Question

我有以下代码，但它不起作用。我正在尝试创建脚本输出：

<script type="text/javascript">
    var files = <?php echo json_encode($files); ?>;
    var len = files.length
    SLIDES = new slideshow("SLIDES");
    for (var i=0; i<len; i++) {
        s = new slide();
            s.src =  "http://www.cashbackflorida.com/wpradmin/modules/wprrets/photos/<?php echo $result ->MLS; ?>/"+ files[i] +;
            s.width = "560";
            s.height = "420";
            s.alt = "";
            s.text = unescape("");
            s.link = "";
            s.target = "";
            s.attr = "";
            s.filter = "";
            SLIDES.add_slide(s);
    }
</script>

Answer 1

您必须将JSON字符串解码为javascript对象。你应该这样做：

<script type="text/javascript">
    var xy = <?php echo json_encode($files); ?>;
    var files = JSON.parse(xy); /*<- this makes a JavaScript object from JSON string*/
    var len = files.length
    SLIDES = new slideshow("SLIDES");
    for (var i=0; i<len; i++) {
        s = new slide();
            s.src =  "http://www.cashbackflorida.com/wpradmin/modules/wprrets/photos/<?php echo $result ->MLS; ?>/"+ files[i] +;
            s.width = "560";
            s.height = "420";
            s.alt = "";
            s.text = unescape("");
            s.link = "";
            s.target = "";
            s.attr = "";
            s.filter = "";
            SLIDES.add_slide(s);
    }
</script>

我有以下代码，但它不起作用。我正在尝试创建脚本输出：

1 个答案: