Bilder wechseln mit Links

  • Kinder

  • Hunde

  • Katzen
  • Dieser Teil des Scripts kommt in den Headbereich

    <SCRIPT LANGUAGE="JavaScript">

    <!-- Begin
    var kinder = new Image();
    var hunde = new Image();
    var katzen = new Image();
    var anfang = new Image();

    kinder.src = "../images/s-sonja.jpg";
    hunde.src = "../images/s-struppi.jpg";
    katzen.src = "../images/s-kica.jpg";
    anfang.src = "../images/s-blank.gif";

    function doButtons(picimage) {
    eval("document['picture'].src = " + picimage + ".src");
    }
    // End -->
    </script>

    Dieser Teil des Scripts kommt in den Bodybereich
    <table border=1 cellpadding="5" cellspacing="5" background="../images/bgr-cell.gif" bgcolor="#FF0000" align="center">
    <tr>
    <td width="90">
    <center>
    </center>
    <p>
    <li><a href = "http://www.google.com/search?hl=de&q=Kinder&btnG=Google-Suche&lr=lang_de" onMouseOver = "doButtons('kinder')" onMouseOut = "doButtons('anfang')" target="_blank"><font size="4"><b>Kinder</b></font>
    </a><br>
    <br>
    <li><a href = "http://www.google.com/search?hl=de&q=Hunde&btnG=Google-Suche&lr=lang_de" onMouseOver = "doButtons('hunde')" onMouseOut = "doButtons('anfang')" target="_blank"><font size="4"><b>Hunde</b></font><br>
    <br>
    </a>
    <li><b><font size="4"><a href = "http://www.google.com/search?hl=de&q=Katzen&btnG=Google-Suche&lr=lang_de" onMouseOver = "doButtons('katzen')" onMouseOut = "doButtons('anfang') "target="_blank">Katzen</a></font></b>
    <td width="302">
    <div align="center"><img name=picture src="../images/s-blank.gif" width=300 height=300 border=0></div>
    </td>
    </tr>
    </table>