[英]Call ajax on submit form in bootstrap modal
我有一个按钮添加与我用引导模态显示的形式,像这样:
$table.on('click', 'button#add', function() {
jQuery('#add_div').modal('show', {backdrop: 'static'});
$('form#add_form').on('click', 'button#submit', function() {
var field1 = (get inserted data on field1);
alert(field1);
});
});
我在模态中显示的形式:
<!-- Modal 6 (Long Modal)-->
<div class="modal fade" id="modal-6">
<div class="modal-dialog">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" aria-hidden="true">×</button>
<h4 class="modal-title">Modal Content is Responsive</h4>
</div>
<div class="modal-body">
<form action="" method="post" id="add_form">
<div class="row">
<div class="col-md-6">
<div class="form-group">
<label for="field-1" class="control-label">Name</label>
<input type="text" class="form-control" id="field1" name="field1" placeholder="">
</div>
</div>
</div>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
<button id="submit" type="button" class="btn btn-info">Save changes</button>
</div>
</form>
</div>
</div>
</div>
问题是我不知道如何使其在javascript中工作。 通过单击此按钮,模式已启动,但是当我提交表单时,即使我注释了此var field1 =(在field1上获取插入的数据),也没有任何阴影。 并将其替换为此var field1 =“ submited”;
更改
<button type="button" class="btn btn-info">Save changes</button>
至
<button id="btnSubmitModal" type="button" class="btn btn-info">Save changes</button>
并在您的JS脚本中添加一个click函数
$( "#btnSubmitModal" ).click(function() {
var field1value = $( "#field1" ).val()
alert( "Modal submitted with text: " + field1value);
});
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