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Home/ Questions/Q 8391781
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Editorial Team
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Editorial Team
Asked: June 9, 20262026-06-09T19:21:08+00:00 2026-06-09T19:21:08+00:00

I’ve written this code using various online sources but I cannot seem to figure

  • 0

I’ve written this code using various online sources but I cannot seem to figure out the last part.

function loadajax (event) {
    event.preventDefault();
    xhr = new XMLHttpRequest();
    xhr.onreadystatechange  = function(){ 
        if(xhr.readyState  == 4){
            if(xhr.status  == 200) 
                document.ajax.dyn="Received:"  + xhr.responseText; 
            else
                document.ajax.dyn="Error code " + xhr.status;
        }
    }; 

    xhr.open('GET', this.href, true);
    var content = document.getElementsByTagName('article')[0];

    content.innerHTML = xhr.responseText;
}

It seems to work until I need to add content to my page. Indeed content.innerHTML = xhr.responseText; returns nothing. I am getting a simple HTML file, how can I post it in my page? what am I doing wrong?

Thanks for your help!

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-09T19:21:09+00:00Added an answer on June 9, 2026 at 7:21 pm

    ajax calls are asynchronous. it will work if you’ll move the content.innerHTML = xhr.responseText; line into the onreadystatechange function like this:

    function loadajax (event) {
    event.preventDefault();
    xhr = new XMLHttpRequest();
     xhr.onreadystatechange  = function() 
        { 
           if(xhr.readyState  == 4)
           {
            if(xhr.status  == 200) 
                document.ajax.dyn="Received:"  + xhr.responseText; 
    
                content.innerHTML = xhr.responseText;
            else
                document.ajax.dyn="Error code " + xhr.status;
            }
        }; 
    
    xhr.open('GET', this.href, true);
    var content = document.getElementsByTagName('article')[0];
    
    }
    
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