我正在使用Firebase来处理我的网站上的Google OAuth登录信息.有谁知道如何限制有权访问该应用程序的用户?例如,我只想要x@gmail.com,y @ gmail.com和z@gmail.com才能成功通过谷歌登录我的应用程序.
我不确定这是Firebase还是Google问题,但是我们非常感谢任何帮助.
我正在尝试创建一个轮廓图,其中x坐标为标签EF,y标记为EB,z为标记为a的函数.它返回下面发布的长错误.任何帮助,将不胜感激.错误是
File "contour.py", line 19, in <module>
c = plt.contour(EF,EB,a)
File "/usr/lib/pymodules/python2.7/matplotlib/pyplot.py", line 2016, in contour
ret = ax.contour(*args, **kwargs)
File "/usr/lib/pymodules/python2.7/matplotlib/axes.py", line 7326, in contour
return mcontour.QuadContourSet(self, *args, **kwargs)
File "/usr/lib/pymodules/python2.7/matplotlib/contour.py", line 1088, in __init__
ContourSet.__init__(self, ax, *args, **kwargs)
File "/usr/lib/pymodules/python2.7/matplotlib/contour.py", line 686, in __init__
self._process_args(*args, **kwargs)
File "/usr/lib/pymodules/python2.7/matplotlib/contour.py", line 1101, in _process_args
x, y, z = self._contour_args(args, kwargs)
File "/usr/lib/pymodules/python2.7/matplotlib/contour.py", line 1148, in _contour_args
x,y,z = self._check_xyz(args[:3], kwargs)
File "/usr/lib/pymodules/python2.7/matplotlib/contour.py", line 1180, in _check_xyz
raise TypeError("Input z must …Run Code Online (Sandbox Code Playgroud) 我有这种形式,允许人们根据他们需要的数量添加地址,它由两个按钮控制,一个是"添加地址",另一个是"删除".我想知道是否有人可以帮我删除"删除"按钮,而是在文本框的右角放置一个"x"作为删除该框的按钮.我已经放置了与下面表格相关的代码.再次感谢您的帮助,感谢您的帮助.
jQuery的
<script>
$(window).load(function(){
$("#add-address").click(function(e){
e.preventDefault();
var numberOfAddresses = $("#form1").find("input[name^='data[address]']").length;
var label = '<label for="data[address][' + numberOfAddresses + ']"></label> ';
var input = '<input type="text" name="data[address][' + numberOfAddresses + ']" id="data[address][' + numberOfAddresses + ']" placeholder= "Address ' + (numberOfAddresses+1) + '"/>';
var removeButton = '<button class="remove-address">Remove</button>';
var html = "<div class='address'>" + label + input + removeButton + "</div>";
$("#form1").find("#add-address").before(html);
});
$(document).on("click", ".remove-address",function(e){
e.preventDefault();
$(this).parents(".address").remove();
//update labels
$("#form1").find("label[for^='data[address]']").each(function(){
//$(this).html("Address " + ($(this).parents('.address').index() + 1));
$(this).next("input").attr("placeholder","Address " + ($(this).parents('.address').index() + …Run Code Online (Sandbox Code Playgroud) contour ×1
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