我想从数据(ajax调用)中绘制谷歌图表
var jsonPieChartData = $.ajax({
url: "ajax.php",
data: {selValue1 : 1,selValue2 : 1} ,
dataType: "json",
async: false
}).responseText;
//create our data table out of json data loaded from server
console.log(jsonPieChartData);
这是返回值
的ajax.php文件function interncompany(){
global $DB;
//query by experience total post
$sql = 'SELECT lc.id, count(ljj.job_id) as count, lc.companyname FROM {local_jobs_job} ljj INNER JOIN {local_companydetail} lc ON ljj.job_company_userid = lc.userid where ljj.job_type = 1 group by lc.companyname';
//get the query into record
$data = $DB->get_records_sql($sql);
//put the query into array
$rows = array();
$rows = array_map(function($item) {
return (object) ['c' => [
(object) ['v' => $item->companyname, 'f' => null],
(object) ['v' => intval($item->count), 'f' => null]
]];
}, array_values($data));
$cols = [
(object) ['id' => '', 'label' => 'LABEL', 'pattern' => '', 'type' => 'string'],
(object) ['id' => '', 'label' => 'TOTAL', 'pattern' => '', 'type' => 'number'],
];
$returndata = new stdClass;
$returndata->cols = $cols;
$returndata->rows = $rows;
echo json_encode($returndata);
}
使用选择框
动态选择数据if ($select1 == '1') {
if ($select2 == '1') {
jobcompany();
}
if ($select2 == '2') {
joblocation();
}
if ($select2 == '3') {
jobcategory();
}
if ($select2 == '4') {
jobsalary();
}
if ($select2 == '5') {
jobexperience();
}
if ($select2 == '6') {
joblevel();
}
}
elseif ($select1 == '2') {
if ($select2 == '1') {
interncompany();
}
if ($select2 == '2') {
internlocation();
}
if ($select2 == '3') {
interncategory();
}
if ($select2 == '4') {
internsalary();
}
if ($select2 == '5') {
internexperience();
}
if ($select2 == '6') {
internlevel();
}
}
我的问题:我想如何创建一个动态获取数据的函数并将结果插入数据:{},以便数据返回到php文件,以便可以通过ajax调用读取以绘制图表。
现在它什么也没有返回。我需要在ajax中硬编码内部数据:{}来绘制图表。
数据动态选择:
// get the select value
$(document).ready(function() {
// for post-filter
$('#post-filter').on('change',function(){
var select1 = $(this).val(); // Post filter value
var select2 = $("#field-filter").val(); // Field Filter value
$.ajax({
type: 'POST',
url: 'ajax.php',
data: {selValue1 : select1,selValue2 :select2 },
success: function(result){
console.log(result);
}
});
});
// field filter value.
$('#field-filter').on('change',function(){
var select2 = $(this).val(); // Field filter value
var select1 = $("#post-filter").val(); // post Filter value
$.ajax({
type: 'POST',
url: 'ajax.php',
data: {selValue1 : select1,selValue2 :select2 },
success: function(result){
console.log(result);
}
});
});
});
这是单击选择框时的结果...只有结果不会返回到ajax。填写内部数据的内容:{}将结果导入drawitem函数ajax调用?
插入标题时的错误
结果数据出现但出现错误,图表不存在。
答案 0 :(得分:0)
您可以尝试以下方法吗?我已经展示了字段过滤器的示例,如果它可以工作,您可以复制后过滤器
var dataVars = {};
dataVars['selValue1'] = $("#post-filter").val(); // Post filter value
dataVars['selValue2'] = $("#field-filter").val(); // Field Filter value
var jsonPieChartData = $.ajax({
url: "ajax.php",
data: JSON.stringify(dataVars) ,
dataType: "json",
async: false
}).responseText;
}
还要确保在php中将内容类型标头设置为JSON
header('Content-Type: application/json');