Reputation: 43
First I display a div container with disabled input fields.
$("#button").click(function() {
$(".disable-input").prop('disabled', true);
})
And with a second button inside the displayed div container I want to enable the change for input fields after the first click and after the second one the enabled input fields should be disabled. For example: Display my customers via div container and inside this div container I can change them with a click on the button. What is my mistake?
var count = 0;
if(count == 0) {
$("#edit").click(function(){
$(".disable-input").prop('disabled', false);})
count = 1;
}
if(count == 1) {
$("edit").click(function(){
$(".disable-input").prop('disabled', true);})
count = 0;
}
Upvotes: 2
Views: 1531
Reputation: 5766
Use attr('disabled', 'disabled') and removeAttr('disabled') to do this
var count = 0;
if(count == 0) {
$("#edit").click(function(){
$(".disable-input").attr('disabled', 'disabled');
count = 1;
});
}
if(count == 1) {
$("edit").click(function(){
$(".disable-input").removeAttr('disabled');
count = 0;
});
}
The reason is that the presence of the 'disabled' attribute (regardless of value) on an input element is enough to cause the input to be disabled, whatever the value - i.e
<input type="text" disabled="false" />
...is disabled.
Upvotes: 2