如何在选中任何复选框的情况下启用提交按钮,而在未选中任何复选框的情况下如何禁用提交按钮?
$('.1').change(function(){
form_valid = false;
if($('.1').prop('checked')==true){
$('#submit-btn').attr('class','btn btn-primary');
form_valid = true;
}
else{
form_valid = false;
}
});
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<input type="checkbox" name="electronic[]" value="101" class="1"> fan<br>
<input type="checkbox" name="electronic[]" value="102" class="1"> tv<br>
<input type="checkbox" name="electronic[]" value="103" class="1"> ac<br>
<input type="checkbox" name="electronic[]" value="104" class="1"> mobile<br>
答案 0 :(得分:0)
function check(){
$('input[type=checkbox]').each(function () {
$("#btnSubmit").attr("disabled", false);
var Val = (this.checked ? "1" : "0");
if(Val=="1"){
$("#button").attr("disabled", false);
return false;
}else{
$("#button").attr("disabled", true);
}
});
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<button id="button" disabled="true" >Button</button>
<br>
<br>
<input type="checkbox" onclick="check()" /> Button Status
<input type="checkbox" onclick="check()" /> Button Status
尝试此代码, 快乐编码:-)
答案 1 :(得分:0)
const checkbox = document.getElementsByTagName("input");
const sub = document.getElementById("sub");
sub.disabled = true;
function check() {
sub.disabled = true;
for (let i = 0; i < checkbox.length; i++) {
if (checkbox[i].type == "checkbox") {
if (checkbox[i].checked == true) {
sub.disabled = false;
}
}
}
}
<form>
<input
type="checkbox"
onclick="check()"
name="firstCheck"
value="First"
/>
<input
type="checkbox"
onclick="check()"
name="secondCheck"
value="Second"
/>
<input type="submit" id="sub" name="sub" />
</form>
这是纯js。但是我建议使用jQuery。