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響應我有一個HTML頁面如何填充在同一div,這是作爲填充AJAX
說main.php,用ajax其中填充「upload.php的」在DIV。
現在,在這個upload.php中,我上傳了一張圖片,並且我想以某種方式在main.php中創建這個div,以向我顯示圖片加載成功與否的響應。
我的代碼如下所示:
的main.php
<script type="text/javascript">
function showHint()
{
var str = document.form1.filenumber.value;
if (str.length==0)
{
document.getElementById("txtHint").innerHTML="";
return;
}
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","upload.php?filenumber="+str,true);
xmlhttp.send();
}
</script>
</head>
<body>
<form name="form1" >
<tr>
<td>File Number </td><td><input type="text" id="11" name="filenumber" /></td>
</form>
<td colspan="2"><input type="submit" onclick='showHint()'></td>
</tr>
</table>
<div id="txtHint"></div>
而upload.php的
<?php
$filenumber = clean($_GET['filenumber']);
if($filenumber != '') {
$qry = "SELECT * FROM profile WHERE filenum='$filenumber'";
$result = mysql_query($qry)
or die(mysql_error());
$row = mysql_fetch_array($result);
if($result) {
if(mysql_num_rows($result) == 0) {
$errmsg = '<div style="width:300px; height:100px;color:red;margin:0px auto;position:relative;top: 30%">No such record found. Redirecting back to the status page. </p>';
$errflag = true;
}
else
{
?>
<form enctype="multipart/form-data" action="uploader.php" method="POST">
Please choose a file: <input name="uploaded" type="file" /><br />
<input type="submit" value="Upload" />
</form>
<?php
}
//@mysql_free_result($result);
}
else {
die("Query111 failed");
}
}
?>
而upload.php的要求uploader.php,其響應我想展示在同一個div
<?php
$target = "images/";
$target = $target . basename($_FILES['uploaded']['name']) ;
$ok=1;
if(move_uploaded_file($_FILES['uploaded']['tmp_name'], $target))
{
echo "The file ". basename($_FILES['uploadedfile']['name']). " has been uploaded";
}
else {
echo "Sorry, there was a problem uploading your file.";
}
?>
你的建議是最受歡迎的,你的幫助表示讚賞。
感謝 Zee的
一個建議是在使用AJAX和DOM操作時使用JQuery。你會讓自己變得更容易。 – GolezTrol