从事件创建提交按钮

时间:2017-05-15 09:52:55

标签: javascript php html

我想在单击提交按钮时出现的元素上创建一个提交按钮

第1步:我在' index.php'然后我点击一个按钮开始一个评论框,该评论框由一个java脚本函数调用,它来自一个评论.php'

步骤2:我想在被调用元素上插入一个提交按钮(如注释按钮)。

第1页



function showcomment(question_ID) {
  var request;
  try {
    request = new XMLHttpRequest();
  } catch (tryMicrosoft) {
    try {
      request = new ActiveXObject("Msxml2.XMLHTTP");
    } catch (otherMicrosoft) {
      try {
        request = new ActiveXObject("Microsoft.XMLHTTP");
      } catch (failed) {
        request = null;
      }
    }
  }
  var php = "./core/show-comment.php";
  var comment_content = document.getElementById("comment_content_" + question_ID).value;
  var vars = "comment_content=" + comment_content + "&user_ID=" + user_ID + "&question_ID=" + question_ID;
  request.open("POST", php, true);

  request.setRequestHeader("Content-type", "application/x-www-form-urlencoded");

  request.onreadystatechange = function() {
    if (request.readyState == 4 && request.status == 200) {
      var return_data = request.responseText;
      document.getElementById(question_ID).innerHTML = return_data;
    }
  }

  request.send(vars);
}

<input type='submit' name='comment' value='show comment' onclick='fonction()'>
&#13;
&#13;
&#13;

第2页

&#13;
&#13;
<type="submit" name="like">

    <?php
    if (isset($_POST['like']))
    {
    	$query_new_like = "UPDATE comments SET approval_rate = approval_rate +1 WHERE comment_ID = '$comment_ID'" ;
    mysqli_query ($connect,$query_new_like) ;
    }
    ?>
&#13;
&#13;
&#13;

0 个答案:

没有答案