jQuery 如果变量 = 数字，做 X，否则什么都不做
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/8956397/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
If variable = number, do X, else do nothing
提问by dylfreak6494
I wrote a jquery code that is going to expand certain elements on hover. I want the code to only expand if not expanded. Here is my code. It doesn't seem to be working correctly.
我写了一个 jquery 代码，它将在悬停时扩展某些元素。我希望代码仅在未扩展时扩展。这是我的代码。它似乎无法正常工作。
$(document).ready(function() {
var labelstatus = 0;
});
$(document).ready(function() {
$("a.rm").hover(function () {
if (labelstatus != 1){
$("#expensereduction,#envimpact,#mtextlft,#initaud,#energaud").fadeOut();
var currentFontSize = $('.initiate').css('font-size');
var currentFontSizeNum = parseFloat(currentFontSize, 10);
var newFontSize = currentFontSizeNum*3;
$('.initiate').delay(500).animate({
fontSize: newFontSize
});
return false;
var labelstatus = 1;
}
else {
}
});
});
采纳答案by Guffa
You are declaring local variables inside the functions, so they are independent of each other. Also, you get a new variable each time, so the previous value doesn't persist.
您在函数内部声明局部变量，因此它们彼此独立。此外，您每次都会获得一个新变量，因此之前的值不会持续存在。
Also, you are calling returnbefore assigning the variable, which will exit out of the function, so the assignment will never happen.
此外，您return在分配变量之前调用，这将退出函数，因此分配永远不会发生。
You need to declare the variable in a scope outside the functions:
您需要在函数之外的范围内声明变量：
$(document).ready(function() {
var labelstatus = 0;
$("a.rm").hover(function () {
if (labelstatus != 1){
$("#expensereduction,#envimpact,#mtextlft,#initaud,#energaud").fadeOut();
var currentFontSize = $('.initiate').css('font-size');
var currentFontSizeNum = parseFloat(currentFontSize);
var newFontSize = currentFontSizeNum * 3;
$('.initiate').delay(500).animate({
fontSize: newFontSize
});
labelstatus = 1;
return false;
}
else {
}
});
});
Also: the parseFloatfunction doesn't have a radix parameter.
另外：该parseFloat函数没有基数参数。
回答by
this...
这个...
return false;
var labelstatus = 1;
should be...
应该...
labelstatus = 1;
return false;
...and this...
...和这个...
$(document).ready(function() {
var labelstatus = 0;
});
should be...
应该...
var labelstatus = 0;
Code that comes after the
returnstatement will not runUsing
varmakeslabelstatuslocal to the functionYou really don't need two
$(document).ready(function() {calls. Put your code in one, and you can makelabelstatuslocal to that handler so that all code inside can use it.
return语句之后的代码将不会运行使用
var使labelstatus函数本地化你真的不需要两个
$(document).ready(function() {电话。将您的代码合二为一，您可以将其labelstatus设为该处理程序的本地代码，以便其中的所有代码都可以使用它。
回答by TimWolla
labelstatuswon't be set to 1as the function is already returned in the line before.
labelstatus不会被设置1为该函数已经在之前的行中返回。
