Question

我想把两个盒子放在一起，但到目前为止第二个对象总是在第一个对象之下。这是我从 getbootstrap.com 使用的代码。

我只需要像这样的盒子，彼此之间的nexto不是在不同的线上。注意：我已尝试在CSS中使用它 float: left;遗憾的是没有发生任何好事。

<div class="row">
    <div class="col-xs-6 col-md-3">
        <a href="#" class="thumbnail">
            <img src="..." alt="...">
        </a>
    </div>
</div>

Answer 1

看来你只有一个盒子。您应该阅读twitter引导网格背后的列概念：http://getbootstrap.com/css/#grid-intro。

我假设您正在尝试为移动设备和桌面设备添加混合行为（通过混合不同单元大小的col-xs和col-md）。

<!-- Columns start at 50% wide on mobile and bump up to 33.3% wide on desktop -->
<div class="row">
  <div class="col-xs-6 col-md-4">.col-xs-6 .col-md-4</div>
  <div class="col-xs-6 col-md-4">.col-xs-6 .col-md-4</div>
  <div class="col-xs-6 col-md-4">.col-xs-6 .col-md-4</div>
</div>

<!-- Columns are always 50% wide, on mobile and desktop -->
<div class="row">
  <div class="col-xs-6">.col-xs-6</div>
  <div class="col-xs-6">.col-xs-6</div>
</div>

.col-xs-12 .col-md-8 .col-xs-6 .col-md-4

%This is the number of bits

m=1000;

%This is a counter needed for the probability calculation

count=0;

%Loop needed for the probability calculation

for i = 1:1000000

%This is a random number generator that generates m (1000 in this case)
%random numbers between 0 and 1

randomseq=rand(1,m);

%This creates a random number between 0 and 1 and then multiplies it by
%0.29 so it can actually be between 0 and 0.29. If n is not a random
%number, then change it with the fixed value that it has.

n=rand(1,1)*0.29; 

%This line is for the following:
%If a number is less than n, then make it 1, else make it 0 (this is
%needed so you can acquire a probability for the number of ones and the
%number of zeros). Here you will have around  n*1000 ones, but not
%equal since randomseq is a random vector with 1000 elements and those
%are too few numbers so it can create a real distribution.

onezerorand=randomseq<n;

%That's why this is needed, just in case if the number of ones go over
%30%. If it goes over 30%, than make the numones coef 0.29

if sum(onezerorand)>=300
   numones=0.29;
else

%This is a coef which represent the number of ones, but in probability
%terms (between 0 and 1). Mean just calculates the mean value between
%the zeros and the ones, which is essentially the probability.

    numones=mean(onezerorand);
end

%The rest of the numbers are zero, so you get the probability by
%subtracting numones from 1

numzeros=1-numones;

%This creates a random number x between 0 and 0.59. If x is not a
%random number, the same comment goes as for n.

x=rand(1,1)*0.59;

%There is a 50% chance that x of the ones will be changed to zeros and
%vice versa, so you create a random number between 0 and 0.5 and
%multiply it by x.

p0=rand(1,1)*0.5;
p1=1-p0;
x_ones=rand(1,1)*p0*x;
x_zeros=rand(1,1)*p1*x;

%You add the changed zeros to the ones and the changed ones to the
%zeros and you subtract the changed ones from the ones and the changed
%zeros from the zeros, so you can get the last value of the number of
%zeros.

new_numones=numones-numones*x_ones+numzeros*x_zeros;
new_numzeros=numzeros-numzeros*x_zeros+numones*x_ones;

%You create a random number between 0.51 and 0.99, so you can compare
%the number of zeros to it. If q is not a random number, the same
%comment as the one for x and n.

q=rand(1,1)*0.48+0.51;

%You compare the number of zeros to the new random number. If it's
%bigger than it, than you raise the counter, if not, you do nothing.
%You repeat this 1000000 times so you can get a good probability
%estimation and the number you get is the probability of this occuring.

if new_numzeros>q
    count=count+1;
end
end

%The probability of this event occuring

prob=count/1000000;

Answer 2

我认为你正在寻找类似的东西 http://www.layoutit.com/build

numbers[i+2]

如何使用Bootstrap正确移动对象

2 个答案: