by Tony
    [root@234571-app2 git]# ./test.py File "./test.py", line 4 with open("/home/git/post-receive-email.log",'a') as log_file: ^ SyntaxError: invalid syntax The code looks like this: [root@234571-app2 git]# more test.py #!/usr/bin/python from __future__ import with_statement with open("/home/git/post-receive-email.log",'a') as log_file: log_file.write("hello world") and I am using Python 2.5.5 [root@234571-app2 git]# python -V Python 2.5.5

    Read the article

  • How to call a scalar function in a stored procedure

    - by Luke101
    I am wacking y head over the problem with this code. DECLARE @root hierarchyid declare @lastchild hierarchyid SELECT @root = NodeHierarchyID from NodeHierarchy where ID = 1 set @lastchild = getlastchild(@root) it says it does not recognize getlastchild function. What am I doing wrong here

    Read the article

  • PHP incomplete code - scan dir, include only if name starts or end with x

    - by Adrian M.
    I posted a question before but I am yet limited to mix the code without getting errors.. I'm rather new to php :( ( the dirs are named in series like this "id_1_1" , "id_1_2", "id_1_3" and "id_2_1" , "id_2_2", "id_2_3" etc.) I have this code, that will scan a directory for all the files and then include a same known named file for each of the existing folders.. the problem is I want to modify a bit the code to only include certain directories which their names: ends with "_1" starts with "id_1_" I want to create a page that will load only the dirs that ends with "_1" and another file that will load only dirs that starts with "id_1_".. <?php include_once "$root/content/common/header.php"; include_once "$root/content/common/header_bc.php"; include_once "$root/content/" . $page_file . "/content.php"; $page_path = ("$root/content/" . $page_file); $includes = array(); $iterator = new RecursiveIteratorIterator( new RecursiveDirectoryIterator($page_path), RecursiveIteratorIterator::SELF_FIRST); foreach($iterator as $file) { if($file->isDir()) { $includes[] = strtoupper($file . '/template.php'); } } $includes = array_reverse($includes); foreach($includes as $file){ include $file; } include_once "$root/content/common/footer.php"; ?> Many Thanks!

    Read the article

  • GoDaddy Subdomain Hosting Issue/Question with Disk Access (C#/ASP.NET 3.5)

    - by Vogel
    This isn't a very complicated scenario really, but as I start to type out the problem I'm realizing how convoluted it can become textually. Let me try and be very clear: First, the set up... I have a C#/ASP.NET web application that is publicly facing on my main domain (www), let's call it www.mysite.com. Nothing fancy, just a front-end that connects to SQL to display records. Then, I have a second C#/ASP.NET web application that is secured using forms authentication running on a subdomain, let's call it admin.mysite.com. This is a very light-weight CMS system to administer the public site. Now, the problem... Both of these sites run fine for basic tasks, however, my problem arises when I try to gain access to the file system for uploading. GoDaddy requires subdomains to run as a virtual directories under the main application in IIS (so the subdomains actually resolve/re-direct to www.mysite.com/admin when you type in admin.mysite.com), but because of this I am unable to write to my website root from the subfolder. Let me explain a little more... The CMS system (running as a virtual directory) gives the admin the ability to upload photos for display on the main site, the target folder of which is www.mysite.com/images - when attempting disk access from the root app, I am able to write to the virtual directory, but cannot do the opposite -- that is, write to the root from the virtual directory, getting security violations. If I can only upload to the /admin/ virtual directory, the entire point is moot because it's a secured folder that the public can't see! The only solution I can think of is to upload the files to the /admin/ virtual directory, then call a URL in the root that moves files from /admin/ back to the root, but that is entirely ghetto. I hope this post makes sense. Anyone else experience anything like this? The bottom line is that it seems virtual directories ONLY have access to themselves, and not their parent directories, no matter what credentials are used. Thanks!

    Read the article

  • Why are mercurial subrepos behaving as unversioned files in eclipse

    - by noam
    I am trying to use the subrepo feature of mercurial, using the mercurial eclipse plugin . I created and added the .hgsub file in the root repo, put all the mappings of the sub repos in it, and committed + pushed. Then, I pulled the root repo in eclipse, using import-mercurial. Now I see that all the subrepos appear as though they are unversioned (no "orange cylinder" icon next to their corresponding folders in the eclipse file explorer). Furthermore, when I right click on one of the subrepos, I don't get all the hg commands in the "team" menu as I usually get, with root projects - no "pull", "push" etc. Also, when I made a change to a file in a subrepo, and then "committed" the root project, it told me there were no changes found.

    Read the article

  • How to minimize total cost of shortest path tree

    - by Michael
    I have a directed acyclic graph with positive edge-weights. It has a single source and a set of targets (vertices furthest from the source). I find the shortest paths from the source to each target. Some of these paths overlap. What I want is a shortest path tree which minimizes the total sum of weights over all edges. For example, consider two of the targets. Given all edge weights equal, if they share a single shortest path for most of their length, then that is preferable to two mostly non-overlapping shortest paths (fewer edges in the tree equals lower overall cost). Another example: two paths are non-overlapping for a small part of their length, with high cost for the non-overlapping paths, but low cost for the long shared path (low combined cost). On the other hand, two paths are non-overlapping for most of their length, with low costs for the non-overlapping paths, but high cost for the short shared path (also, low combined cost). There are many combinations. I want to find solutions with the lowest overall cost, given all the shortest paths from source to target. Does this ring any bells with anyone? Can anyone point me to relevant algorithms or analogous applications? Cheers!

    Read the article

  • ASP.NET Deployment under IIS7/VS2010 as Web Application

    - by adchased
    I transformed my VS2008 ASP.NET Website to a "Web Application" today using VS2010. So now it's possible to build a Deployment Package. A Zip Package which can be direclty imported into IIS7. Usually I added a website in IIS7 called mydomain.com and put everything in its root dir. That worked. However, since I converted to an Web Application, this Application is added beneath my "Website container". Now I'm confused, this is how it actually looks now when I try to open the website: Browsing to mydomain.com says 404 ERROR. Browsing to mydomain.com/mydomain.com opens the actual website, but in a subfolder instead of the root directory. (The Application is named after the Domain) How to make this application the root of the website now? I want the application to run under the mydomain.com ROOT and not some subfolder. Thanks a lot!

    Read the article

  • Problem building a complete binary tree of height 'h' in Python

    - by Jack
    Here is my code. The complete binary tree has 2^k nodes at depth k. class Node: def __init__(self, data): # initializes the data members self.left = None self.right = None self.data = data root = Node(data_root) def create_complete_tree(): row = [root] for i in range(h): newrow = [] for node in row: left = Node(data1) right = Node(data2) node.left = left node.right = right newrow.append(left) newrow.append(right) row = copy.deepcopy(newrow) def traverse_tree(node): if node == None: return else: traverse_tree(node.left) print node.data traverse_tree(node.right) create_complete_tree() print 'Node traversal' traverse_tree(root) The tree traversal only gives the data of root and its children. What am I doing wrong?

    Read the article

  • How to set a global before PHPUnit's skeleton-test is run

    - by ministerOfPower
    We set a global in our prepend file used to form the path for our require_once calls. For example: require_once($GLOBALS['root'].'/library/particleboard/JsonUtil.php'); Problem is, when I run PHPUnit's skeleton test builder, the prepend file is not run, so the global is never set. When I run cd /company/trunk/queue/process; phpunit --skeleton-test QueueProcessView PHPUnit tries to resolve a require_once in QueueProcessView, but since the $GLOBALS['root'] is never set, I get a fatal error when including the required file. For example, to PHPUnit, what should be require_once(/code/trunk/library/particleboard/JsonUtil.php) is resolved as require_once(/library/particleboard/JsonUtil.php) Notice the missing root. Does anyone know if the skeleton-test code has some way to call PHP file before it is run? In this I could set my GLOBAL['root'] in this file. Any other creative solutions would be appreciated.

    Read the article

  • Reloading Rails Directories on Change for Dev: Not Lib!

    - by yar
    I have checked out several questions on this, including all of those you see next to the question. Unfortunately, I'm not working with a plugin, and I don't want to work in lib. I have a directory called File.join(Rails.root, 'classes') and I'd like the classes in this directory to reload automatically in dev. In my environment.rb I have this line config.load_paths << File.join(Rails.root, 'classes') which works fine and blows up if the path isn't there. The reloading line in my development.rb also works fine require_dependency File.join(Rails.root, 'classes', 'blah.rb') which blows up if the file is not there (a good sign). However, the file doesn't reload. This all works if the file is in the root of lib and I use the require_dependency line, but my whole point is to get stuff out of lib as suggested here.

    Read the article

  • Overloading assignment operator in C++

    - by jasonline
    As I've understand, when overloading operator=, the return value should should be a non-const reference. A& A::operator=( const A& ) { // check for self-assignment, do assignment return *this; } It is non-const to allow non-const member functions to be called in cases like: ( a = b ).f(); But why should it return a reference? In what instance will it give a problem if the return value is not declared a reference, let's say return by value?

    Read the article

  • converting a Tree to newick format. java

    - by Esmond
    I'm having problems converting a binary rooted tree to newick format. The full explanation for such a format can be found: http://code.google.com/p/mrsrf/wiki/NewickTree An example of a newick format would be as follows: for a tree T such as http://www.cs.mcgill.ca/~cs251/OldCourses/1997/topic8/images/completetreetwo.gif the newick representation would be: (((8,9),(10,11)),((12,13),(14,15))) the internal node will become the commas while the leaves will be retained. such trees have internal nodes which will always have 2 children. I have a problem using recursion to come out with this newick format. The output contains far too many nodes and braces. Any comments to resolve this problem is appreciated or even an iterative algorithm would be welcomed import java.util.Stack; public class Tree { .... public String inOrderNewick(Node root, String output) throws ItemNotFoundException { if (root.hasChild()) { output += "("; output += inOrderNewick(root.child1, output); output += ","; output += inOrderNewick(root.child2, output); output += ")"; return output; } else { output += root.getSeq(); return output; } } }

    Read the article

  • for ps aux what are Ss Sl Ssl proccess types UNIX

    - by JiminyCricket
    when doing a "ps aux" command I get some process listed as Ss, Ssl and Sl what do these mean? root 24653 0.0 0.0 2256 8 ? Ss Apr12 0:00 /bin/bash -c /usr/bin/python /var/python/report_watchman.py root 24654 0.0 0.0 74412 88 ? Sl Apr12 0:01 /usr/bin/python /var/python/report_watchman.py root 21976 0.0 0.0 2256 8 ? Ss Apr14 0:00 /bin/bash -c /usr/bin/python /var/python/report_watchman.py root 21977 0.0 0.0 73628 88 ? Sl Apr14 0:01 /usr/bin/python /var/python/report_watchman.py

    Read the article

  • java.io.FileNotFoundException (Permission denied) When trying to write to the Android sdcard

    - by joefischer1
    I am trying to select an image file from the photo gallery and write to the sdcard. Below is the code that results in an exception. It appears to throw this exception when trying to create the FileOutputStream. I have the following line added to the manifest file nested inside the application element. I can't find a solution to the problem: <uses-permission android:name="android.permission.WRITE_EXTERNAL_STORAGE" /> public boolean saveSelectedImage( Uri selectedImage, int imageGroup, int imageNumber ) { boolean exception = false; InputStream input = null; OutputStream output = null; if( externalStorageIsWritable() ) { try { ContentResolver content = ctx.getContentResolver(); input = content.openInputStream( selectedImage ); if(input != null) Log.v( CLASS_NAME, "Input Stream Opened successfully"); File outFile = null; File root = Environment.getExternalStorageDirectory( ); if(root == null) Log.v(CLASS_NAME, "FAILED TO RETRIEVE DIRECTORY"); else Log.v(CLASS_NAME, "ROOT DIRECTORY is:"+root.toString()); output = new FileOutputStream( root+"/Image"+ imageGroup + "_" + imageNumber + ".png" ); if(output != null) Log.e( CLASS_NAME, "Output Stream Opened successfully"); // output = new FileOutputStream // ("/sdcard/Image"+imageGroup+"_"+imageNumber+".png"); byte[] buffer = new byte[1000]; int bytesRead = 0; while ( ( bytesRead = input.read( buffer, 0, buffer.length ) ) >= 0 ) { output.write( buffer, 0, buffer.length ); } } catch ( Exception e ) { Log.e( CLASS_NAME, "Exception occurred while moving image: "); e.printStackTrace(); exception = true; } finally { // if(input != null)input.close(); // if(output != null)output.close(); // if (exception ) return false; } return true; } else return false; }

    Read the article

  • can't login to phpmyadmin

    - by user574383
    Hi, i am new at linux but i need phpmyadmin on my centos server. I did this: cd /var/www/html/ (document root of apache) wget http://sourceforge.net/projects/phpmyadmin/path/to/latest/version tar xvfz phpMyAdmin-3.3.9-all-languages.tar.gz mv phpMyAdmin-3.3.9-all-languages phpmyadmin rm phpMyAdmin-3.3.9-all-languages.tar.gz cd phpmyadmin/ cp config.sample.inc.php config.inc.php Ok so then i just got to a webbrowser and go to www.$ip/phpmyadmin and i am presented with a login screen asking for username and password. How can i get these credentials to log in? I'd like to log in as root i guess. But i don't know how to setup a root account and create a password for root using the cli and mysql. Please help? Thanks.

    Read the article

< Previous Page | 175 176 177 178 179 180 181 182 183 184 185 186  | Next Page >