自个儿写了一个跳转函数一直在报错

自己写了一个跳转函数一直在报错

<?php<br />/**<br /> * ShowMsg()返回信息定向函数<br /> */<br />function ShowMsg($_info,$_url)<br />	If $_url="" Then <br />		echo "<script type='text/javascript'>alert('$_info');history.back();</script>"<br />	Else <br />		If $_info="" Then <br />			header('Location:'.$_url);<br />		else<br />			echo "<script type='text/javascript'>alert('$_info');location.href='$_url';</script>";<br />		End If<br />	End If<br /><br /><br />?>

------解决方案--------------------

function ShowMsg($_info,$_url){<br />    If ($_url==""  ){<br />        echo "<script type='text/javascript'>alert('$_info');history.back();</script>";<br />    }Else{ <br />        If ($_info=="") <br />            header('Location:'.$_url);<br />        else<br />            echo "<script type='text/javascript'>alert('$_info');location.href='$_url';</script>";<br />        <br />    }<br />}