Question

我正忙着用ajax表单传递隐藏输入的一些值。我试过可能的方法，但看起来价值没有通过。

HTML

<form id="upload_form" enctype="multipart/form-data" method="post">
    <input type="hidden" name="sid" value="$rec_sStdi[std_sid]" />
    <input type="hidden" name="stid" value="$rec_sStdi[stdi_sid]" />
    <input type="text" name="title" class="form-control" size="50" maxlength="128" autocomplete="off" placeholder="ชื่อไฟล์ เช่น แบบประเมินครู เป็นต้น"/>
    <progress id="progressBar" value="0" max="100" style="width:100%;"></progress>                          
    <button class="btn btn-primary btn-xs pull-right" type="button" value="Upload File" onclick="uploadFile()"><i class="glyphicon glyphicon-upload"></i>&nbsp;อัพโหลด</button>
    <input type="file" name="file1" id="file1" class="btn btn-danger btn-xs pull-right">
  <b id="status"></b>
  <p id="loaded_n_total"></p>
</form>

的Javascript

function _(el){
    return document.getElementById(el);
}
function uploadFile(){
    var file = _("file1").files[0];
    //alert(file.name+" | "+file.size+" | "+file.type);
    var formdata = new FormData();
    formdata.append("file1", file);
    var ajax = new XMLHttpRequest();
    ajax.upload.addEventListener("progress", progressHandler, false);
    ajax.addEventListener("load", completeHandler, false);
    ajax.addEventListener("error", errorHandler, false);
    ajax.addEventListener("abort", abortHandler, false);
    ajax.open("POST", "inc/eval_file_upload.php");
    ajax.send(formdata);
}
function progressHandler(event){
    _("loaded_n_total").innerHTML = "Uploaded "+event.loaded+" bytes of "+event.total;
    var percent = (event.loaded / event.total) * 100;
    _("progressBar").value = Math.round(percent);
    _("status").innerHTML = Math.round(percent)+"% uploaded... please wait";
}
function completeHandler(event){
    _("status").innerHTML = event.target.responseText;
    _("progressBar").value = 0;
}
function errorHandler(event){
    _("status").innerHTML = "Upload Failed";
}
function abortHandler(event){
    _("status").innerHTML = "Upload Aborted";
}

PHP（eval_file_upload.php）

<?
$sid=$_POST['sid'];
$stid=$_POST['stid'];
$title=$_POST['title'];

$fileName = $_FILES["file1"]["name"]; // The file name
$fileTmpLoc = $_FILES["file1"]["tmp_name"]; // File in the PHP tmp folder
$fileType = $_FILES["file1"]["type"]; // The type of file it is
$fileSize = $_FILES["file1"]["size"]; // File size in bytes
$fileErrorMsg = $_FILES["file1"]["error"]; // 0 for false... and 1 for true
if (!$fileTmpLoc) { // if file not chosen
    echo "ERROR: Please browse for a file before clicking the upload button.";
    exit();
}
if(move_uploaded_file($fileTmpLoc, "../files/$fileName")){

    echo "\"$fileName\" upload is complete (sid:$sid , stid:$stid, title:$title)";  
} else {
    echo "move_uploaded_file function failed";
}
?>

我试图通过回显隐藏的值来测试传递。但没有任何表现。参考：http://www.developphp.com/view.php?tid=1351

Answer 1

尝试在ajax调用中添加params：

sid = document.getElementById("sid").value;
stid = document.getElementById("stid").value;
formdata.append("sid", sid);
formdata.append("stid", stid);

并为隐藏的输入类型添加一个ID。

<input type="hidden" id="sid" name="sid" value="$rec_sStdi[std_sid]" />
<input type="hidden" id="stid" name="stid" value="$rec_sStdi[stdi_sid]" />

如何使用ajax上传表单传递隐藏的值

1 个答案: