Am trying to generate multiple line charts at one go using the code below. However, it isn't working. What would be the best way to generate graphs using a for/while or any other looping mechanism? I have many charts to generate.
var db_query = Array();
db_query[1] = <?php echo json_encode($db_query_1) ?>;
db_query[2] = <?php echo json_encode($db_query_2) ?>;
var chartConfig1 = clone(chartConfigLineOne);
var chartConfig2 = clone(chartConfigLineOne);
for(var i=1, len=2; i <= len; i++) {
/* chart initialization */
var chart_num = "chart" + i.toString();
var plot_num = "plot" + i.toString();
var chartConfig_num = "chartConfig" + i.toString();
/*alert(chart_num);
alert(plot_num);
alert(chartConfig_num);*/
chart_num = AmCharts.makeChart(plot_num, chartConfig_num);
$.ajax({
type: 'POST',
url: "query_db.php",
data: {'db_que': db_query[i]},
dataType: 'json',
context: document.body,
global: false,
async: true,
success: function(data) {
//alert(data);
chart_num.dataProvider = data;
chart_num.validateNow();
}
});
}
UPDATED CODE
<script type="text/javascript">
var chartNameList = ['chart1','chart2'];
var divId = ['plot1','plot2'];
var configList = ['chartConfig1','chartConfig2'];
var dbQuery = [];
var chartConfig1 = clone(chartConfigLineOne);
var chartConfig2 = clone(chartConfigLineOne);
dbQuery[0] = <?php echo json_encode($db_query_1) ?>;
dbQuery[1] = <?php echo json_encode($db_query_2) ?>;
/* chart initialization */
for(var i =0; i < 2; i += 1) {
//window["chart"+i] = createChartObj(divId[i],configList[i]);
execDbQuery(divId[i],configList[i],dbQuery[i],chartNameList[i]);
}
</script>
/**
* Execute of DB query
*/
function execDbQuery(divId, configObj, dbQuery, chartObj) {
chartObj = AmCharts.makeChart(divId, configObj);
$.ajax({
type: 'POST',
url: "query_db.php",
data: {'db_que': dbQuery},
dataType: 'json',
context: document.body,
global: false,
async: true,
success: function(data) {
//alert(data);
chartObj.dataProvider = data;
chartObj.validateNow();
}
});
}
JavaScriptwithPHPlikedb_query[1] = <?php echo json_encode($db_query_1) ?>;. It is terribly awful. And try not to use multiplevar a; var b; var c;, but instead just usevar a = 123, b = 345, .... If you don't change your style of programming, some day your code will look like spagetti.alert(data)useconsole.log(data). The style you are programming has become obsolete several years ago. So, you'd better read some books about professional programming inJavaScript.PHPdata insideJavascript. This is actually whatajaxdoes in your example.