jQuery 选择器按名称获取表单
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jQuery selector to get form by name
提问by Oscar Jara
I have the following HTML:
我有以下 HTML：
<form name="frmSave">...</form>
Just to know, I am not ableto modify the HTML in order to add an idor something else.
只是要知道，我无法修改 HTML 以添加id或其他内容。
This is what I tried to get the formelement by it's name:
这就是我试图form通过名称获取元素的方法：
var frm = $('form[name="frmSave"]');
console.log(frm);
(but I believe the above code will try to get a children element with the frmSavename inside the formwhich is not what I need).
（但我相信上面的代码将尝试获取一个frmSave名称在里面的子元素，form这不是我需要的）。
How can I achieve this, is it possible to get the formby just the nameand using a selector?
我怎样才能做到这一点，是否可以form通过name使用选择器来获得？
Update:
更新：
I was doing the wrong way in my original code (I was using a space aka "combinator" in the selector) and not in the code snippet from here so jQuery was trying to get any children element with the name needed and I was asking myself what was wrong since nothing was returned.
我在我的原始代码中做错了（我在选择器中使用了一个空格，也就是“组合器”）而不是在这里的代码片段中，所以 jQuery 试图获取任何具有所需名称的子元素，我在问自己出了什么问题，因为什么都没有返回。
The accepted answer gives a better explanation about this so a little space could change everything - I will sleep properly next time :-)
接受的答案对此给出了更好的解释，因此一点空间可以改变一切-下次我会好好睡觉:-)
回答by Christian
$('form[name="frmSave"]')is correct. You mentioned you thought this would get all children with the name frmsaveinside the form; this would only happen if there was a space or other combinator between the form and the selector, eg: $('form [name="frmSave"]');
$('form[name="frmSave"]')是正确的。你提到你认为这会让所有孩子frmsave在表格中都有名字；只有在表单和选择器之间有空格或其他组合器时才会发生这种情况，例如：$('form [name="frmSave"]');
$('form[name="frmSave"]')literally means find all forms with the name frmSave, because there is no combinator involved.
$('form[name="frmSave"]')字面意思是查找所有带有 name 的形式frmSave，因为不涉及组合器。
回答by Arun Pratap Singh
// this will give all the forms on the page.
$('form')
// If you know the name of form then.
$('form[name="myFormName"]')
// If you don't know know the name but the position (starts with 0)
$('form:eq(1)') // 2nd form will be fetched.
回答by Niet the Dark Absol
You have no combinator (space, >, +...) so no children will get involved, ever.
你有没有组合子（空间>，+...），所以没有孩子会涉足，直到永远。
However, you could avoid the need for jQuery by using an IDand getElementById, or you could use the old getElementsByName("frmSave")[0]or the even older document.forms['frmSave']. jQuery is unnecessary here.
但是，您可以通过使用IDand来避免对 jQuery 的需求getElementById，或者您可以使用旧的getElementsByName("frmSave")[0]或更旧的document.forms['frmSave']. 这里不需要 jQuery。
回答by Mike Irving
For detecting if the form is present, I'm using
为了检测表单是否存在，我正在使用
if($('form[name="frmSave"]').length > 0) {
//do something
}
