如何使用 jQuery DataTables 提交所有页面的复选框
我正在尝试为每一行获取第一个单元格 (td
) 并获取它,但仅限于当前页面.如果我导航到下一页,则不会发送上一页上选中的复选框.
I'm trying to get first cell (td
) for each row and getting it but only for current page. If I navigate to next page then the checkbox checked on the previous page is not being sent.
<table class="table" id="example2">
<thead><tr>
<th>Roll no</th><th>Name</th></tr><thead>
<?php
$sel = "SELECT * FROM `st`";
$r = mysqli_query($dbc, $sel);
while ($fet = mysqli_fetch_array($r)) {
?>
<tr>
<td><?php echo $fet['trk'] ?></td>
<td><input type="text" value="<?php echo $fet['ma'] ?>" id="man" class="form-control"></td>
<td><input type="checkbox" id="check" name="myCheckbox" class="theClass"></td></tr>
<?php } ?>
</table>
<input type="submit" id="sub_marks" class="btn btn-info" value="Submit & Continue">
<script src="plugins/datatables/jquery.dataTables.min.js" type="text/javascript"></script>
<script src="plugins/datatables/dataTables.bootstrap.min.js" type="text/javascript"></script>
<script type="text/javascript">
$(function () {
$('#example2').DataTable({
"paging": true,
"lengthChange": false,
"searching": false,
"ordering": true,
"info": true,
"autoWidth": false,
})
});
</script>
<script>
$('#sub_marks').click(function () {
var values = $("table #check:checked").map(function () {
return $(this).closest("tr").find("td:first").text();
}).get();
alert(values);
})
</script>
推荐答案
CAUSE
jQuery DataTables 出于性能原因从 DOM 中删除了不可见的行.提交表单时,仅将可见复选框的数据发送到服务器.
CAUSE
jQuery DataTables removes non-visible rows from DOM for performance reasons. When form is submitted, only data for visible checkboxes is sent to the server.
你需要把元素<input type="checkbox">
被选中的并且在DOM中不存在的元素转成<input type="hidden">
表单提交时.
You need to turn elements <input type="checkbox">
that are checked and don't exist in DOM into <input type="hidden">
upon form submission.
var table = $('#example').DataTable({
// ... skipped ...
});
$('form').on('submit', function(e){
var $form = $(this);
// Iterate over all checkboxes in the table
table.$('input[type="checkbox"]').each(function(){
// If checkbox doesn't exist in DOM
if(!$.contains(document, this)){
// If checkbox is checked
if(this.checked){
// Create a hidden element
$form.append(
$('<input>')
.attr('type', 'hidden')
.attr('name', this.name)
.val(this.value)
);
}
}
});
});
解决方案 2:通过 Ajax 发送数据
var table = $('#example').DataTable({
// ... skipped ...
});
$('#btn-submit').on('click', function(e){
e.preventDefault();
var data = table.$('input[type="checkbox"]').serializeArray();
// Include extra data if necessary
// data.push({'name': 'extra_param', 'value': 'extra_value'});
$.ajax({
url: '/path/to/your/script.php',
data: data
}).done(function(response){
console.log('Response', response);
});
});
演示
参见 jQuery 数据表:如何提交所有页面表单数据以获取更多详细信息和演示.
DEMO
See jQuery DataTables: How to submit all pages form data for more details and demonstration.
- 每个复选框都应该有一个
value
属性,并分配有唯一值. - 避免对多个元素使用
id
属性check
,这个属性应该是唯一的. - 您不需要为 jQuery DataTables 显式启用
paging
、info
等选项,这些选项是默认启用的. - 考虑使用
htmlspecialchars()
函数来正确编码 HTML 实体.例如,<?php echo htmlspecialchars($fet['trk']);?>
.
- Each checkbox should have a
value
attribute assigned with unique value. - Avoid using
id
attributecheck
for multiple elements, this attribute is supposed to be unique. - You don't need to explicitly enable
paging
,info
, etc. options for jQuery DataTables, these are enabled by default. - Consider using
htmlspecialchars()
function to properly encode HTML entities. For example,<?php echo htmlspecialchars($fet['trk']); ?>
.
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