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PHP 中 构造js中alert() 提示框

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2016-05-26 08:19:17967browse

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/**
 * 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 = "
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