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  • Free Photoshop Plugin or Software to auto remove backgrounds

    - by Rogue
    I'm looking for a background removal plug-in or software that automates or atleast eases the process of removing backgrounds from pictures / digital photos. I have seen a few like Mask Pro 4, Snap and BackGround Remover all these are paid software. I would like to know if there are any free solutions available before I invest in any of the above plug-ins / software.

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  • Completely remove MySQL from macbook pro

    - by mike
    Im prety sure i completely removed mysql from my system, except for one thing. When i type mysql in the command line i get this bash: /opt/local/bin/mysql5: No such file or directory How is it still recognizing where it thinks mysql should be? I'm trying to build it myself in /usr/local, and when i do install it there, i still get that error message for it looking for it in opt/local.

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  • How to safely copy an object?

    - by Prog
    This question is going to be a little long. Please bear with me. Something that happened in a project of mine made me think about how to safely copy objects. I'll present the situation I had and then ask a question. There was a class SomeClass: class SomeClass{ Thing[] things; public SomeClass(Thing[] things){ this.things = things; } // irrelevant stuff omitted public SomeClass copy(){ return new SomeClass(things); } } There was another class Processor that takes SomeClass objects, copies them (via someClassInstance.copy()), manipulates the copy's state, and returns the copy. Here it is: class Processor{ public SomeClass processObject(SomeClass object){ SomeClass copy = object.copy(); manipulateTheCopy(copy); return copy; } // irrelevant stuff omitted } I ran this, and it had bugs. I looked into these bugs, and it turned out that the manipulations Processor does on copy actually affect not only the copy, but also the original SomeClass object that was passed into processObject. I found out that it was because the original and the copy shared state - because the original passed it's field things into the copy when creating it. This made me realize that copying objects is harder than simply instantiating them with the same fields as the original. For the two objects to be completely disconnected, without any shared state, each of the fields passed to the copy also has to be copied. And if that object contains other objects - they have to be copied too. And so on. So basically, in order to be able to actually copy an object, each class in the system must have a copy() method, that also invokes copy() on all of it's fields, and so on. So for example, for copy() in SomeClass to work, it needs to look like this: public SomeClass copy(){ Thing[] copyThings = new Thing[things.length]; for(int i=0; i<things.length; i++) copyThings[i] = things[i].copy(); return new SomeClass(copyThings); } And if Thing has object fields of it's own, than it's own copy() method must be appropriate: class Thing{ Apple apple; Pencil pencil; int number; public Thing(Apple apple, Pencil pencil, int number){ this.apple = apple; this.pencil = pencil; this.number = number; } public Thing copy(){ // 'number' is a primitve. return new Thing(apple.getCopy(), pencil.getCopy(), number); } } And so on. Of course, instead of all classes having a copy() method, the copying mechanism can happen in all of the getters and the constructors of classes (unless places where it isn't suitable, for example when the field points to an external object, not to an object that 'is part' of this object). Still, that means that in order to be able to safely copy an object - most classes would have to have copying mechanisms in their getters. My question is divided into two parts: How frequently do you need to get a copy of an object? Is this a regular issue? Is the technique described common and/or reasonable? Or is there a better way to make safe copies of objects? Or is there an easier way to safely copy objects, without them sharing any state?

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  • php remove button keeps retreiving posted information

    - by Doug
    Hi every1! im quickly finding im quite the beginner to php, so please bare with me! I have a quick problem, I have a remove button for a shopping cart and the code works for it everytime except for the last product thats in the cart. If a person pushes the addtoCart button the URL is reloaded with ?buyproduct=$productNumber and when the remove button is pushed the product is removed. Right so everything is good, but when you try to remove the last item, it keeps reading the product thats in the URL. so the quantity remains 1 for the current $productNumber. i've tried adding the action in the form method tag so that the page reloads without the ?buyproduct=$productNumber, which does work however there are page numbers and sections that were also in the URL and these get reset also. i know the remove is working because once the ?buyproduct=$productNumber is gone from the URL (which can happen for example if they go to another section in the catalog) then the cart can be completely emptied. Thank you so much ahead of time for any1's help, I've used this site before and you guys are genius's!

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  • Jquery Append() and remove() element

    - by BandonRandon
    Hi, I have a form where I'm dynamically adding the ability to upload files with the append function but I would also like to be able to remove unused fields. Here is the html markup <span class="inputname">Project Images: <a href="#" class="add_project_file"><img src="images/add_small.gif" border="0"></a></span> <span class="project_images"> <input name="upload_project_images[]" type="file" /><br/> </span> Right now if they click on the "add" gif a new row is added with this jquery $('a.add_project_file').click(function() { $(".project_images").append('<input name="upload_project_images[]" type="file" class="new_project_image" /> <a href="#" class="remove_project_file" border="2"><img src="images/delete.gif"></a><br/>'); return false; }); To remove the input box i've tried to add the class "remove_project_file" then add this function. $('a.remove_project_file').click(function() { $('.project_images').remove(); return false;}); I think there should be a much easier way to do this. Maybe i need to use the $(this) function for the remove. Another possible solution would be to expand the "add project file" to do both adding and removing fields. Any of you JQuery wizards have any ideas that would be great

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  • remove 2 subviews in one go.

    - by Pavan
    hi, I am trying to remove two viewcontrollers (that have been added on top of each other) with one method. I have made the views in interfacebuilder. they all have their own .h and .m files to go with it. Scenario I am in: I have a main menu which has the view2 header file imported. In a method I add the second view on top of the superview like so view2ViewController * view2 = [[view2ViewController alloc] initWithNibName:@"view2ViewController" bundle:nil]; [self.view addSubview:view2.view]; then in view 2 I have added the view 3 header file so i can add view 3 as a subview ontop of view2. i have another method which is connected again to interface builder to a UIButton so upon button press a method gets called in view2 which adds view 3 on top in exactly the same way like so: view3ViewController * view3 = [[view3ViewController alloc] initWithNibName:@"view3ViewController" bundle:nil]; [self.view addSubview:view3.view]; What im trying to solve: I have a button in view 3 which should remove view 3.... and then it should also remove view 2 aswell so the main screen is visible. How can this be achieved? What I have so far: [self.view removeFromSuperview]; This however only removes View 3... but leaves view 2 in place. What needs to be modified so that i can remove view 2 as well?? Any help is appreciated.

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  • Python beautifulsoup trying to remove html tags 'span'

    - by Michelle Jun Lee
    I am trying to remove [<span class="street-address"> 510 E Airline Way </span>] and I have used this clean function to remove the one that is in between < > def clean(val): if type(val) is not StringType: val = str(val) val = re.sub(r'<.*?>', '',val) val = re.sub("\s+" , " ", val) return val.strip() and it produces [ 510 E Airline Way ]` i am trying to add within "clean" function to remove the char '[' and ']' and basically i just want to get the "510 E Airline Way". anyone has any clue what can i add to clean function? thank you

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  • Remove trailing slash from comment form

    - by Sergio Vargott
    and i really need a code to remove the ending slash when a user put their link. for example i need them to put their url to grab their avatar, but in some cases they put their url ending with a slash (.com/) how can i remove that slash automatically? because when they put their url like that the avatar doesn't show i need them to end like this (.com) in order to show their avatar. I was looking for a remove trailing slash php code, but any solution will be appreciated. i tried to use this code but didn't work $string = rtrim($string, '/');

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  • jquery: remove table row while iterating through table rows

    - by deostroll
    #exceptions is a html table. I try to run the code below, but it doesn't remove the table row. $('#exceptions').find('tr').each(function(){ var flag=false; var val = 'excalibur'; $(this).find('td').each(function(){ if($(this).text().toLowerCase() == val) flag = true; }); if(flag) $(this).parent().remove($(this)); }); What is the correct way to do it?

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  • jquery: How to completely remove a dialog on close

    - by Svish
    When an ajax operation fails, I create a new div with the errors and then show it as a dialog. When the dialog is closed I would like to completely destroy and remove the div again. How can I do this? My code looks something like this at the moment: $('<div>We failed</div>') .dialog( { title: 'Error', close: function(event, ui) { $(this).destroy().remove(); } }); When I run this the dialog box shows up correctly, but when I close it the dialog is still visible in the html (using FireBug). What am I missing here? Something I have forgotten? Update: Just noticed my code gives me an error in the firebug console. $(this).destroy is not a function Anyone able to help me out? Update: If I do just $(this).remove() instead, the item is removed from the html. But is it completely removed from the DOM? Or do I somehow need to call that destroy function first as well?

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  • How To Safely Eject Your USB Devices From the Desktop Context Menu

    - by Taylor Gibb
    If you are one of those people who don’t safely remove their USB Devices just because you’re lazy, here’s a neat trick to do it from the context menu on your desktop. Even if you are not lazy and just forget, the icon will serve as a mental reminder. So let’s take a look. How to Run Android Apps on Your Desktop the Easy Way HTG Explains: Do You Really Need to Defrag Your PC? Use Amazon’s Barcode Scanner to Easily Buy Anything from Your Phone

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  • how to safely Uninstall Ubuntu 12.04 and reinstall 11.10

    - by Jmais
    I upgraded from 11.10 to 12.04 following the instructions here. https://help.ubuntu.com/community/PreciseUpgrades. But after the upgrade a lot of my image processing packages don't work some of their dependencies have been removed. Honestly there is not that much difference between 11.10 and 12.04. So please can someone tell me how to safely remove 12.04 and go back to 11.10 without losing documents, and source packages on my old Ubuntu. Thanks

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  • Javascript - remove part of cookie with a split('|') like array

    - by MgS
    I'm abit stuck. I'm using jQuery. I have a cookie which has the value such as : 1,something|2,somethingg|1,something We are treating it as [0] as id and [1] as name. I need to remove ONE where the id == '1' for example, this will leave the cookie like this: 1,something|2,somethingg How do I go about this, it will probally be in a loop, but not sure how to remove one of them. I have this so far: function removeItem(id){ var cookieName = 'myCookie'; var cookie = $.cookie(cookieName); if(cookie){ var cookie = cookie.split('|'); $(cookie).each(function(index){ var thisCookieData = this.split(','); if(thisCookieData[0] == id ){ } }); }

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  • JQuery: Remove duplicate elements?

    - by Keith Donegan
    Say I have a list of links with duplicate values as below: <a href="#">Book</a> <a href="#">Magazine</a> <a href="#">Book</a> <a href="#">Book</a> <a href="#">DVD</a> <a href="#">DVD</a> <a href="#">DVD</a> <a href="#">Book</a> How would I, using JQuery, remove the dups and be left with the following for example: <a href="#">Book</a> <a href="#">Magazine</a> <a href="#">DVD</a> Basically I am looking for a way to remove any duplicate values found and show 1 of each link.

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  • jQuery.remove() - Is there a way to get the object back after you remove it?

    - by Jack Marchetti
    I basically have the same problem in this questions: Flash Video still playing in hidden div I've used the .remove jquery call and this works. However, I have previous/next buttons when a user scrolls through hidden/non-hidden divs. What I need to know is, once I remove the flash object, is there a way to get it back other than refreshing the page? Basically, can this be handled client side or am I going to need to implement some server side handling. detach() won't work because the flash video continues to play. I can't just hide it because the video continues to play as well.

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  • jquery remove selected element and append to another

    - by KnockKnockWhosThere
    I'm trying to re-append a "removed option" to the appropriate select option menu. I have three select boxes: "Categories", "Variables", and "Target". "Categories" is a chained select, so when the user selects an option from it, the "Variables" select box is populated with options specific to the selected categories option. When the user chooses an option from the "Variables" select box, it's appended to the "Target" select box. I have a "remove selected" feature so that if a user "removes" a selected element from the "Target" select box, it's removed from "Target" and put back into the pool of "Variables" options. The problem I'm having is that it appends the option to the "Variables" items indiscriminately. That is, if the selected category is "Age" the "Variables" options all have a class of "age". But, if the removed option is an "income" item, it will display in the "Age Variables" option list. Here's the HTML markup: <select multiple="" id="categories" name="categories[]"> <option class="category" value="income">income</option> <option class="category" value="gender">gender</option> <option class="category" value="age">age</option> </select> <select multiple="multiple" id="variables" name="variables[]"> <option class="income" value="10">$90,000 - $99,999</option> <option class="income" value="11">$100,000 - $124,999</option> <option class="income" value="12">$125,000 - $149,999</option> <option class="income" value="13">Greater than $149,999</option> <option class="gender" value="14">Male</option> <option class="gender" value="15">Female</option> <option class="gender" value="16">Ungendered</option> <option class="age" value="17">Ages 18-24</option> <option class="age" value="18">Ages 25-34</option> <option class="age" value="19">Ages 35-44</option> </select> <select height="60" multiple="multiple" id="target" name="target[]"> </select> And, here's the js: /* This determines what options are display in the "Variables" select box */ var cat = $('#categories'); var el = $('#variables'); $('#categories option').click(function() { var class = $(this).val(); $('#variables option').each(function() { if($(this).hasClass(class)) { $(this).show(); } else { $(this).hide(); } }); }); /* This adds the option to the target select box if the user clicks "add" */ $('#add').click(function() { return !$('#variables option:selected').appendTo('#target'); }); /* This is the remove function in its current form, but doesn't append correctly */ $('#remove').click(function() { $('#target option:selected').each(function() { var class = $(this).attr('class'); if($('#variables option').hasClass(class)) { $(this).appendTo('#variables'); sortList('variables'); } }); });

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  • Git: Remove specific commit

    - by Joshua Cheek
    I was working with a friend on a project, and he edited a bunch of files that shouldn't have been edited. Somehow I merged his work into mine, either when I pulled it, or when I tried to just pick the specific files out that I wanted. I've been looking and playing for a long time, trying to figure out how to remove the commits that contain the edits to those files, it seems to be a toss up between revert and rebase, and there are no straightforward examples, and the docs assume I know more than I do. So here is a simplified version of the question: Given the following scenario, how do I remove commit 2? $ mkdir git_revert_test && cd git_revert_test $ git init Initialized empty Git repository in /Users/josh/deleteme/git_revert_test/.git/ $ echo "line 1" > myfile $ git add -A $ git commit -m "commit 1" [master (root-commit) 8230fa3] commit 1 1 files changed, 1 insertions(+), 0 deletions(-) create mode 100644 myfile $ echo "line 2" >> myfile $ git commit -am "commit 2" [master 342f9bb] commit 2 1 files changed, 1 insertions(+), 0 deletions(-) $ echo "line 3" >> myfile $ git commit -am "commit 3" [master 1bcb872] commit 3 1 files changed, 1 insertions(+), 0 deletions(-) The expected result is $ cat myfile line 1 line 3 Here is an example of how I have been trying to revert $ git revert 342f9bb Automatic revert failed. After resolving the conflicts, mark the corrected paths with 'git add <paths>' or 'git rm <paths>' and commit the result.

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  • jquery disable/remove option on select?

    - by SoulieBaby
    Hi all, I have two select lists, I would like jquery to either remove or disable a select option, depending on what is selected from the first select list: <select name="booking" id="booking"> <option value="3">Group Bookings</option> <option value="2" selected="selected">Port Phillip Bay Snapper Charters</option> <option value="6">Portland Tuna Fishing</option> <option value="1">Sport Fishing</option> </select> Here's the second list (which will only ever have two values): <select name="charterType" id="charterType"> <option value="1">Individual Booking</option> <option value="2">Group Booking</option> </select> If Group Bookings or Port Phillip Bay Snapper Charters are selected, I need only "Group Booking" to be displayed. (To basically hide "Individual Booking") but I can't seem to get it to work.. If someone could help me that'd be great!! :) I've also tried using a switch, but that doesnt work either.. /* select list */ switch (jQuery('#booking :selected').text()) { case 'Sport Fishing': alert('AA'); break; case 'Port Phillip Bay Snapper Charters': jQuery("#charterType option[value=1]").remove(); alert('BB'); break; case 'Portland Tuna Fishing': alert('CC'); break; case 'Group Bookings': alert('DD'); break; }; It alerts, but doesn't do anything else..

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  • UIViewController remove subview

    - by ghiboz
    Hi all! I add a view as subview of my uiviewcontroller like this: // into my ViewController: UIImageView *imView =[[UIImageView alloc] initWithImage:[UIImage imageNamed:@"img.jpg"]]; imView.frame = CGRectMake(2, 46, 1020, 720); [self.view addSubview:imView]; now, with another button I wish remove the imView from the subview chain.. how can I do to do this?? thanks

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