我有下面的数据表脚本工作。单击复选框时,我需要其他脚本功能来激活。这是行不通的,有人知道我在做什么错吗?我检查了控制台,没有帖子被发送,也没有错误日志。
数据表脚本
<script type="text/javascript">
var checkCol = 4; //checkbox column
$(document).ready(function() {
$('#accountTable').dataTable({
rowId: 'id',
"processing": true,
select: true,
"ajax": "selectdatatable_allacc.php",
"columns": [
{data: 'name'},
{data: 'email'},
{data: 'password'},
{data: 'rang'},
{data: 'do_lead'},
{data: 'notes',
render: function ( data, type, row, meta ) {
return '<font color='+row.cat_color+'>'+data+'</font>';
}
},
{ 'data': null, title: 'Dead', wrap: true, "render": function (item) { return '<input type="submit" value="Dead" style="width:57px !important;" class="example_e" onClick="mobreset(' + item.id + ') "/>' } },
{ 'data': null, title: 'Login', wrap: true, "render": function (item) { return '<input type="submit" value="Login" style="width:57px !important;" class="example_e" onClick="mobLogin(' + item.id + ') "/>' } },
{ 'data': null, title: 'Action 2', wrap: true, "render": function (item) { return '<input type="button" value=" Edit " id= ' + item.id + ' style="width:57px !important;" class="example_c edit_data" />' } },
],
columnDefs: [{targets: [checkCol],
render: function ( data, type, row ) {
if ( type === 'display' ) { //if column data is 1 then set attr to checked, use row id as input id (plus prefix)
return '<input type="checkbox" ' + ((data == 1) ? 'checked' : '') + ' value="' + row.id + '" class="active" />';
}
return data;
},
className: "dt-body-center"
},
{targets: 0, className: "dt-body-center"},
],
select: {
style: 'os',
selector: 'td:not(:nth-child(2))'
},
});
});
</script>
选中或取消选中复选框时调用的脚本
<script type="text/javascript">
$(document).ready(function(){
$("input.active").click(function() {
// store the values from the form checkbox, then send via ajax below
var check_active = $(this).is(':checked') ? 1 : 0;
var check_id = $(this).attr('value');
$.ajax({
type: "POST",
url: "dolead.php",
data: {id: check_id, active: check_active},
});
return true;
});
});
</script>
您可以尝试使用jqueryon
函数来完成工作
<script type="text/javascript">
$(document).ready(function () {
$("#accountTable").on("click", "input.active", function () { // notice the on function
// store the values from the form checkbox, then send via ajax below
console.log("checkbox clicked"); // to check if the event is working
var check_active = $(this).is(':checked') ? 1 : 0;
var check_id = $(this).attr('value');
$.ajax({
type: "POST",
url: "dolead.php",
data: {
id: check_id,
active: check_active
},
});
return true;
});
});
</script>
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