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PHP 实现类似js中alert() 提示框_PHP

WBOY
WBOYOriginal
2016-06-01 11:06:131061browse

主要应用于添加判断提示,跳转,返回,刷新。

代码如下:


/**
 * JS提示跳转
 * @param  $tip  弹窗口提示信息(为空没有提示)
 * @param  $type 设置类型 close = 关闭 ,back=返回 ,refresh=提示重载,jump提示并跳转url
 * @param  $url  跳转url
 */
function alert($tip = "", $type = "", $url = "") {
    $js = "<script>";<br /> if ($tip)<br /> $js .= "alert('" . $tip . "');";<br /> switch ($type) {<br /> case "close" : //关闭页面<br /> $js .= "window.close();";<br /> break;<br /> case "back" : //返回<br /> $js .= "history.back(-1);";<br /> break;<br /> case "refresh" : //刷新<br /> $js .= "parent.location.reload();";<br /> break;<br /> case "top" : //框架退出<br /> if ($url)<br /> $js .= "top.location.href='" . $url . "';";<br /> break;<br /> case "jump" : //跳转<br /> if ($url)<br /> $js .= "window.location.href='" . $url . "';";<br /> break;<br /> default :<br /> break;<br /> }<br /> $js .= "</script>";
    echo $js;
    if ($type) {
        exit();
    }
}

以上所述就是本文的全部内容了,希望对大家学习php能有所帮助。

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