Question

<table class="table table-bordered table-striped table-hover datatable">
      <thead>
        <tr>
          <th>ID</th>
          <th>organisation Name</th>
          <th>User name</th>
          <th>Email</th>
          <th>Contact No</th>
          <th>IP</th>
          <th>Date</th>
          <th>Status</th>
        </tr>
      </thead>
      <tbody>
          <tr class="success"> <td>3</td>
        <td>Harshit </td>
        <td>AtulSaini</td>
        <td>arpitkumar@gmail.com</td>
        <td>786048</td>
        <td>::1</td>
        <td>14/03/2015</td><td><button id="status"><span class="label label-success">Active</span></button></td></tr><tr class="none"> <td>4</td>
        <td>Meghaa.co.edu</td>
        <td>megha</td>
        <td>meghaa16@gmail.com</td>
        <td>786048</td>
        <td>::1</td>
        <td>14/03/2015</td><td><button id="status"><span class="label label-success">Active</span></button></td></tr>            

      </tbody>
    </table>

如果要在服务器响应上执行条件以改变res变量，那么我想要的最后一个问题是当我把整个if代码块放在ajax成功然后变量＆＃34;这个＆＃34;值得到改变，禁用代码来改变tr 如果我把它保持在外面，就像我在这里所做的那样，变量res在成功函数中被破坏

<script type="text/javascript">
$(document).ready(function() {
$(".label").click(function() {
    var idx = $(this).parents("tr").find('td:first').html();
    var status = $(this).parents("tr").find('span').html();
    var cname=$(this).parents('tr').attr("class");
    alert(this);

    $.ajax({
        url : "scripts/update.php",
        type : 'GET',
        data : {
            "status" : status,
            "id" : idx
        },
        error : function(xhr, status, error) {
            alert(error);
        },
        success : function(entry) {
            var res = entry;    

        }
    });
  alert(status);
    if (status == 'Active') {
                $(this).parents("tr").removeClass(cname);
                $(this).parents("tr").addClass("warning");
                $(this).removeClass("label-success");
                $(this).addClass("label-warning");
                $(this).text("Deactive");
                alert("Account Deacivated");
            } else {
                $(this).parents('tr').removeClass(cname);
                $(this).parents('tr').addClass("success");
                $(this).removeClass("label-warning");
                 $(this).addClass("label-success");
                $(this).text("Active");
                alert("Account Acivated");

         }
   });
   });   
   </script>

Thanx提前帮助你......

Answer 1

我不确定您要实现的目标（例如，res / entry中存储了哪种数据）。但也许以下内容对您有所帮助：

<script type="text/javascript">
$(document).ready(function() {
  $(".label").click(function() {
     // store reference for later use (and to avoid creating the same object multiple times)
    var item   = $(this);

    // use the previously stored jQuery item instead of recreating it every time with $(...)
    var idx    = item.parents("tr").find('td:first').html();
    var status = item.parents("tr").find('span').html();

    // ...

    $.ajax({
      url: "scripts/update.php",
      type: 'GET',
      data: {
        "status": status,
        "id": idx
      },
      error: function(xhr, status, error) {
        alert(error);
      },
      success: function(entry) {
        // `item` will still reference to the label you clicked on
      }
    });

  });
});   
</script>

Answer 2

试试这个：

...
var $this = $(this);
$.ajax({
        url : "scripts/update.php",
        type : 'GET',
        data : {
            "status" : status,
            "id" : idx
        },
        error : function(xhr, status, error) {
            alert(error);
        },
        success : function(entry) {
            changeStatuts(entry, $this);

        }
    });

    var changeStatuts = function(status, ele){

           if (status == 'Active') {
                ele.parents("tr").removeClass(cname);
                ele.parents("tr").addClass("warning");
                ele.removeClass("label-success");
                ele.addClass("label-warning");
                ele.text("Deactive");
                alert("Account Deacivated");
            } else {
                ele.parents('tr').removeClass(cname);
                ele.parents('tr').addClass("success");
                ele.removeClass("label-warning");
                ele.addClass("label-success");
                ele.text("Active");
                alert("Account Acivated");

         }
    };

Answer 3

好的，在阅读完评论之后，或许类似以下内容可以解决您的问题。但是：您的PHP代码似乎容易受到sql injection

的攻击

<script type="text/javascript">
$(document).ready(function() {
  $(".label").click(function() {
    // store reference for later use (and to avoid creating the same object multiple times)
    var item   = $(this);

    // use the previously stored jQuery item instead of recreating it every time with $(...)
    var parentRow = item.parents("tr");
    var idx       = parentRow.find('td:first').html();
    var status    = parentRow.find('span').html();

    $.ajax({
      url: "scripts/update.php",
      type: 'GET',
      data: {
        "status": status,
        "id": idx
      },
      error: function(xhr, status, error) {
        alert(error);
      },
      success: function(status) {
        if (status == 'Active') {
          parentRow.removeClass(cname);
          parentRow.addClass("warning");
          item.removeClass("label-success").addClass("label-warning").text("Deactive");
          alert("Account Deacivated");
        } else {
          parentRow.removeClass(cname);
          parentRow.addClass("success");
          item.removeClass("label-warning").addClass("label-success").text("Active");
          alert("Account Acivated");
        }
      }
    });
  });
});   
</script>

根据服务器响应vai php更改ajax成功的变量值

3 个答案: