我正在构建一个Web应用程序,用户在其中搜索相机(位于具有dyndns域的远程网络上),并显示一个按钮来触发相机上的警报器。当用户单击按钮时,他们应该看不到页面刷新或发生任何事情(可能已禁用按钮)。该按钮正在向网站发送HTTP GET请求:
function n_points = BlueDotsDist()
I = imread('BlueDots.png');
%Fix the image uploaded to the site - each dot will be a pixel.
I = imresize(imclose(I, ones(3)), 1/6, 'nearest');
%Convert each color channel to a bynary image:
R = imbinarize(I(:,:,1));G = imbinarize(I(:,:,2));B = imbinarize(I(:,:,3));
%figure;imshow(cat(3, double(R), double(G), double(B)));impixelinfo
%Find blue dost:
[blue_y, blue_x] = find((B == 1) & (R == 0));
%Compute distance map from first blue point
P = CalcDistMap(R, blue_y(1), blue_x(1));
%Compute distance map from second blue point
Q = CalcDistMap(R, blue_y(2), blue_x(2));
%Get 3x3 pixels around second blue point.
A = P(blue_y(2)-1:blue_y(2)+1, blue_x(2)-1:blue_x(2)+1);
dist_p = min(A(:)); %Minimum value is the shortest distance from first point (but not the number of white points).
T = max(P, Q); %Each element of T is maximum distance from both points.
T(T > dist_p) = 10000; %Remove points that are more far than dist_p.
%Return number of white points between blue points.
n_points = sum(T(:) < 10000);
function P = CalcDistMap(R, blue_y, blue_x)
%Each white pixel in R gets the distance from the blue dot in coordinate (blue_x, blue_y).
%Initialize array with values of 10000 (high value - marks infinite distance).
P = zeros(size(R)) + 10000; %P - distance from blue dot
P(blue_y, blue_x) = 0; %Distance from itself.
prvPsum = 0;
while (sum(P(:) < 10000) ~= prvPsum)
prvPsum = sum(P(:) < 10000); %Sum of "marked" dots.
for y = 2:size(R, 1)-1
for x = 2:size(R, 2)-1
p = P(y, x);
if (p < 10000) %If P(y,x) is "marked"
A = P(y-1:y+1, x-1:x+1); %3x3 neighbors.
A = min(A, p+1); %Distance is the minimum of current distance, and distance from p pixel + 1.
A(R(y-1:y+1, x-1:x+1) == 0) = 10000; %Ignore black pixels.
A(P(y-1:y+1, x-1:x+1) == 0) = 0; %Restore 0 in blue point.
P(y-1:y+1, x-1:x+1) = A; %Update distance map.
end
end
end
end
。
但是,我不确定用户刷新或单击后退按钮时如何处理,以及浏览器如何在历史记录中存储请求。我不希望用户能够按下“后退”按钮再次激活该请求,或者刷新页面并激活它。用户也不能将按钮请求另存为书签。本质上,只有在单击按钮后才能触发此事件,并且应该在整个网站上将按钮禁用30秒(所有用户都应该看到该按钮被禁用)。
这引发了一个问题:这应该是GET还是POST请求,如何通过按钮将其合并?
以下是相关代码:
example.com:port/file.xml?relay2State=2
<!-- siren_search.html -->
{% if cameras %}
<button id="trigger-all-btn" type="button" class="btn btn-outline-danger">Trigger all sirens</button>
<table id="camera-table" class="table table-hover">
<thead>
<tr class="header">
<th scope="col" style="width:33%;">Asset Name</th>
<th scope="col" style="width:33%;">Job Site</th>
<th scope="col" style="width:33%;"></th>
</tr>
</thead>
<tbody>
{% for camera in cameras %}
<tr>
<td>{{ camera.asset_name }}</td>
<td>{{ camera.site.name }}</td>
<td><button id="{{ camera.asset_name }}-btn" type="button" class="btn siren-btn btn-outline-danger">Siren</button></td>
</tr>
{% endfor %}
</tbody>
</table>
{% endif %}
答案 0 :(得分:0)
您要问的不是一个简单的解决方案,实际上您有两个问题:
1)如何在不更改浏览器的情况下触发URL?
答案:异步执行。这需要Javascript,您可能会找到类似JQuery的库来简化此解决方案的创建。例如,考虑一下JQuery的$.get使用GET或POST对解决方案不重要
2)如果单击一个按钮,您如何与任何其他浏览器通信并禁用其按钮?
答案:这并不简单。浏览器到服务器的通信是由浏览器拉动的,而不是由服务器推动的。一种选择是使用异步Javascript轮询服务器,以便服务器轮询并提供有关按钮状态的更新。但是,然后您必须处理以下情况:在另一个浏览器上单击某个按钮之后,但在第二次单击的浏览器上,该按钮在最新的民意调查中获得了更新的按钮状态信息。
希望这个答案提供了一些有用的信息。