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  • What the best way to convert from String to HashMap?

    - by eugenn
    I would like to serialize a Java HashMap to string representation. The HashMap will contains only primitive values like string and integer. After that this string will be stored to db. How to restore back the HashMap? Is it make sense to use BeanUtils and interface Converter or use JSON? For example: List list = new ArrayList(); list.add(new Long(1)); list.add(new Long(2)); list.add(new Long(4)); Map map = new HashMap(); map.put("cityId", new Integer(1)); map.put("name", "test"); map.put("float", new Float(-3.2)); map.put("ids", list); map.toString() -> {float=-3.2,ids=[1, 2, 4],name=test,cityId=1} map.toJSON -> {"float":-3.2,"ids":[1,2,4],"name":"test","cityId":1}

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  • How can I put the results of Cursor into String[]. For more detail, please refer to the code below

    - by Hicen
    public class DisplayCustomersActivity extends Activity implements Button.OnClickListener { private SalesDB sdb; private ListView lvDCList; @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.displaycustomers); lvDCList = (ListView) findViewById(R.id.lvDCList); sdb = new SalesDB(this); SQLiteDatabase db = sdb.getReadableDatabase(); Cursor results = db.query(sdb.TABLE_CUSTOMER, new String[] {sdb.CUSTOMER_ID, sdb.CUSTOMER_NAME, sdb.CUSTOMER_GENDER}, null, null, null, null, null); int resultCount = results.getCount(); String[] customers = new String[resultCount]; if (resultCount == 0 || !results.moveToFirst()) { customers = null; } else { for(int i=0; i<resultCount; i++) { //Process results to String array here ... ... results.moveToNext(); } } }

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  • how do I check if a c++ string is an int?

    - by user342231
    when I use getline, I would input a bunch of strings or numbers, but I only want the while loop to output the "word" if it is not a number. so is there any way to check if "word" is a number or not, I know I could use atoi() for c-strings but how about for strings of the string class int main () { stringstream ss (stringstream::in | stringstream::out); string word; string str; getline(cin,str); ss<<str; while(ss>>word) { //if( ) cout<<word<<endl; } }

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  • How can I make my Google Maps api v3 address search bar work by hitting the enter button on the keyboard?

    - by Gavin
    I'm developing a webpage and I would just like to make something more user friendly. I have a functional Google Maps api v3 and an address search bar. Currently, I have to use the mouse to select search to initialize the geocoding function. How can I make the map return a placemark by hitting the enter button on my keyboard? I just want to make it as user-friendly as possible. Here is the javascript and div, respectively, I created for the address bar: var geocoder; function initialize() { geocoder = new google.maps.Geocoder (); function codeAddress () { var address = document.getElementById ("address").value; geocoder.geocode ( { 'address': address}, function(results, status) { if (status == google.maps.GeocoderStatus.OK) { map.setCenter(results [0].geometry.location); marker.setPosition(results [0].geometry.location); map.setZoom(14); } else { alert("Geocode was not successful for the following reason: " + status); } }); } <div id="geocoder"> <input id="address" type="textbox" value=""> <input type="button" value="Search" onclick="codeAddress()"> </div> Thank you in advance for your help

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  • How to get rid of `deprecated conversion from string constant to ‘char*’` warnings in GCC?

    - by Josh Matthews
    So I'm working on an exceedingly large codebase, and recently upgraded to gcc 4.3, which now triggers this warning: warning: deprecated conversion from string constant to ‘char*’ Obviously, the correct way to fix this is to find every declaration like char *s = "constant string"; or function call like void foo(char *s); foo("constant string"); and make them const char pointers. However, that would mean touching 564 files, minimum, which is not a task I wish to perform at this point in time. The problem right now is that I'm running with -werror, so I need some way to stifle these warnings. How can I do that?

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  • How to parse out html links from a huge string with html links and other text (Java).

    - by Robert
    Hello, my question is how would i be able to go through a string and take out only the links and erase all the rest? I thought about using some type of delemiter, but wouldnt know how to go about using it in java. an example of what i am trying to do: this is my String: String myString = "The file is http: // www. .com/hello.txt and the second file is " + "http: // www. .com/hello2.dat"; I would want the output to be: "http: // www. .com/hello.txt http: // www. .com/hello2.dat" or each could be added to an array, separately. I just want some ideas, id like to write the code myself but am having trouble on how to do it. Any help would be awesome.

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  • C: Why does gcc allow char array initialization with string literal larger than array?

    - by Ashwin
    int main() { char a[7] = "Network"; return 0; } A string literal in C is terminated internally with a nul character. So, the above code should give a compilation error since the actual length of the string literal Network is 8 and it cannot fit in a char[7] array. However, gcc (even with -Wall) on Ubuntu compiles this code without any error or warning. Why does gcc allow this and not flag it as compilation error? gcc only gives a warning (still no error!) when the char array size is smaller than the string literal. For example, it warns on: char a[6] = "Network"; [Related] Visual C++ 2012 gives a compilation error for char a[7]: 1>d:\main.cpp(3): error C2117: 'a' : array bounds overflow 1> d:\main.cpp(3) : see declaration of 'a'

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  • sorting set of five string in alphabetical oder throwing warning?

    - by rost rost
    sorting set of five string in alphabetical oder throwing warning ?my code is below please help me to fix it #include<stdio.h> #include<string.h> int main() { char a[5][20],t[20]; int i,j; printf("enter 5 string\n") scanf("%s",a); for(i=1;i<5;i++) { for(j=1;j<5;j++) { if(strcmp(a[j-1],a[j])>0) { strcpy (t,a[j-1]); strcpy (a[j-1],a[j]); strcpy(a[j],t); } } } for(i=1;i<5;i++) printf("%s\n",a[i]); } ~

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  • search a collection for a specific keyword

    - by icelated
    What i want to do is search a hashset with a keyword.. I have 3 classes... main() Library Item(CD, DVD,Book classes) In library i am trying to do my search of the items in the hashsets.. In Item class is where i have the getKeywords function.. here is the Items class... import java.io.PrintStream; import java.util.Collection; import java.util.*; class Item { private String title; private String [] keywords; public String toString() { String line1 = "title: " + title + "\n" + "keywords: " + Arrays.toString(keywords); return line1; } public void print() { System.out.println(toString()); } public Item() { } public Item(String theTitle, String... theKeyword) { this.title = theTitle; this.keywords = theKeyword; } public String getTitle() { return title; } public String [] getKeywords() { return keywords; } } class CD extends Item { private String artist; private String [] members; // private String [] keywords; private int number; public CD(String theTitle, String theBand, int Snumber, String... keywords) { super(theTitle, keywords); this.artist = theBand; this.number = Snumber; // this.keywords = keywords; } public void addband(String... member) { this.members = member; } public String getArtist() { return artist; } public String [] getMembers() { return members; } // public String [] getKeywords() // { // return keywords; //} public String toString() { return "-Music-" + "\n" + "band: " + artist + "\n" + "# songs: " + number + "\n" + "members: " + Arrays.toString(members) + "\n" + super.toString() // + "keywords: " + Arrays.toString(keywords) + "\n" + "\n" ; } public void print() { System.out.println(toString()); } } class DVD extends Item { private String director; private String [] cast; private int scenes; // private String [] keywords; public DVD(String theTitle, String theDirector, int nScenes, String... keywords) { super(theTitle, keywords); this.director = theDirector; this.scenes = nScenes; // this.keywords = keywords; } public void addmoviecast(String... members) { this.cast = members; } public String [] getCast() { return cast; } public String getDirector() { return director; } // public String [] getKeywords() // { // return keywords; // } public String toString() { return "-Movie-" + "\n" + "director: " + director + "\n" + "# scenes: " + scenes + "\n" + "cast: " + Arrays.toString(cast) + "\n" + super.toString() // + "keywords: " + Arrays.toString(keywords) + "\n" + "\n" ; } public void print() { System.out.println(toString()); } } class Book extends Item { private String author; private int pages; public Book(String theTitle, String theAuthor, int nPages, String... keywords) { super(theTitle, keywords); this.author = theAuthor; this.pages = nPages; // this.keywords = keywords; } public String getAuthor() { return author; } //public String [] getKeywords() // { // return keywords; //} public void print() { System.out.println(toString()); } public String toString() { return "-Book-" + "\n" + "Author: " + author + "\n" + "# pages " + pages + "\n" + super.toString() // + "keywords: " + Arrays.toString(keywords) + "\n" + "\n" ; } } I hope i didnt confuse you? I need help with the itemsForKeyword(String keyword) function.. the first keyword being passed in is "science fiction" and i want to search the keywords in the sets and return the matches.. What am i doing so wrong? Thank you

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  • How to find the insertion point in an array using binary search?

    - by ????
    The basic idea of binary search in an array is simple, but it might return an "approximate" index if the search fails to find the exact item. (we might sometimes get back an index for which the value is larger or smaller than the searched value). For looking for the exact insertion point, it seems that after we got the approximate location, we might need to "scan" to left or right for the exact insertion location, so that, say, in Ruby, we can do arr.insert(exact_index, value) I have the following solution, but the handling for the part when begin_index >= end_index is a bit messy. I wonder if a more elegant solution can be used? (this solution doesn't care to scan for multiple matches if an exact match is found, so the index returned for an exact match may point to any index that correspond to the value... but I think if they are all integers, we can always search for a - 1 after we know an exact match is found, to find the left boundary, or search for a + 1 for the right boundary.) My solution: DEBUGGING = true def binary_search_helper(arr, a, begin_index, end_index) middle_index = (begin_index + end_index) / 2 puts "a = #{a}, arr[middle_index] = #{arr[middle_index]}, " + "begin_index = #{begin_index}, end_index = #{end_index}, " + "middle_index = #{middle_index}" if DEBUGGING if arr[middle_index] == a return middle_index elsif begin_index >= end_index index = [begin_index, end_index].min return index if a < arr[index] && index >= 0 #careful because -1 means end of array index = [begin_index, end_index].max return index if a < arr[index] && index >= 0 return index + 1 elsif a > arr[middle_index] return binary_search_helper(arr, a, middle_index + 1, end_index) else return binary_search_helper(arr, a, begin_index, middle_index - 1) end end # for [1,3,5,7,9], searching for 6 will return index for 7 for insertion # if exact match is found, then return that index def binary_search(arr, a) puts "\nSearching for #{a} in #{arr}" if DEBUGGING return 0 if arr.empty? result = binary_search_helper(arr, a, 0, arr.length - 1) puts "the result is #{result}, the index for value #{arr[result].inspect}" if DEBUGGING return result end arr = [1,3,5,7,9] b = 6 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9,11] b = 6 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9] b = 60 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9,11] b = 60 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9] b = -60 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9,11] b = -60 arr.insert(binary_search(arr, b), b) p arr arr = [1] b = -60 arr.insert(binary_search(arr, b), b) p arr arr = [1] b = 60 arr.insert(binary_search(arr, b), b) p arr arr = [] b = 60 arr.insert(binary_search(arr, b), b) p arr and result: Searching for 6 in [1, 3, 5, 7, 9] a = 6, arr[middle_index] = 5, begin_index = 0, end_index = 4, middle_index = 2 a = 6, arr[middle_index] = 7, begin_index = 3, end_index = 4, middle_index = 3 a = 6, arr[middle_index] = 5, begin_index = 3, end_index = 2, middle_index = 2 the result is 3, the index for value 7 [1, 3, 5, 6, 7, 9] Searching for 6 in [1, 3, 5, 7, 9, 11] a = 6, arr[middle_index] = 5, begin_index = 0, end_index = 5, middle_index = 2 a = 6, arr[middle_index] = 9, begin_index = 3, end_index = 5, middle_index = 4 a = 6, arr[middle_index] = 7, begin_index = 3, end_index = 3, middle_index = 3 the result is 3, the index for value 7 [1, 3, 5, 6, 7, 9, 11] Searching for 60 in [1, 3, 5, 7, 9] a = 60, arr[middle_index] = 5, begin_index = 0, end_index = 4, middle_index = 2 a = 60, arr[middle_index] = 7, begin_index = 3, end_index = 4, middle_index = 3 a = 60, arr[middle_index] = 9, begin_index = 4, end_index = 4, middle_index = 4 the result is 5, the index for value nil [1, 3, 5, 7, 9, 60] Searching for 60 in [1, 3, 5, 7, 9, 11] a = 60, arr[middle_index] = 5, begin_index = 0, end_index = 5, middle_index = 2 a = 60, arr[middle_index] = 9, begin_index = 3, end_index = 5, middle_index = 4 a = 60, arr[middle_index] = 11, begin_index = 5, end_index = 5, middle_index = 5 the result is 6, the index for value nil [1, 3, 5, 7, 9, 11, 60] Searching for -60 in [1, 3, 5, 7, 9] a = -60, arr[middle_index] = 5, begin_index = 0, end_index = 4, middle_index = 2 a = -60, arr[middle_index] = 1, begin_index = 0, end_index = 1, middle_index = 0 a = -60, arr[middle_index] = 9, begin_index = 0, end_index = -1, middle_index = -1 the result is 0, the index for value 1 [-60, 1, 3, 5, 7, 9] Searching for -60 in [1, 3, 5, 7, 9, 11] a = -60, arr[middle_index] = 5, begin_index = 0, end_index = 5, middle_index = 2 a = -60, arr[middle_index] = 1, begin_index = 0, end_index = 1, middle_index = 0 a = -60, arr[middle_index] = 11, begin_index = 0, end_index = -1, middle_index = -1 the result is 0, the index for value 1 [-60, 1, 3, 5, 7, 9, 11] Searching for -60 in [1] a = -60, arr[middle_index] = 1, begin_index = 0, end_index = 0, middle_index = 0 the result is 0, the index for value 1 [-60, 1] Searching for 60 in [1] a = 60, arr[middle_index] = 1, begin_index = 0, end_index = 0, middle_index = 0 the result is 1, the index for value nil [1, 60] Searching for 60 in [] [60]

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  • Mysql optimization question - How to apply AND logic in search and limit on results in one query?

    - by sandeepan-nath
    This is a little long but I have provided all the database structures and queries so that you can run it immediately and help me. Run the following queries:- CREATE TABLE IF NOT EXISTS `Tutor_Details` ( `id_tutor` int(10) NOT NULL auto_increment, `firstname` varchar(100) NOT NULL default '', `surname` varchar(155) NOT NULL default '', PRIMARY KEY (`id_tutor`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=41 ; INSERT INTO `Tutor_Details` (`id_tutor`,`firstname`, `surname`) VALUES (1, 'Sandeepan', 'Nath'), (2, 'Bob', 'Cratchit'); CREATE TABLE IF NOT EXISTS `Classes` ( `id_class` int(10) unsigned NOT NULL auto_increment, `id_tutor` int(10) unsigned NOT NULL default '0', `class_name` varchar(255) default NULL, PRIMARY KEY (`id_class`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=229 ; INSERT INTO `Classes` (`id_class`,`class_name`, `id_tutor`) VALUES (1, 'My Class', 1), (2, 'Sandeepan Class', 2); CREATE TABLE IF NOT EXISTS `Tags` ( `id_tag` int(10) unsigned NOT NULL auto_increment, `tag` varchar(255) default NULL, PRIMARY KEY (`id_tag`), UNIQUE KEY `tag` (`tag`), KEY `id_tag` (`id_tag`), KEY `tag_2` (`tag`), KEY `tag_3` (`tag`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=18 ; INSERT INTO `Tags` (`id_tag`, `tag`) VALUES (1, 'Bob'), (6, 'Class'), (2, 'Cratchit'), (4, 'Nath'), (3, 'Sandeepan'), (5, 'My'); CREATE TABLE IF NOT EXISTS `Tutors_Tag_Relations` ( `id_tag` int(10) unsigned NOT NULL default '0', `id_tutor` int(10) default NULL, KEY `Tutors_Tag_Relations` (`id_tag`), KEY `id_tutor` (`id_tutor`), KEY `id_tag` (`id_tag`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1; INSERT INTO `Tutors_Tag_Relations` (`id_tag`, `id_tutor`) VALUES (3, 1), (4, 1), (1, 2), (2, 2); CREATE TABLE IF NOT EXISTS `Class_Tag_Relations` ( `id_tag` int(10) unsigned NOT NULL default '0', `id_class` int(10) default NULL, `id_tutor` int(10) NOT NULL, KEY `Class_Tag_Relations` (`id_tag`), KEY `id_class` (`id_class`), KEY `id_tag` (`id_tag`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1; INSERT INTO `Class_Tag_Relations` (`id_tag`, `id_class`, `id_tutor`) VALUES (5, 1, 1), (6, 1, 1), (3, 2, 2), (6, 2, 2); Following is about the tables:- There are tutors who create classes. Tutor_Details - Stores tutors Classes - Stores classes created by tutors And for searching we are using a tags based approach. All the keywords are stored in tags table (while classes/tutors are created) and tag relations are entered in Tutor_Tag_Relations and Class_Tag_Relations tables (for tutors and classes respectively)like this:- Tags - id_tag tag (this is a a unique field) Tutors_Tag_Relations - Stores tag relations while the tutors are created. Class_Tag_Relations - Stores tag relations while any tutor creates a class In the present data in database, tutor "Sandeepan Nath" has has created class "My Class" and "Bob Cratchit" has created "Sandeepan Class". 3.Requirement The requirement is to return tutor records from Tutor_Details table such that all the search terms (AND logic) are present in the union of these two sets - 1. Tutor_Details table 2. classes created by a tutor in Classes table) Example search and expected results:- Search Term Result "Sandeepan Class" Tutor Sandeepan Nath's record from Tutor Details table "Class" Both the tutors from ... Most importantly, there should be only one mysql query and a LIMIT applicable on the number of results. Following is a working query which I have so far written (It just applies OR logic of search key words instead of the desired AND logic). SELECT td . * FROM Tutor_Details AS td LEFT JOIN Tutors_Tag_Relations AS ttagrels ON td.id_tutor = ttagrels.id_tutor LEFT JOIN Classes AS wc ON td.id_tutor = wc.id_tutor INNER JOIN Class_Tag_Relations AS wtagrels ON td.id_tutor = wtagrels.id_tutor LEFT JOIN Tags AS t ON t.id_tag = ttagrels.id_tag OR t.id_tag = wtagrels.id_tag WHERE t.tag LIKE '%Sandeepan%' OR t.tag LIKE '%Nath%' GROUP BY td.id_tutor LIMIT 20 Please help me with anything you can. Thanks

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  • Search files blazing fast

    If you know there is a file somewhere on your machine, but you cannot find it with the default Windows Search Tools (that why they tend to call it Windows Search and not Windows Find ) then switch to a tool that really works. Go to http://www.voidtools.com/ to download your copy of Everything. The download is only small (350KB), it indexes fast (within 5 mins) and searches my complete computer even faster then I can type. I only blame David Carpenter for not spreading the word more aggressively and for not developing this earlier.

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  • Supporting Windows Search with MFC

    Windows 7 brings a new level of maturity to Windows Search, and by taking advantage of new MFC functionality first publicly unveiled with the Beta 2 release of Visual Studio 2010, writing a Search filter handler for a MFC application can be easily accomplished.

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  • Ubuntu Preseed set Norwegian Keyboard?

    - by Vangelis Tasoulas
    It's been a couple of days now that I am trying to make a fully automated unattended installation. I managed to make it work with Ubuntu/Cobbler and a preseed file, but I cannot set the correct keyboard layout which is Norwegian in this case. I am doing the tests on a virtual machine and when I am going with a normal manual installation (no preseed) everything is working fine. When I am using the preseed file, I always end up with an "English (US)" keyboard no matter the many different options I have tried. I can change it manually with the "dpkg-reconfigure keyboard-configuration" command, but that's not the case. It should be handled automatically using the preseed file. I am using DEBCONF_DEBUG=5 when the grub is loading, and as I see in "/var/log/installer/syslog" file after the installation has finished, the preseeding commands are accepted. Can anyone help on this? The preseed file I am using is following: d-i debian-installer/country string NO d-i debian-installer/language string en_US:en d-i debian-installer/locale string en_US.UTF-8 d-i console-setup/ask_detect boolean false d-i keyboard-configuration/layout select Norwegian d-i keyboard-configuration/variant select Norwegian d-i keyboard-configuration/modelcode string pc105 d-i keyboard-configuration/layoutcode string no d-i keyboard-configuration/xkb-keymap select no d-i netcfg/choose_interface select auto d-i netcfg/get_hostname string myhostname d-i netcfg/get_domain string simula.no d-i hw-detect/load_firmware boolean true d-i mirror/country string manual d-i mirror/http/hostname string ftp.uninett.no d-i mirror/http/directory string /ubuntu d-i mirror/http/proxy string http://10.0.1.253:3142/ d-i mirror/codename string precise d-i mirror/suite string precise d-i clock-setup/utc boolean true d-i time/zone string Europe/Oslo d-i clock-setup/ntp boolean true d-i clock-setup/ntp-server string 10.0.1.254 d-i partman-auto/method string lvm partman-auto-lvm partman-auto-lvm/new_vg_name string vg0 d-i partman-auto/purge_lvm_from_device boolean true d-i partman-lvm/device_remove_lvm boolean true d-i partman-md/device_remove_md boolean true d-i partman-lvm/confirm boolean true d-i partman-lvm/confirm_nooverwrite boolean true d-i partman-auto-lvm/guided_size string max d-i partman-auto/choose_recipe select 30atomic d-i partman/default_filesystem string ext4 d-i partman-partitioning/confirm_write_new_label boolean true d-i partman/choose_partition select finish d-i partman/confirm boolean true d-i partman/confirm_nooverwrite boolean true d-i partman/mount_style select uuid d-i passwd/root-login boolean false d-i passwd/make-user boolean true d-i passwd/user-fullname string vangelis d-i passwd/username string vangelis d-i passwd/user-password-crypted password $6$asdafdsdfasdfasdf d-i passwd/user-uid string d-i user-setup/allow-password-weak boolean false d-i passwd/user-default-groups string adm cdrom dialout lpadmin plugdev sambashare d-i user-setup/encrypt-home boolean false d-i apt-setup/restricted boolean true d-i apt-setup/universe boolean true d-i apt-setup/backports boolean true d-i apt-setup/services-select multiselect security d-i apt-setup/security_host string security.ubuntu.com d-i apt-setup/security_path string /ubuntu tasksel tasksel/first multiselect Basic Ubuntu server, OpenSSH server d-i pkgsel/include string build-essential htop vim nmap ntp d-i pkgsel/upgrade select safe-upgrade d-i pkgsel/update-policy select none d-i pkgsel/updatedb boolean true d-i grub-installer/only_debian boolean true d-i grub-installer/with_other_os boolean true d-i finish-install/keep-consoles boolean false d-i finish-install/reboot_in_progress note d-i cdrom-detect/eject boolean true d-i debian-installer/exit/halt boolean false d-i debian-installer/exit/poweroff boolean false

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  • Search Engine Ranking Competition

    Search engine ranking competition just got tougher. With individuals and businesses pooling a team of SEO experts to update their website, SEO software, working on intensive keyword research, as well as tapping into social media marketing, continuous marketing is necessary to improve and maintain search engine ranking competitiveness.

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  • The ABC's of Search Engine Optimization

    If you're trying to learn more about Search Engine Optimization you're not alone. Many business owners have heard this term, but may not know exactly what it mean. Simply put, SEO is a way to help improve your web site so that it can be found easily by search engines, allowing it to be ranked higher in their results, and ultimately be found by your customers.

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  • Chrome causing 404's ending with "/cache/[hex-string]/"?

    - by Jan Fabry
    Since the last weeks we see many 404's on our sites caused by Chrome adding /cache/[hex-string]/ to the current page URL. The hex strings we have seen are: e9c5ecc3f9d7fa1291240700c8da0728 1d292296547f895c613a210468b705b7 408cfdf76534ee8f14657ac884946ef2 9b0771373b319ba4f132b9447c7060a4 b8cd4270356f296f1c5627aa88d43349 If you search for these strings you get matches from different sites, but they are most likely auto-generated (/search/cache/e9c5ecc3f9d7fa1291240700c8da0728/ for example). Is this a known issue with Chrome (or an extension)?

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  • Don't Let Someone Else Optimize Your Search Results

    I build websites for a wide variety of clients, and every single client asks me to get their website placed highly within search engine results. However, they don't know that there's more to ranking high in the search engines than just launching a site; this is especially true when the website has just been created.

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  • Designing Search Engine Keyword Optimization Friendly Websites and Blogs

    To generate free targeted traffic, you must know how to go about search engine keyword optimization. This is because optimizing your website or blog for placement on the first page of Google or Yahoo is the best online money making secret. Google, MSN and Yahoo are used by most people to search for information thus if you can optimize your website pages for common keywords you will definitely attract traffic.

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  • Search engine optimization Links

    - by Michael Freidgeim
    Below there are a few links, that I used for my Search engine optimization research:     http://websearch.about.com/od/designforsearch/ss/tendesigntips.htm     Keyword Selection Guidelines   Where To Use Keywords  Google Search Engine Optimization http://websearch.about.com/od/keywordsandphrases/a/sitedesign.htm     http://en.wikipedia.org/wiki/Search_engine_optimization       http://www.google.com/support/webmasters/bin/answer.py?answer=35291

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  • Search Engine Placement Optimization and Link Popularity

    The increased visibility of your website due to high link popularity and search engine placement optimization can mean so much, especially if you are promoting a product or service through your website. If you are new to the business of link building, you might be wondering how to get started with it and how search engine placement optimization can help you. Knowledge of link popularity basics is essential even if you are planning on hiring someone to do link building tasks for you.

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  • Heating Up the Search Results With Local SEO

    With the help of an SEO agencies and local SEO, local businesses are dominating the first page of search results. There are millions of websites in existence today and With so much competition, website owners must have high rankings with the search engines in order to succeed online.

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