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  • How to setup Mercurial on Mac OS X 10.5.8 for remote access

    - by Abhic
    Hello I have a stable hackintosh box running 10.5.8 at home with py 2.5.1 and mercurial 1.5.2 setup successfully. Now I do not have any ports opened from this box and the associated router nor do I have any web servers running on this system. What are the steps that I have to take to setup this machine to be a remotely available as a mercurial repository? Thank you folks.

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  • Linux software Raid 10 no superblock

    - by Shoshomiga
    I have a software raid 10 with 6 x 2tb hard drives (raid 1 for /boot), ubuntu 10.04 is the os. I had a raid controller failure that put 2 drives out of sync, crashed the system and initially the os didnt boot up and went into initramfs instead, saying that drives were busy but I eventually managed to bring the raid up by stopping and assembling the drives. The os booted up and said that there were filesystem errors, I chose to ignore because it would remount the fs in read-only mode if there was a problem. Everything seemed to be working fine and the 2 drives started to rebuild, I was sure that it was a sata controller failure because I had dma errors in my log files. The os crashed soon after that with ext errors. Now its not bringing up the raid, it says that there is no superblock on /dev/sda2. I tried to reassemble manually with all the device names but it still would not bring up the raid 10 complaining about the missing superblock on sda2, and sda1 was also dropped from the raid 1. When I did examine on the raid10 it says that 1 of the initially failed drives is a spare, the other is spare rebuilding and sda2 is removed. It seems that sda decided to fail right when the system was vulnerable to it because when I boot up a live cd it spews out sda unrecoverable read failures. I have been trying to fix this all week but I'm not sure where to go with this now, I ordered more hard drives because I didn't have a complete backup, but its too late for that now and the only thing I could do is mirror all the hard drives onto the new ones (I'm not sure whether sda was mirrored without errors). On the internet I read that you can recover from this by recreating the array with the same options as when it was made, however because sda is failing I cant use it and I don't want to risk using its mirror instead, so I'm waiting to get another hard drive. I'm also not sure whether to include the out of sync drives or if I can actually use those instead to recover the array. Sorry if this is a mess to read but I've been trying to fix this all day and its late at night now, any thoughts on this would be greatly appreciated. I also did a memtest and changed the motherboard in addition to everything else. EDIT: This is my partition layout Disk /dev/sdb: 2000.4 GB, 2000398934016 bytes 255 heads, 63 sectors/track, 243201 cylinders, total 3907029168 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 4096 bytes I/O size (minimum/optimal): 4096 bytes / 4096 bytes Disk identifier: 0x0009c34a Device Boot Start End Blocks Id System /dev/sdb1 * 2048 511999 254976 83 Linux /dev/sdb2 512000 3904980991 1952234496 83 Linux /dev/sdb3 3904980992 3907028991 1024000 82 Linux swap / Solaris

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  • Solaris SPARC 10 32bit mode

    - by TM.
    I'm looking for a definitive answer, does Solaris 10 running on a SPARC machine support booting into 32bit mode? I've found one site that states Solaris 8 was the last version that supported booting in a 32bit mode for SPARC. I've read multiple items that explain how to boot Solaris into 32bit mode, however they did not list the Solaris version. We've tried all the ways specified, but the system keeps booting into 64bit mode.

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  • Snow Leopard 10.6.3 Freezes Frequently

    - by Abhishek
    Snow Leopard 10.6.3 on my Macbook Pro freezes quite frequently now. It freezes for few seconds and then works fine. During that time trackpad does not work and keyboard works partially (missing keystrokes while I type). Initially it was once or twice in a day but now it has become quite frequent. Is somebody else facing similar issue?

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  • Tracing slow Java I/O in Solaris 10

    - by antispam
    We have a Java application which is significantly slower in Solaris 10 server than in Windows PC. We have profiled (-Xprof) the application and observed that UnixFileSystem.getBooleanAttributes0 method consumes about 40% CPU time with native calls. How can we follow our search to identify which is the cause of the slow behaviour? Thank you very much.

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  • LDAP hangs for 10-15 minutes if user put wrong credentials

    - by danny
    Hi: I am using a windows 2003 server .I am using LDAP to allow my wireless clients connect to the network. I can connect fine to the network. But whenever I enter a wrong credential my LDAP server freezes and no new clients can log in to the wireless network for 10 -15 minutes. i am using a cisco wlc and its doing web-authentication.

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  • Solaris 10 very slow ssh file transfers

    - by user133080
    Trying to copy a few TBs betweek Solaris 10 u9 systems A single scp only seems to be able to transfer around 120MB/min, over a 1GB network. If I run multiple scp copies, each one will do 120MB/min, so it is not the network as far as I can see. Any hints on how to tweak the Solaris settings to open a bigger pipe. Have the same problem with another piece of software that unfortunately does not seem to be able to be split into separate processes.

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  • DirectX 9 or DirectX 10 for starters ??

    - by numerical25
    I want to do projects to make my resume more appealing to game companies. So I am going to start buying books. But I don't know rather to read DirectX 9 or 10 api books to start off with. DirectX10 is great, but it seems the industry is moving slow to 10. so should I use 9 or go with 10 ??

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  • Weird Java Math ,10 ^ 1 = 11?

    - by Simon
    For an exercise I was writing a loop that turns a string into an integer without using the built in functions by multiplying each individual value by its numerical position. 75 would be 7*(10 ^ 1) + 5*(10 ^ 0), for example. However, for some reason (10 ^ 1) keeps coming back as 11. Is there some mistake I have made or an explanation for this?

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  • Kill a 10 minute old zombie process in linux bash script

    - by Steve
    I've been tinkering with a regex answer by yukondude with little success. I'm trying to kill processes that are older than 10 minutes. I already know what the process IDs are. I'm looping over an array every 10 min to see if any lingering procs are around and need to be killed. Anybody have any quick thoughts on this? Thanks, Steve ps -eo uid,pid,etime 3233332 | egrep ' ([0-9]+-)?([0-9]{2}:?){3}' | awk '{print $2}' | xargs -I{} kill {} I've been tinkering with the answer posted by yukondude with little success. I'm trying to kill processes that are older than 10 minutes. I already know what the process IDs are. I'm looping over an array every 10 min to see if any lingering procs are around and need to be killed. Anybody have any quick thoughts on this? Thanks, Steve

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  • If I want to play the same sound 10 times per second, must I have 10 copies of that sound in memory?

    - by mystify
    I have a sound that needs to get played 10 times per second. The sound is 1 second long. So it does overlap like 10 times. However, as far as I understand the Finch sound library, I would need 10 different instances of a sound in place so that I can play it 10 times at almost the same time. When I have just one instance, the sound would stop and play from the beginning on every iteration, but not overlap with itself. How to do that?

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  • How to convert <font size="10"> to px?

    - by marknt15
    Hi, I need to convert <font size="10"> to px. Example only(not correct): <font size="10"> is equivalent to 12px. Is there any formula or table conversion out there to convert <font size="10"> to px? Thanks :) Kind Regards, Mark PHP Developer from Philippines

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  • How to match ColdFusion encryption with Java 1.4.2?

    - by JohnTheBarber
    * sweet - thanks to Edward Smith for the CF Technote that indicated the key from ColdFusion was Base64 encoded. See generateKey() for the 'fix' My task is to use Java 1.4.2 to match the results a given ColdFusion code sample for encryption. Known/given values: A 24-byte key A 16-byte salt (IVorSalt) Encoding is Hex Encryption algorithm is AES/CBC/PKCS5Padding A sample clear-text value The encrypted value of the sample clear-text after going through the ColdFusion code Assumptions: Number of iterations not specified in the ColdFusion code so I assume only one iteration 24-byte key so I assume 192-bit encryption Given/working ColdFusion encryption code sample: <cfset ThisSalt = "16byte-salt-here"> <cfset ThisAlgorithm = "AES/CBC/PKCS5Padding"> <cfset ThisKey = "a-24byte-key-string-here"> <cfset thisAdjustedNow = now()> <cfset ThisDateTimeVar = DateFormat( thisAdjustedNow , "yyyymmdd" )> <cfset ThisDateTimeVar = ThisDateTimeVar & TimeFormat( thisAdjustedNow , "HHmmss" )> <cfset ThisTAID = ThisDateTimeVar & "|" & someOtherData> <cfset ThisTAIDEnc = Encrypt( ThisTAID , ThisKey , ThisAlgorithm , "Hex" , ThisSalt)> My Java 1.4.2 encryption/decryption code swag: package so.example; import java.security.*; import javax.crypto.Cipher; import javax.crypto.spec.IvParameterSpec; import javax.crypto.spec.SecretKeySpec; import org.apache.commons.codec.binary.*; public class SO_AES192 { private static final String _AES = "AES"; private static final String _AES_CBC_PKCS5Padding = "AES/CBC/PKCS5Padding"; private static final String KEY_VALUE = "a-24byte-key-string-here"; private static final String SALT_VALUE = "16byte-salt-here"; private static final int ITERATIONS = 1; private static IvParameterSpec ivParameterSpec; public static String encryptHex(String value) throws Exception { Key key = generateKey(); Cipher c = Cipher.getInstance(_AES_CBC_PKCS5Padding); ivParameterSpec = new IvParameterSpec(SALT_VALUE.getBytes()); c.init(Cipher.ENCRYPT_MODE, key, ivParameterSpec); String valueToEncrypt = null; String eValue = value; for (int i = 0; i < ITERATIONS; i++) { // valueToEncrypt = SALT_VALUE + eValue; // pre-pend salt - Length > sample length valueToEncrypt = eValue; // don't pre-pend salt Length = sample length byte[] encValue = c.doFinal(valueToEncrypt.getBytes()); eValue = Hex.encodeHexString(encValue); } return eValue; } public static String decryptHex(String value) throws Exception { Key key = generateKey(); Cipher c = Cipher.getInstance(_AES_CBC_PKCS5Padding); ivParameterSpec = new IvParameterSpec(SALT_VALUE.getBytes()); c.init(Cipher.DECRYPT_MODE, key, ivParameterSpec); String dValue = null; char[] valueToDecrypt = value.toCharArray(); for (int i = 0; i < ITERATIONS; i++) { byte[] decordedValue = Hex.decodeHex(valueToDecrypt); byte[] decValue = c.doFinal(decordedValue); // dValue = new String(decValue).substring(SALT_VALUE.length()); // when salt is pre-pended dValue = new String(decValue); // when salt is not pre-pended valueToDecrypt = dValue.toCharArray(); } return dValue; } private static Key generateKey() throws Exception { // Key key = new SecretKeySpec(KEY_VALUE.getBytes(), _AES); // this was wrong Key key = new SecretKeySpec(new BASE64Decoder().decodeBuffer(keyValueString), _AES); // had to un-Base64 the 'known' 24-byte key. return key; } } I cannot create a matching encrypted value nor decrypt a given encrypted value. My guess is it's something to do with how I'm handling the initial vector/salt. I'm not very crypto-savvy but I'm thinking I should be able to take the sample clear-text and produce the same encrypted value in Java as ColdFusion produced. I am able to encrypt/decrypt my own data with my Java code (so I'm consistent) but I cannot match nor decrypt the ColdFusion sample encrypted value. I have access to a local webservice that can test the encrypted output. The given ColdFusion output sample passes/decrypts fine (of course). If I try to decrypt the same sample with my Java code (using the actual key and salt) I get a "Given final block not properly padded" error. I get the same net result when I pass my attempt at encryption (using the actual key and salt) to the test webservice. Any Ideas?

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  • Vim hanging after parsing .vimrc (even a blank one) file on Solaris 10

    - by Seamus
    Hello all, I am having a problem with vim 7.2 hanging (for about 10 seconds) after it parses the .vimrc file. I had a similar issue in the past with tcsh on linux, but it was resolved by setting TERM to xterm-color. The same does not resolve the issue here. Any idea what may be causing this? $ env USER=redacted LOGNAME=redacted HOME=/home/redacted PATH=redacted MAIL=/var/spool/mail/redacted SHELL=/bin/tcsh TZ=redacted LC_COLLATE=C SSH_CLIENT=redacted SSH_CONNECTION=redacted SSH_TTY=/dev/pts/11 TERM=dtterm HOSTTYPE=sun4 VENDOR=sun OSTYPE=solaris MACHTYPE=sparc SHLVL=1 PWD=/home/redacted GROUP=redacted HOST=redacted REMOTEHOST=redacted QUOTA_CHECKED=1 WHOAMI=redacted HOSTNAME=redacted EDITOR=vim PRINTER=redacted INFOPATH=/software/gnu/gcc/2.8.1/sun4os5.10/info:/software/gnu/sun4os5/info:/software/gnu/emacs/20.3.1/sun4os5/info:/software/gnuish/sun4os5/info:/usr/local/gnu/info MANPATH=/software/gnu/gcc/2.8.1/sun4os5.10/man:/software/gnu/sun4os5/man:/software/gnu/emacs/20.3.1/sun4os5/man:/opt/rational/clearcase/doc/man:/usr/openwin/man:/usr/share/man:/usr/local/man:/usr/dt/man:/software/gnuish/sun4os5/man H_ARCH=sun4 H_ARCHOS=sun4os5 H_ARCHOS_SUB=sun4os5.10 H_OSTYPE=SUNOS H_OSREV=51000 T_ARCH=sun4 T_ARCHOS=sun4os5 T_ARCHOS_SUB=sun4os5.10 T_OSTYPE=SUNOS T_OSREV=51000 X11HOME=/usr/local/x11/sun4os5 OPENWINHOME=/usr/openwin LD_LIBRARY_PATH=/usr/dt/lib:/usr/openwin/lib MOTIFHOME=/usr/dt XINITRC=/usr/openwin/lib/Xinitrc GCC_REV=281

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  • Fastest way to find the largest power of 10 smaller than x

    - by peoro
    Is there any fast way to find the largest power of 10 smaller than a given number? I'm using this algorithm, at the moment, but something inside myself dies anytime I see it: 10**( int( math.log10(x) ) ) # python pow( 10, (int) log10(x) ) // C I could implement simple log10 and pow functions for my problems with one loop each, but still I'm wondering if there is some bit magic for decimal numbers.

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  • Select top/latest 10 in couchdb?

    - by drozzy
    How would I execute a query equivalent to "select top 10" in couch db? For example I have a "schema" like so: title body modified and I want to select the last 10 modified documents. As an added bonus if anyone can come up with a way to do the same only per category. So for: title category body modified return a list of latest 10 documents in each category. I am just wandering if such a query is possible in couchdb.

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  • installation of mac 10.6.2

    - by mahesh
    hi i installed vmware workstation 7.0 in windows,and it working well.i installed mac os 10.4.8 in it as a guest it is alo working well,but while installin mac os 10.6.2 it dont installed and shows an error as "invalid front-side bus freqency 66000000 hz. disablin the cpu." please help me how to solve this and please suggest any link to easyly understand installation if mac os 10.6.2

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  • New Skool Crosstabbing

    - by Tim Dexter
    A while back I spoke about having to go back to BIP's original crosstabbing solution to achieve a certain layout. Hok Min has provided a 'man' page for the new crosstab/pivot builder for 10.1.3.4.1 users. This will make the documentation drop but for now, get it here! The old, hand method is still available but this new approach, is more efficient and flexible. That said you may need to get into the crosstab code to tweak it where the crosstab dialog can not help. I had to do this, this week but more on that later. The following explains how the crosstab wizard builds the crosstab and what the fields inside the resulting template structure are there for. To create the crosstab a new XDO command "<?crosstab:...?>" has been created. XDO Command: <?crosstab: ctvarname; data-element; rows; columns; measures; aggregation?> Parameter Description Example Ctvarname Crosstab variable name. This is automatically generated by the Add-in. C123 data-element This is the XML data element that contains the data. "//ROW" Rows This contains a list of XML elements for row headers. The ordering information is specified within "{" and "}". The first attribute is the sort element. Leaving it blank means the sort element is the same as the row header element. The attribute "o" means order. Its value can be "a" for ascending, or "d" for descending. The attribute "t" means type. Its value can be "t" for text, and "n" for numeric. There can be more than one sort elements, example: "emp-full-name {emp-lastname,o=a,t=n}{emp-firstname,o=a,t=n}. This will sort employee by last name and first name. "Region{,o=a,t=t}, District{,o=a,t=t}" In the example, the first row header is "Region". It is sort by "Region", order is ascending, and type is text. The second row header is "District". It is sort by "District", order is ascending, and type is text. Columns This contains a list of XML elements for columns headers. The ordering information is specified within "{" and "}". The first attribute is the sort element. Leaving it blank means the sort element is the same as the column header element. The attribute "o" means order. Its value can be "a" for ascending, or "d" for descending. The attribute "t" means type. Its value can be "t" for text, and "n" for numeric. There can be more than one sort elements, example: "emp-full-name {emp-lastname,o=a,t=n}{emp-firstname,o=a,t=n}. This will sort employee by last name and first name. "ProductsBrand{,o=a,t=t}, PeriodYear{,o=a,t=t}" In the example, the first column header is "ProductsBrand". It is sort by "ProductsBrand", order is ascending, and type is text. The second column header is "PeriodYear". It is sort by "District", order is ascending, and type is text. Measures This contains a list of XML elements for measures. "Revenue, PrevRevenue" Aggregation The aggregation function name. Currently, we only support "sum". "sum" Using the Oracle BI Publisher Template Builder for Word add-in, we are able to construct the following Pivot Table: The generated XDO command for this Pivot Table is as follow: <?crosstab:c547; "//ROW";"Region{,o=a,t=t}, District{,o=a,t=t}"; "ProductsBrand{,o=a,t=t},PeriodYear{,o=a,t=t}"; "Revenue, PrevRevenue";"sum"?> Running the command on the give XML data files generates this XML file "cttree.xml". Each XPath in the "cttree.xml" is described in the following table. Element XPath Count Description C0 /cttree/C0 1 This contains elements which are related to column. C1 /cttree/C0/C1 4 The first level column "ProductsBrand". There are four distinct values. They are shown in the label H element. CS /cttree/C0/C1/CS 4 The column-span value. It is used to format the crosstab table. H /cttree/C0/C1/H 4 The column header label. There are four distinct values "Enterprise", "Magicolor", "McCloskey" and "Valspar". T1 /cttree/C0/C1/T1 4 The sum for measure 1, which is Revenue. T2 /cttree/C0/C1/T2 4 The sum for measure 2, which is PrevRevenue. C2 /cttree/C0/C1/C2 8 The first level column "PeriodYear", which is the second group-by key. There are two distinct values "2001" and "2002". H /cttree/C0/C1/C2/H 8 The column header label. There are two distinct values "2001" and "2002". Since it is under C1, therefore the total number of entries is 4 x 2 => 8. T1 /cttree/C0/C1/C2/T1 8 The sum for measure 1 "Revenue". T2 /cttree/C0/C1/C2/T2 8 The sum for measure 2 "PrevRevenue". M0 /cttree/M0 1 This contains elements which are related to measures. M1 /cttree/M0/M1 1 This contains summary for measure 1. H /cttree/M0/M1/H 1 The measure 1 label, which is "Revenue". T /cttree/M0/M1/T 1 The sum of measure 1 for the entire xpath from "//ROW". M2 /cttree/M0/M2 1 This contains summary for measure 2. H /cttree/M0/M2/H 1 The measure 2 label, which is "PrevRevenue". T /cttree/M0/M2/T 1 The sum of measure 2 for the entire xpath from "//ROW". R0 /cttree/R0 1 This contains elements which are related to row. R1 /cttree/R0/R1 4 The first level row "Region". There are four distinct values, they are shown in the label H element. H /cttree/R0/R1/H 4 This is row header label for "Region". There are four distinct values "CENTRAL REGION", "EASTERN REGION", "SOUTHERN REGION" and "WESTERN REGION". RS /cttree/R0/R1/RS 4 The row-span value. It is used to format the crosstab table. T1 /cttree/R0/R1/T1 4 The sum of measure 1 "Revenue" for each distinct "Region" value. T2 /cttree/R0/R1/T2 4 The sum of measure 1 "Revenue" for each distinct "Region" value. R1C1 /cttree/R0/R1/R1C1 16 This contains elements from combining R1 and C1. There are 4 distinct values for "Region", and four distinct values for "ProductsBrand". Therefore, the combination is 4 X 4 è 16. T1 /cttree/R0/R1/R1C1/T1 16 The sum of measure 1 "Revenue" for each combination of "Region" and "ProductsBrand". T2 /cttree/R0/R1/R1C1/T2 16 The sum of measure 2 "PrevRevenue" for each combination of "Region" and "ProductsBrand". R1C2 /cttree/R0/R1/R1C1/R1C2 32 This contains elements from combining R1, C1 and C2. There are 4 distinct values for "Region", and four distinct values for "ProductsBrand", and two distinct values of "PeriodYear". Therefore, the combination is 4 X 4 X 2 è 32. T1 /cttree/R0/R1/R1C1/R1C2/T1 32 The sum of measure 1 "Revenue" for each combination of "Region", "ProductsBrand" and "PeriodYear". T2 /cttree/R0/R1/R1C1/R1C2/T2 32 The sum of measure 2 "PrevRevenue" for each combination of "Region", "ProductsBrand" and "PeriodYear". R2 /cttree/R0/R1/R2 18 This contains elements from combining R1 "Region" and R2 "District". Since the list of values in R2 has dependency on R1, therefore the number of entries is not just a simple multiplication. H /cttree/R0/R1/R2/H 18 The row header label for R2 "District". R1N /cttree/R0/R1/R2/R1N 18 The R2 position number within R1. This is used to check if it is the last row, and draw table border accordingly. T1 /cttree/R0/R1/R2/T1 18 The sum of measure 1 "Revenue" for each combination "Region" and "District". T2 /cttree/R0/R1/R2/T2 18 The sum of measure 2 "PrevRevenue" for each combination of "Region" and "District". R2C1 /cttree/R0/R1/R2/R2C1 72 This contains elements from combining R1, R2 and C1. T1 /cttree/R0/R1/R2/R2C1/T1 72 The sum of measure 1 "Revenue" for each combination of "Region", "District" and "ProductsBrand". T2 /cttree/R0/R1/R2/R2C1/T2 72 The sum of measure 2 "PrevRevenue" for each combination of "Region", "District" and "ProductsBrand". R2C2 /cttree/R0/R1/R2/R2C1/R2C2 144 This contains elements from combining R1, R2, C1 and C2, which gives the finest level of details. M1 /cttree/R0/R1/R2/R2C1/R2C2/M1 144 The sum of measure 1 "Revenue". M2 /cttree/R0/R1/R2/R2C1/R2C2/M2 144 The sum of measure 2 "PrevRevenue". Lots to read and digest I know! Customization One new feature I discovered this week is the ability to show one column and sort by another. I had a data set that was extracting month abbreviations, we wanted to show the months across the top and some row headers to the side. As you may know XSL is not great with dates, especially recognising month names. It just wants to sort them alphabetically, so Apr comes before Jan, etc. A way around this is to generate a month number alongside the month and use that to sort. We can do that in the crosstab, sadly its not exposed in the UI yet but its doable. Go back up and take a look a the initial crosstab command. especially the Rows and Columns entries. In there you will find the sort criteria. "ProductsBrand{,o=a,t=t}, PeriodYear{,o=a,t=t}" Notice those leading commas inside the curly braces? Because there is no field preceding them it means that the crosstab should sort on the column before the brace ie PeriodYear. But you can insert another column in the data set to sort by. To get my sort working how I needed. <?crosstab:c794;"current-group()";"_Fund_Type_._Fund_Type_Display_{_Fund_Type_._Fund_Type_Sort_,o=a,t=n}";"_Fiscal_Period__Amount__._Amt_Fm_Disp_Abbr_{_Fiscal_Period__Amount__._Amt_Fiscal_Month_Sort_,o=a,t=n}";"_Execution_Facts_._Amt_";"sum"?> Excuse the horribly verbose XML tags, good ol BIEE :0) The emboldened columns are not in the crosstab but are in the data set. I just opened up the field, dropped them in and changed the type(t) value to be 'n', for number, instead of the default 'a' and my crosstab started sorting how I wanted it. If you find other tips and tricks, please share in the comments.

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  • management network to a network port for additional ones munin and monit

    - by paolo
    management network to a network port for additional ones munin and monit I want to build a separate Netzwek for server management. I have several network cards a linux / debian / ubuntu with computer. Set both network cards sin in the /etc/network/interfaces. # The primary network interface #allow-hotplug eth0 #iface eth0 inet dhcp auto eth0 iface eth0 inet static address 10.0.0.240 netmast 255.255.255.0 network 10.0.0.0 brodacast 10.0.0.255 gateway 10.0.0.254 auto eth1 iface eth1 inet static address 10.0.10.240 netmast 255.255.255.0 network 10.0.10.0 brodacast 10.0.10.255 post-up ip route add 10.0.0.0/24 dev eth0 src 10.0.0.240 table eth0-WAN post-up ip route add default via 10.0.0.254 table eth0-WAN post-up ip route add 10.0.10.0/24 dev eth1 src 10.0.10.240 table eth1-LAN post-up ip route add default via 10.0.10.200 table eth1-LAN post-up ip rule add from 10.0.0.240 table eth0-WAN post-up ip rule add from 10.0.10.240 table eth1-LAN still i adjusted / etc/iproute2/rt_tables and following routes set up in the /etc/network/interfaces I want to have both applications and the network interface separately as munin and monit only on eth1 and not have to eth0. it goes to the reboot but sometimes not always. # Traceroute-i eth1 10.0.10.200 not go what am I doing wrong?

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  • No client internet access when setting up these iptables rules

    - by Siriss
    I have read many other posts but cannot figure this out. eth0 is my external connected to a Comcast modem. The server has internet access with no issues. eth1 is internal and running DHCP for the clients. I have DHCP working just fine, all my clients can get an IP and ping the server but they cannot access the internet. I am using ISC-DHCP-SERVER and have set /etc/default/isc-dhcp-server to INTERFACE="eht1" Here is my dhcpd.conf file located in /etc/dhcp/dhcpd.conf ddns-update-style interim; ignore client-updates; subnet 10.0.10.0 netmask 255.255.255.0 { range 10.0.10.10 10.0.10.200; option routers 10.0.10.2; option subnet-mask 255.255.255.0; option domain-name-servers 208.67.222.222, 208.67.220.220; #OpenDNS # option domain-name "example.com"; default-lease-time 21600; max-lease-time 43200; authoritative; } I have made the *net.ipv4.ip_forward=1* change in /etc/sysctl.conf here is my interfaces file: auto lo iface lo inet loopback auto eth0 iface eth0 inet dhcp iface eth1 inet static address 10.0.10.2 netmask 255.255.255.0 network 10.0.10.0 auto eth1 And finally- here is my iptables.conf file: # Firewall configuration written by system-config-firewall # Manual customization of this file is not recommended. *nat :PREROUTING ACCEPT [0:0] :OUTPUT ACCEPT [0:0] :POSTROUTING ACCEPT [0:0] -A POSTROUTING -s 10.0.10.0/24 -o eth0 -j MASQUERADE #-A PREROUTING -i eth0 -p tcp --dport 59668 -j DNAT --to-destination 10.0.10.2:59668 COMMIT *filter :INPUT ACCEPT [0:0] :FORWARD ACCEPT [0:0] :OUTPUT ACCEPT [0:0] -A INPUT -m state --state ESTABLISHED,RELATED -j ACCEPT -A INPUT -p icmp -j ACCEPT -A INPUT -i lo -j ACCEPT -A INPUT -i eth1 -j ACCEPT -A INPUT -m state --state NEW -m tcp -p tcp --dport 22 -j ACCEPT -A INPUT -m state --state NEW -m tcp -p tcp --dport 80 -j ACCEPT -A INPUT -m state --state NEW -m tcp -p tcp --dport 443 -j ACCEPT -A INPUT -m state --state NEW -m tcp -p tcp --dport 53 -j ACCEPT -A INPUT -m state --state NEW -m udp -p udp --dport 53 -j ACCEPT -A FORWARD -s 10.0.10.0/24 -o eth0 -j ACCEPT -A FORWARD -d 10.0.10.0/24 -m state --state ESTABLISHED,RELATED -i eth0 -j ACCEPT -A FORWARD -p icmp -j ACCEPT -A FORWARD -i lo -j ACCEPT -A FORWARD -i eth1 -j ACCEPT #-A FORWARD -i eth0 -m state --state NEW -m tcp -p tcp -d 10.0.10.2 --dport 59668 -j ACCEPT -A INPUT -j REJECT --reject-with icmp-host-prohibited -A FORWARD -j REJECT --reject-with icmp-host-prohibited COMMIT I am completely stuck. I cannot figure out why the clients cannot access the internet. Am I missing a service? Is a service not running? Any help would be greatly appreciated. I tried to be as thorough as possible but please let me know if I have missed something. Thank you!

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