通過javascript用php上傳圖像

Question

carica.html

<td>
    <input type="file" size="30" onchange="preview()" id="upload_immagine">
</td>
<td>
    <div id="divImmagine" > </div>
</td>

文件javascript.js

function preview() 
{
    var xmlhttp;
    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("divImmagine").innerHTML=xmlhttp.responseText;
    }
    image=request.getParameter("upload_immagine");
    document.getElementById("divImmagine").innerHTML=image;
    xmlhttp.open("POST","stampaAnteprima.php", false);
    xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded");
    xmlhttp.send("image="+image);
}

stampaAnteprima.php

<?php
    $file_temp=($_FILES['image']['tmp_name']);
    echo"$file_temp";
?>

javascript代碼行image=request.getParameter("upload_immagine"); 什么也不返回。 我如何獲取值傳遞給php，然后通過$_FILES['image']['tmp_name']讀取文件，實際上它是圖像嗎？ 您有什么建議嗎？

Answer 1

這將允許您將文件發送到php。

您可以添加更多事件偵聽器以顯示上載進度和錯誤處理。

function UploadPhoto(){
    var file = document.getElementById('upload_immagine').files[0];
    var formdata = new FormData();
    formdata.append("image", file);
    var req = new XMLHttpRequest();
    req.addEventListener("load", function(event) { uploadcomplete(event); }, false);
    req.open("POST", "stampaAnteprima.php");
    req.send(formdata);
}
function uploadcomplete(event){
// Your php reply = event.target.responseText 
}