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AJAX頁面上無法正常顯示!!!

請問大大們,想利用AJAX做動態即時顯示,利用下面這篇教學延展!!!
https://dotblogs.com.tw/jhsiao/archive/2014/08/30/146411.aspx
這是改成radio單選選項,就不能顯示...不過get網址頁,是有顯示get變數...
是我程式哪邊需要改嗎?
AJAX首頁

<!DOCTYPE html>
<html>
<SCRIPT type="text/javascript">


var xmlhttp = false;


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


function serverconnect(divID, datalocation)
{
var1=$clickvalue;
if(xmlhttp)
{
	
var obj = document.getElementById(divID);
var disc = document.getElementById("select_op");
//xmlhttp.open("GET", datalocation + "?disc="+disc);
xmlHTTP.open("GET","6666.php?disc="+disc,true);


    xmlhttp.onreadystatechange = function()
    {
        if (xmlhttp.readyState == 4 &&
        xmlhttp.status == 200)
        {
        //obj.innerHTML = xmlhttp.responseText;
        var str=xmlHTTP.responseText;
        document.getElementById("message").innerHTML=str;		
        }
    }
    xmlhttp.send(null); 
}

}
</script>
    <head>
        <meta http-equiv="Content-Type" content="text/html; charset=utf-8">
        <script type="text/javascript" src="./ajax_select.js"></script>
        <title>未面談確認清單(查詢)</title>   
</head>
<body>   
   <font size="5"><strong>未面談確認清單(查詢)</strong></font>
   <BR><BR>
   
 <form name="form1" id="form1" onsubmit="serverconnect('Div1', '6666.php'); return false;">
    <fieldset>
        <legend>未面談確認清單</legend>
    
    <p>請選擇:</p>
    <p>
        <label><input type="radio" name="select_op" id="select_op" value="10" checked  /><!--checked-->  未面談確認(受評者)</label>
        <label><input type="radio" name="select_op" id="select_op" value="20" /> 未面談確認(直屬主管)</label>
        <label><input type="radio" name="select_op" id="select_op" value="1000" /> 兩者已面談確認</label>
     <!--<input type="submit" value="搜尋" />-->
    </p>
	
	<p>
    <div id='message'></div>
    </p>
   
    <BR>
    </fieldset>
</form> 
</body>
</html>

PHP頁面

<?php  
$select_op=$_GET['select_op'];
if($select_op != "")
{
	if($select_op == "10")
	{
      echo "未面談確認(受評者)";
    }


	else if($select_op == "20")
	{
      echo "未面談確認(直屬主管)";
    }

	else if($select_op == "1000")
	{
      echo "兩者已面談確認";
    }
	
}	

else
{
  echo "請選擇一個選項";
}
?>

1 個回答

1

id 是不能重覆的…

小哈 iT邦新手 4 級 ‧ 2018-08-22 10:48:49 檢舉

感謝前輩回覆~
已經用以下方法取得到值,再利用前例的教學套用就OK了~
https://blog.csdn.net/magi1201/article/details/44465557
/images/emoticon/emoticon08.gif

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