jQuery检查选择的输入是否为空

How in a simples way would I test, if any of selected inputs is empty?

如果选择的任何输入都是空的，那么如何以简单的方式进行测试呢?

I get my inputs this way:

我的输入是这样的:

$myinputs = $('.inputstotest');

and would test single input with:

测试单个输入:

if ( $.trim($this.val()) == '' )

but how to test a group and return true if all are not empty or false if any (at least one) is empty?

但是，如果一个组(至少一个)为空，那么如何测试一个组并返回true ?

6 个解决方案

#1

I'd suggest:

我建议:

var someEmpty = $('.inputstotest').filter(function(){
    return $.trim(this.value).length === 0;
}).length > 0;

With the above someEmpty will be a Boolean, true if there are inputs of that class-name with a trimmed-value equal to zero, false if no inputs of that class-name have a trimmed-value of a length equal to zero.

上面的某个空值将是一个布尔值，如果这个类名的输入值为零，则为false，如果这个类名的输入没有一个长度为零的trimmed值，则为false。

References:

引用:

filter().
filter()。

#2

You can do

你可以做

var $myinputs = $('.inputstotest');
if($myinputs.filter(function() { return $.trim(this.value) == ''; }).length > 0) {
    //There is at least one empty input
}

#3

var invalidInputs = $('.inputstotest').filter(function() {
    return $.trim(this.value) == '';
});

if (invalidInputs.length > 0) {
   //some were invalid
   invalidInputs.after('<span>Please enter a value!</span>');
}

#4

var result = true;
$('.inputstotest').each(function() {
    if ($.trim($(this).val())) == '') {
        result = false;
        return false; // Terminate the .each loop
    }
});
return result;

#5

What if you had each input have the same class eg. class="checkInput" the used jquery to cycle through them using the code below:

如果每个输入都有相同的类呢?class="checkInput"使用的jquery循环使用:

var anyBlank = false;
$('.checkInput').each(function(){
    if($(this).val() == ''){
        anyBlank = true;
    }
});

if(anyBlank){
    alert('There is at least 1 input that is blank');
}

#6

You can use jQuery each function

您可以使用jQuery每个函数

$(".inputstotest").each(function(){});

Heres a fiddle for you: http://jsfiddle.net/Beyyd/1/

这里有一个小提琴给你:http://jsfiddle.net/Beyyd/1/

#1