Java Variables and Javascript Variables in jsp page

Question

I have a piechart from google that I am trying to populate with data from my database. I can get a result set with my desired results but I am unable to get this to work because I am not sure how to take my java variables and make them into javascript variables so that they work with the google code. When I try the code below I get an error on this line var zipie = <%= zip1 %>; saying that zip1 cannot be resolved to a variable Trying to call my java variable in the javascript directly doesn't work

<script type="text/javascript">
google.load('visualization', '1', {
    'packages' : [ 'corechart' ]
});
<%
try {
    PieChartDAO pDao = new PieChartDAO();
    String sql = PieChartSQLCreation. getTopZipCode();
    ResultSet zipr = pDao.getTopZip(sql);
    int[] zip = new int[5];
    String[] zips = new String[5];
    for(int i = 0; i <5; i++) {
        zip[i] = zipr.getInt("num");
        zips[i] = zipr.getString("zip_code");
        zipr.next();
    }
    int zip1 = zip[0];
}
catch (Exception e) {
    e.printStackTrace();
}
%>      

var zipie = <%= zip1 %> ;
google.setOnLoadCallback(drawVisualization);

function drawVisualization() {
    visualization_data = new google.visualization.DataTable();
    visualization_data.addColumn('string', 'Zip Code');
    visualization_data.addColumn('number', 'Number of Students');
    visualization_data.addRow([ 'Work', zipie ]);
    visualization_data.addRow([ 'Eat', 2 ]);
    visualization_data.addRow([ 'Commute', 2 ]);
    visualization_data.addRow([ 'Watch TV', 2 ]);
    visualization_data.addRow([ 'Sleep', 7 ]);
    visualization = new google.visualization.PieChart(
            document.getElementById('piechart'));
    visualization.draw(visualization_data, {
            title : 'Zip Codes',
            height : 260
    });
}
</script>

Similar to stackoverflow.com/questions/4803906/…

aberry
– aberry

2015-03-01 19:47:18 +00:00
Commented Mar 1, 2015 at 19:47 — aberry
– aberry, Commented Mar 1, 2015 at 19:47

Beau Grantham · Accepted Answer · 2015-03-01 19:55:20Z

1

zip1 is local to the try block. Declare this beforehand for it to be in the correct scope.

int zip1 = -1; // some default value

try {
    // [...]
}

answered Mar 1, 2015 at 19:55

Beau Grantham

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