<img>没有使用jQuery刷新

Question

I am developing with xhtml, LAMP and jQuery 2.0.2 on Debian 7.0.0.

I am using the following php-generated element

echo "<img id=\"displayImageID\" src=\"" . $DisplayFileName . "\" alt=\"Background\" style=\"margin:10px 0px 10px 5px\"/>";

I am hoping to change it to a different color look up table when the used clicks on a radio button. In order to do this, I use the following code

    jQuery.ajax({
    type: "POST",
    url: "DisplayParameters.php",
    data: { LUT:lut }
    }).done(function( result ) {
    $("#msg").html( " Ajax called" );
    });        

with the following code in DisplayParameters.php

    $LUT=$_POST['LUT'];
    $Executable="Executables/changeLUT";
    $DisplayImageName="images/display.png";
    $Str=$Executable . " -t " . $LUT" . " -o " . $DisplayImageName;

    $output=exec($Str, $dummy, $returnValue);

    echo "<script language=\"JavaScript\">";
    echo "d = new Date();";
    echo "jQuery('#displayImageID').attr('src', '" . $DisplayImageName . "?'+d.getTime());";
    echo "</script>";
}

to update the image and to reload it into the image element but nothing happens until I refresh the whole page with the Firefox "Reload current page".

I had thought that maybe, due to the asynchronous nature of Ajax, the image is getting loaded before it is replaced by a new image with the same name but different color LUT. However, if this were the case, I would expect the new LUT image to eventually be loaded if I keep clicking on radio buttons after the image that is saved to the disk is changed. This is not the case. I do not get a new image until I reload the page.

Answer 1

Try this:

JS

jQuery.ajax({
       type: "POST",
       url: "DisplayParameters.php",
       data: { LUT:lut },
       cache: false                               //CHANGED
}).done(function( result ) {
       $("#msg").html( " Ajax called... and returned " + result );
       //('#displayImageID').attr('src', result);   
       ('#displayImageID').attr('src', result+'?'+$.now());   //CHANGED
});    

PHP

$LUT=$_POST['LUT'];
$Executable="Executables/changeLUT";
$DisplayImageName="images/display.png";
$Str=$Executable . " -t " . $LUT" . " -o " . $DisplayImageName;

$output=exec($Str, $dummy, $returnValue);

echo $DisplayImageName;                         //CHANGED