Question

我想使用ajax在不关闭模式的情况下以bootstrap模式显示用户当前插入的注释（在注释提交成功之后：）。我该怎么办？

[图片示例->>：] [https://i.stack.imgur.com/qjpM5.jpg]

process.php：

   <script type="text/javascript">
       $('#button').click(function(){
         var content_id = $('.comment_id').attr("id");
         var comment = $('#comment').val();
         $.ajax({
              url:"process.php",
              method:"post",
              data:{
                content_id:content_id,
                comment:comment
              },
              success:function(data){
// after success I want to display user's current inserted comment in bootstrap modal (just after comment submit) using ajax without closing modal. How can I do this?
              }
         });
       });
      </script>
<?php
 if (isset($_POST['content_id']) && isset($_POST['comment'])) {
   $c_id = $_POST['content_id'];
   $c = $_POST['comment'];
   $connect = mysqli_connect("localhost", "root", "", "database");
   $query = mysqli_query($connect,"INSERT INTO `comments`(content_id,comments) values ('$c_id','$c')");

 }
?>

Answer 1

在AJAX成功回调函数中，使用JavaScript复制文本并将其显示在想要的元素中。

例如：

function showHint(str) {
    if (str.length == 0) {
        document.getElementById("txtHint").innerHTML = "";
        return;
    } else {
        var xmlhttp = new XMLHttpRequest();
        xmlhttp.onreadystatechange = function() {
            if (this.readyState == 4 && this.status == 200) {
                // THIS LINE //
                document.getElementById("txtHint").innerHTML = this.responseText;
            }
        };
        xmlhttp.open("GET", "gethint.php?q=" + str, true);
        xmlhttp.send();
    }
}

参考：https://www.w3schools.com/js/js_ajax_php.asp

如何使用Ajax从mysql数据库以引导方式显示当前插入的数据？

1 个答案: