CODEWITHSUNDEEP
javascriptgoogle-chrome
Matt Fenlon
Matt Fenlon

Reputation: 688

blocked a frame of origin "null" from accessing a cross-origin frame - chrome

I'm new to Javascript and am learning the basics via a textbook that focuses on its applications in IE 7+ and Firefox 2+. However, I am using Chrome and am getting the following error when running the program given in the book: "blocked a frame of origin 'null' from accessing a cross-origin frame." Can anyone tell me what is causing the error and how I can fix it? The two programs are below.

//This is the program being loaded into the browser
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Example</title>

<script type="text/javascript">

function calcFactorial(factorialNumber){
    var factorialResult = 1;
    for(;factorialNumber>0;factorialNumber--) factorialResult *= factorialNumber;
    return factorialResult;
}

</script>

</head>

<frameset cols="100%,*">
<frame name="fraCalcFactorial" src="calcfactorial.htm"/>
</frameset>

</html>

Below is the src file

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Example</title>
<script type="text/javascript">
function butCalculate_onclick(){
    try{
        if (window.top.calcFactorial == null)
            throw "This page is not loaded within the correct frameset";
        if (document.form1.txtNum1.value == "")
            throw "!Please enter a value before you calculate its factorial";
        if (isNaN(document.form1.txtNum1.value))
            throw "!Please enter a valid number";
        if (document.form1.txtNum1.value < 0)
            throw "!Please enter a positive number";
    }
    catch(exception){
        if (typeof(exception) == "string"){
            if (exception.charAt(0) == "!"){
                alert(exception.substr(1));
                document.form1.txtNum1.focus();
                document.form1.txtNum1.select();
            }
            else alert(exception);
        }
        else alert("The following error occurred: " + exception.message);
    }
}
</script>
</head>
<body>
<form action="" name="form1">
    <input type="text" name="txtNum1" size="3" /> factorial is
    <input type="text" name="txtResult" size="25" /><br/>
    <input type="button" value="Calculate Factorial"
        name="butCalculate" onclick="butCalculate_onclick()" />
</form>
</body>
</html>

Upvotes: 53

Views: 145593

Answers (4)

Cees Timmerman
Cees Timmerman

Reputation: 19644

Save the following code as same_server_source.html, run python -m http.server in the same folder, and browse to http://localhost:8000/same_server_source.html

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <meta name="viewport" content="width=device-width, initial-scale=1.0">
    <title>View same server source</title>
</head>
<body>
    <iframe id="iframe" src="/" sandbox="allow-scripts allow-same-origin allow-modal"></iframe><br>
    <button onclick="alert(iframe.contentWindow.document.body.innerHTML)">View home source</button>
</body>
</html>

Upvotes: 1

Олексій Сивак
Олексій Сивак

Reputation: 101

If you use Visual Studio Code, you can install an extension named "Live Server". It helped me when I had the same problem.

Upvotes: 10

Alex P.
Alex P.

Reputation: 1158

If you don't want to use a local web server as suggested in the accepted answer you can run the browser with cross domain web security / same origin policy disabled.

For Chrome:

Disable same origin policy in Chrome

For Firefox:

Disable cross domain web security in Firefox

https://addons.mozilla.org/en-US/firefox/addon/access-control-allow-origin/

Disable firefox same origin policy

Upvotes: 3

amulya349
amulya349

Reputation: 1238

This happens because Chrome doesn't allow frames from your hard disk to access each others' content. Which, technically we term as Cross-origin request.

Solution of the above problem is:

1. Either you host your webpage on a local web server. See the following link:
What is a faster alternative to Python's http.server (or SimpleHTTPServer)?

2. Use any other browser like Firefox

Upvotes: 45

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