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Editorial Team
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Asked: 2026-06-08T21:36:43+00:00

I’m trying to get JSON data from a website using the following code, but

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I’m trying to get JSON data from a website using the following code, but I get the error Uncaught SyntaxError: Unexpected identifier at file:///android_asset/www/projectName.html:1
If I save the data from the website as a .json file I can access it fine, but not if I access it from the website. How do I make this work from the website?

HTML:

<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js">
    </script>
<script type="text/javascript">
        $(document).ready(function(){
            $("button").click(function(){
            $.get('http://example.ca/log_sheets.json', function(data) {
                alert('get performed');
                var obj = eval ("(" + data + ")");
                $("p").html(obj[0].log_sheet.activity_type);
            });
        });
    });
</script>
</head>

<body>
    <h2>Heading</h2>
    <p>Display</p>
    <button>Click me</button>
</body>
</html>

Sample of JSON data:

[{"log_sheet":{"activity_type":"Normal","activity_value":null,"carbs_value":25.0, ect... }}]

Code to run html page with phonegap:

package com.example.projectName;

import android.os.Bundle;
import android.app.Activity;
import android.view.Menu;
import android.view.MenuItem;
import android.support.v4.app.NavUtils;
import org.apache.cordova.*;

public class MainActivity extends DroidGap {

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        super.loadUrl("file:///android_asset/www/projectName.html");
    }

    @Override
    public boolean onCreateOptionsMenu(Menu menu) {
        getMenuInflater().inflate(R.menu.activity_main, menu);
        return true;
    }


}
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1 Answer

  1. Editorial Team

    If jQuery thinks you’re retrieving JSON (via server-provided mime-type, a final argument of $.get(...,'json'), or by using $.getJSON), the data passed to the callback will already be a JavaScript object.

    In that case, you won’t want to eval it.

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