javascript - Fancybox问题，需要帮助

Question

我用从数据库获取的图像制作了一个画廊，并使用Fancybox来显示它。我有一些相同的文章，只是颜色不同，我只显示一个和下面的颜色框来更改颜色。

The problem is that when I click on the button to change the color, picture changes, Fancybox on click displays first picture, not the Current one

部分代码:

JS:

function changeImage(element,id){
    var img=document.getElementById(id).src=element;
    return false;
}

PHP:

while($row = mysql_fetch_array($sql)){

    $prikaz =$row['prikaz'];
    $id = $row['id'];
    $ime = $row['ime'];
    $thumb = $row['thumbs'];
    $boja = $row['boja_id'];
    $slicka = $row['slika'];
    $spec = $row['tekst'];

    if ($prikaz == 1){

        echo "<table style ='display: inline' align='center'>";
        echo "<tr>";
        echo "<td><a class='fancybox-effects-a' href='$slicka' ><img id='$id' src='$thumb' alt='' /></a></td>";
        echo "</tr>";
        echo "<tr><td>";
        echo "Boja: ";
        $bsql = mysql_query ("SELECT muski.tekst, muski.id,muski.thumbs,boja.bslika FROM boja INNER JOIN muski ON muski.boja_id = boja.id WHERE muski.ime = '$ime' " );                  

        while($res = mysql_fetch_array($bsql)){ 

            $slicica = $res['thumbs'];    
            $muid = $res['id'];
            $kockica = $res['bslika'];
            echo "<button id = 'boja' onclick =changeImage('$slicica','$id')><img src= $kockica ></button>";

        }

        echo "</br>";
        echo nl2br($spec);

}

echo "</td>";
echo "</tr>";    
echo "</table>";

Answer 1

我找到了解决方案....我修改了js脚本来更改img的父href...它有效...

function changeImage(element,id,staza) {
    var img = document.getElementById(id);
        img.src= element;
        img.parentNode.href=staza;
    return false;
}