在几天之前,我从jsfiddle的代码中获得了一个静态json,并进行了一些更改以从中创建动态json。 jsfiddle from which i got code
现在我的代码正在运行。但我想要对它进行一些更改。下面是我当前的代码和图像,我使用json从mysql获取数据。
$(document).ready(function() {
var data = [];
$.ajax({
url: "<?php echo base_url('responsible/get_pay_slips'); ?>",
datatype: 'json',
success: function(response) {
response = $.parseJSON(response);
x = 0;
$.each(response, function(i, value) {
console.log(value.PAY_YEAR, value.PAY_MONTH);
var data1 = [];
data[x++] = {
text: value.PAY_YEAR,
data: {},
children: [{
text: value.PAY_MONTH,
data: {
price: 0.1,
quantity: 20
}
}, ],
'state': {
'opened': false
}
};
});
console.log(data);
// load jstree
$("div#jstree").jstree({
plugins: ["table", "dnd", "contextmenu", "sort"],
core: {
data: data
},
// configure tree table
table: {
columns: [{
width: 500,
header: "YEAR"
},
{
width: 300,
value: "ACTION",
header: "ACTION",
format: function(v) {
if (v) {
return 'Print | Save'
}
}
},
],
resizable: true,
draggable: true,
contextmenu: true,
}
});
}
});
});
那么,我们如何在单个树中创建相同PAY_MONTH
的{{1}}。
我只想要一个父PAY_YEAR
,并且在其中只需要PAY_YEAR
{。}}。
我正在使用codeigniter。
我的模特:
PAY_MONTH
我的控制器:
public function json_get_slips()
{
$empid ='34';
$where = "where EMPLOYEE_ID='".$empid."' ";
$query = $this->db->query("select PAY_MONTH,PAY_YEAR from india_salary_slip_details ".$where." group by pay_month order by pay_year");
return $query->result_array();
}
我不太了解jquery,在这里我想用它创建动态数据。我没有希望。请帮助我。
答案 0 :(得分:1)
使用此查询获取月份,您的查询将只显示一个月
SELECT CASE
WHEN salary_amt IS NULL THEN 0 ELSE salary_amt
END as salary_amt
FROM
(
SELECT 1 AS MONTH
UNION SELECT 2 AS MONTH
UNION SELECT 3 AS MONTH
UNION SELECT 4 AS MONTH
UNION SELECT 5 AS MONTH
UNION SELECT 6 AS MONTH
UNION SELECT 7 AS MONTH
UNION SELECT 8 AS MONTH
UNION SELECT 9 AS MONTH
UNION SELECT 10 AS MONTH
UNION SELECT 11 AS MONTH
UNION SELECT 12 AS MONTH
) as meses
LEFT JOIN salaryData e ON meses.MONTH = MONTH(STR_TO_DATE(SALARY_DATE,'%d-%b-%y')) GROUP BY meses.MONTH