我想通过单击添加按钮()动态添加下拉菜单。下拉选项将来自我的mysql数据库。
我在jquery中尝试过以下方法,但是无法显示“ items”表中的项目记录。
<script type="text/javascript">
$(document).ready(function() {
var x = 0;
$('#save_btn').hide();
$('#add_btn').click(function(e) {
e.preventDefault();
appendRow(); // appen dnew element to form
x++; // increment counter for form
$('#save_btn').show(); // show save button for form
});
$('#input_wrapper').on('click', '.deleteBtn', function(e) {
e.preventDefault();
var id = e.currentTarget.id; // set the id based on the event 'e'
$('div[id='+id+']').remove(); //find div based on id and remove
x--; // decrement the counter for form.
if (x === 0) {
$('#save_btn').hide(); // hides the save button if counter is equal to zero
}
});
$('#save_btn').click(function(e) {
e.preventDefault();
var formData = $('#test_form').serializeObject(); // serialize the form
var obj; //obj can be use to send to ajax call
if(Array.isArray(formData.fn)) {
for(var i = 0; i < formData.fn.length; i++) {
obj = {};
// set the obj for submittion
obj.firstName = formData.fn[i];
obj.lastName = formData.ln[i];
// This object could be push into an array
console.log('object from form array ', obj);
};
} else {
obj = {};
obj.firstName = formData.fn;
obj.lastName = formData.ln;
console.log('single obj from form ', obj);
}
});
function appendRow() {
$('#input_wrapper').append(
'<div id="'+x+'" class="form-group" style="display:flex;">' +
'<div>' +
'<select name="items" class="form-control" placeholder="First
Name">**<option> Item one</option><option> Item 2</option>**
`enter code here`</select>' +
'</div>' +
'<div>'+
'<input type="text" id="'+x+'" class="form-control" name="ln" placeholder="Last Name"/>'+
'</div>' +
'<div>'+
'<button id="'+x+'" class="btn btn-danger deleteBtn"><i class="glyphicon glyphicon-trash"></i></button>' +
'</div>' +
'</div>');
}
});
</script>
我希望项目1部分包含显示项目表中记录的数据,而不是上面写的统计文本和值。
答案 0 :(得分:0)
在控制器中,您需要将数据传递到视图,例如:
// $dropdown_options could be an associative array of value and label
$data['dropdown_options'] = $dropdown_options;
// Call your view with the packed data
return $this->load->view('your_form', $data);
在表单中,您需要将php数组转换为Javascript。
var dropdownOptions = <?php echo json_encode($dropdown_options) ?>;
使用Javascript数组后,就可以连接每个选项。
var optionString = "";
for (var i in dropdownOptions) {
optionString += "<option>"+dropdownOptions[i]['label']+"</option>";
}
var selectString = "<select>"+optionString+"</select>";