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Home/ Questions/Q 6644179
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T00:08:13+00:00 2026-05-26T00:08:13+00:00

I am not getting my image correctly. I am unsure what is happening. First

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I am not getting my image correctly. I am unsure what is happening. First things first, the image is coming out of a mysql query. Little confused about how to make that image ready for a ajax call?

here is how I get the image out mysql

    if(mysql_query("insert into Personal_Photos (Email, Pics) values('$email', '$data')"))
            {
                $query="select Pics, MAX(ID) from Personal_Photos where Email='$email'";
                $result=mysql_query($query) or die("Error: ".mysql_error());
                $row=mysql_fetch_array($result);
                //$mime = 'image/yourtype';
                //$base64 = base64_encode($contents);
                //$uri = "data:$mime;base64,$base64";
                //header("Content-type: image/jpg");
                echo '<img src="data:image/jpeg;base64'.base64_encode($row['Pics']).'"/>';
            }

the jquery that I use is like so

$('#profilepicbutton').live('change', function(){
    $("#preview").html('');
    $("#preview").html('<img src="loader.gif" alt="Uploading...."/>');
        $("#registerpt3").ajaxForm({
                target: '#preview',
                success: function(data)
                {                                   
                    $("#preview").html('');
                    $("#preview").append(data);
                }
            }).submit();
 })
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-26T00:08:14+00:00Added an answer on May 26, 2026 at 12:08 am

    Your are missing the comma for the data URI

    <img src="data:image/jpeg;base64,###############"/>
    
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