te')); return $arr; } /* 遍历用户所有主题 * @param $uid 用户ID * @param int $page 页数 * @param int $pagesize 每页记录条数 * @param bool $desc 排序方式 TRUE降序 FALSE升序 * @param string $key 返回的数组用那一列的值作为 key * @param array $col 查询哪些列 */ function thread_tid_find_by_uid($uid, $page = 1, $pagesize = 1000, $desc = TRUE, $key = 'tid', $col = array()) { if (empty($uid)) return array(); $orderby = TRUE == $desc ? -1 : 1; $arr = thread_tid__find($cond = array('uid' => $uid), array('tid' => $orderby), $page, $pagesize, $key, $col); return $arr; } // 遍历栏目下tid 支持数组 $fid = array(1,2,3) function thread_tid_find_by_fid($fid, $page = 1, $pagesize = 1000, $desc = TRUE) { if (empty($fid)) return array(); $orderby = TRUE == $desc ? -1 : 1; $arr = thread_tid__find($cond = array('fid' => $fid), array('tid' => $orderby), $page, $pagesize, 'tid', array('tid', 'verify_date')); return $arr; } function thread_tid_delete($tid) { if (empty($tid)) return FALSE; $r = thread_tid__delete(array('tid' => $tid)); return $r; } function thread_tid_count() { $n = thread_tid__count(); return $n; } // 统计用户主题数 大数量下严谨使用非主键统计 function thread_uid_count($uid) { $n = thread_tid__count(array('uid' => $uid)); return $n; } // 统计栏目主题数 大数量下严谨使用非主键统计 function thread_fid_count($fid) { $n = thread_tid__count(array('fid' => $fid)); return $n; } ?>html - multipartform-data possible to send with javascript? - Stack Overflow

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    html - multipartform-data possible to send with javascript? - Stack Overflow

    programmeradmin4浏览0评论

    I am sending a file over POST together with text "name" using this form:

    <form enctype="multipart/form-data" action="https://site[DOT]net/upload" method="post">
  <input id="name" type="text" />
  <input id="data" type="file" />
  <button type="submit" name="submit" />
</form>

    I would like to do the exect same using javascript. In addition I dont want to be redirected. I want to stay on the html page and just show a popup "Upload done". How can I do this in javascript (no jquery)?

    EDIT:

    I tried this code but the POST is not working:

    <script>
function uploader {
  var formData = new FormData();
  formData.append("name", "Smith");
  formData.append("data", fileInputElement.files[0]);
  var request = new XMLHttpRequest();

  request.open("POST", "https://site[DOT]net/upload");
  request.send(formData);
}
</script>

<form>
    <input id="name" type="text" />
    <input id="data" type="file" />
    <button type="submit" name="submit" />
    <button onclick="uploader()">Click</button>
</form>

    I am sending a file over POST together with text "name" using this form:

    <form enctype="multipart/form-data" action="https://site[DOT]net/upload" method="post">
  <input id="name" type="text" />
  <input id="data" type="file" />
  <button type="submit" name="submit" />
</form>

    I would like to do the exect same using javascript. In addition I dont want to be redirected. I want to stay on the html page and just show a popup "Upload done". How can I do this in javascript (no jquery)?

    EDIT:

    I tried this code but the POST is not working:

    <script>
function uploader {
  var formData = new FormData();
  formData.append("name", "Smith");
  formData.append("data", fileInputElement.files[0]);
  var request = new XMLHttpRequest();

  request.open("POST", "https://site[DOT]net/upload");
  request.send(formData);
}
</script>

<form>
    <input id="name" type="text" />
    <input id="data" type="file" />
    <button type="submit" name="submit" />
    <button onclick="uploader()">Click</button>
</form>
    • javascript
    • html
    • ajax
    Share Improve this question edited Feb 19, 2016 at 18:17 haramash asked Feb 19, 2016 at 17:55 haramashharamash 1151 gold badge1 silver badge4 bronze badges
    Add a ment  | 

    2 Answers 2

    Reset to default 5

    In case any wants to do this with the new fetch instead of xhr this is the equivalent. Also see: How do I POST with multipart form data using fetch?

    var form = document.getElementById('formid');

form.onsubmit = async (e) => {
  e.preventDefault();
  const form = e.currentTarget;
  const url = form.action;

  try {
      const formData = new FormData(form);
      const response = await fetch(url, {
          method: 'POST',
          body: formData
      });

      console.log(response);
  } catch (error) {
      console.error(error);
  }

}
    <form id="formid" enctype="multipart/form-data" action="#" method="post">
  <input id="name" type="text" />
  <input id="data" type="file" />
  <button type="submit" name="submit">Submint</button>
</form>

    Uploading the entire form with javascript, including the files, can be done by using the FormData API and XMLHttpRequest

    var form = document.getElementById('myForm'); // give the form an ID
var xhr  = new XMLHttpRequest();              // create XMLHttpRequest
var data = new FormData(form);                // create formData object


xhr.onload = function() {
    console.log(this.responseText); // whatever the server returns
}

xhr.open("post", form.action);      // open connection
xhr.send(data);                     // send data

    If you need to support IE10 and below, it gets increasingly plicated, and would in some cases require posting through iFrames etc.

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