Question

我有一个以表格格式显示信息列表。现在我在此表中添加了一列，其中包含下拉列表，其中包含2个选项批准＆amp;拒绝。现在我想从下拉列表中选择选项时更新单个记录的状态，我想通过使用AJAX来实现。

PHP代码：

<form name="post_action" method="post" action="" id="post_action">
<select name="post_status" id="<?=$row['id']; ?>" style="width:175px; float:left; height:25px;" >
<option selected="selected" disabled="disabled">--Select--</option>
<option value="1">Approved</option>
<option value="2">Rejected</option>
<option value="3">Pending</option>
</select>

如果您有任何解决方案，请与我分享。

Answer 1

    function change_status(drop_id){
    var status=document.getElementById(drop_id).value;
    try
    {
    // Firefox, Opera 8.0+, Safari
    xmlHttp=new XMLHttpRequest();
    }
    catch (e)
    {
    // Internet Explorer
    try
      {
      xmlHttp=new ActiveXObject("Msxml2.XMLHTTP");
      }
    catch (e)
      {
      try
    {
    xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");
    }
      catch (e)
    {
    alert("Your browser does not support AJAX!");
    return false;
    }
      }
    }
    if(xmlHttp.readyState != 0){
        xmlHttp.abort(); 
    }
    xmlHttp.onreadystatechange=function()
      {
      if(xmlHttp.readyState==4)
        {
          //Do something to notify the user that the status has been updated.
        }
    }
    url="change_status.php?id="+drop_id+"&status="+status;
    xmlHttp.open("GET",url,true);
    xmlHttp.send(null);
}

并且选择是这样的：

<select name="post_status" id="<?=$row['id']; ?>" onchange="change_status(<?=$row['id']; ?>)" style="width:175px; float:left; height:25px;" >
<option selected="selected" disabled="disabled">--Select--</option>
<option value="1">Approved</option>
<option value="2">Rejected</option>
<option value="3">Pending</option>
</select>

然后创建文件change_status.php 从$ _GET收到id和新状态的地方，你可以适当地更新数据库。

Answer 2

JQuery

var serializedData = $form.serialize();
$("input[type=select]").change(function() {
    $.ajax({//Submit the value of the drop_down
        url: "/updateDB.php",
        type: "post",
        data: serializedData
        }).done(function() {
        alert('Updated!');
    });
});

PHP

<?php
     if(isset($_POST['id']){
        $id = $_POST['id'];
        $post_status = $_POST['post_status'];
        $sql = "UPDATE mytable SET myvalue = '$post_status' WHERE id = $id";
        //Connect to database
        $link = mysqli_connect("myhost","myuser","mypassw","mydb") or die("Error " . mysqli_error($link));
        //Execute the Query
        $result = $link->query($sql);
        //Using the $result object you can determine if the update was successful or not and alert the user of success/failure

?>

我认为这应该足以让你开始，这主要是在餐巾编码的背面写的，但我认为你会得到主要的想法。

使用AJAX / PHP / MySQL批准/拒绝记录

2 个答案: