php同页面传值如何写 _PHP

php同页面传值怎么写？
我想要的效果是
<table>
<?php $i=1;
do { ?>
<tr>
//标题
<td><a href="default_task_edit.php?editID=<?php echo $row_Recordset_task['TID']; ?>" ><?php echo $row_Recordset_task['csa_title']; ?></a></td>
</tr>
<?php $i=$i+1;
} while ($row_Recordset_task = mysql_fetch_assoc($Recordset_task));
?>
</table>
</td>
<td rowspan="2" width="80%" align="center"><?php echo $multilingual_neirong; ?>
//在这里显示标题所相关的内容，内容在在、数据库的值为$row_Recordset_task['csa_text']
</td>

------解决方案--------------------
你是不要页面刷新么那就用ajax 或者笨点的方法就是做个跟原来页面样式一样的新页面跳转到这个新页面跟原来的界面差不多视觉上好像是没刷新一样
------解决方案--------------------
同意楼上用ajax！！！
------解决方案--------------------

PHP code


<?php
/* Created on [2012-5-16] Author[yushuai.niu] */
#查询标题信息
$sql="select * from table";
    $res=mysql_query($sql);
    if(!$res) die("SQL: {$sql} <br>Error:".mysql_error());
    if(mysql_affected_rows() > 0){
        $titles = array();
        while($rows = mysql_fetch_array(MYSQL_ASSOC)){
            array_push($titles,$rows);
        }
    }
?>
<table border=1>
<?php foreach($titles as $row_Recordset_task){ ?>
    <tr>
        <td>
            <a href="javascript:void(0)" onclick="record(<?=$row_Recordset_task['TID']?>)" >
                <?=$row_Recordset_task['csa_title']?>
            </a>
        </td>
    </tr>
<?php } ?>
</table>
<div id="show"></div>
<script>
//Ajax
var xmlHttp;

    function createXMLHttpRequest() {
        if(window.XMLHttpRequest) {
            xmlHttp = new XMLHttpRequest();
        } else if (window.ActiveXObject) {
            xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");
        }
    }

    function record(id){
        createXMLHttpRequest();
        url = "action.php?id="+id+"&ran="+Math.random();
        method = "GET";
        xmlHttp.open(method,url,true);
        xmlHttp.onreadystatechange = showList;
        xmlHttp.send(null);
    }

    function show(){
        if (xmlHttp.readyState == 4){
            if (xmlHttp.status == 200){
                var text = xmlHttp.responseText;
                document.getElementById("show").innerHTML = text;
            }else {
                alert("response error code:"+xmlHttp.status);
            }
        }
    }
</script>
<?php
#action.php
if(isset($_GET['id'])){
    $sql="select * from table where id=".$_GET['id'];
    $res=mysql_query($sql);
    if(!$res) die("SQL: {$sql} <br>Error:".mysql_error());
    if(mysql_affected_rows() > 0){
        $rows = mysql_fetch_array(MYSQL_ASSOC);
    }
    print_r($rows);
    mysql_close();
}

?>

------解决方案--------------------
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PHP code

<?php
/* Created on [2012-5-16] Author[yushuai.niu] */
#查询标题信息
$sql="select * from table";
    $res=mysql_query($sql);
    if(!$res) die("SQL: {$sql} <br>Error:".mysql_error());
    if(mys……