获得两次ajax回调结果

Question

代码：

Sub Demo()
Application.ScreenUpdating = False
Dim StrFnd As String, Rng As Range, i As Long
StrFnd = InputBox("What is the Text to Find")
If Trim(StrFnd) = "" Then Exit Sub
With ActiveDocument.Range
  With .Find
    .ClearFormatting
    .Replacement.ClearFormatting
    .Text = StrFnd
    .Replacement.Text = ""
    .Forward = True
    .Wrap = wdFindStop
    .Format = False
    .MatchCase = True
    .MatchWholeWord = True
    .MatchWildcards = False
    .MatchSoundsLike = False
    .MatchAllWordForms = False
    .Execute
  End With
  Do While .Find.Found
    i = .Information(wdActiveEndAdjustedPageNumber)
    Set Rng = ActiveDocument.GoTo(What:=wdGoToPage, Name:=i)
    Set Rng = Rng.GoTo(What:=wdGoToBookmark, Name:="\page")
    If Rng.Tables.Count > 0 Then
      MsgBox Chr(34) & StrFnd & Chr(34) & " and table found on page " & i & "."
      With Rng.Tables(1)
        'process this table
      End With
    Else
      MsgBox Chr(34) & StrFnd & Chr(34) & " found on page " & i & " but no table."
    End If
    .Start = Rng.End
    .Find.Execute
  Loop
End With
Application.ScreenUpdating = True
End Sub

在这段代码中，我创建了一个ajax jquery脚本，我正在使用onclick函数。现在当我点击

时会发生什么

<script>
    $(document).ready(function(){
        $(".apply").click(function(){
            idd = this.id;
            fname = $("#fname").val();
            company_name = $(".company_name_"+idd).val();
            phone = $("#phone").val();
            alert(fname);
            alert(company_name);
            alert(phone);
            $.ajax({
                type:"POST",
                data:{"fname":fname, "company_name":company_name, "phone":phone},
                url:"get_data.php",
                success:function(data){
                    $("#search").html(data);
                }
            });
        });
    });
</script>

<?php
    $sql = "select * from company";
    $results = mysqli_query($con,$sql);
    while($rows = mysqli_fetch_array($results))
    {
        echo "<input type='text' name='company_name' class='company_name_".$rows['id']."' value='".$rows['company_name']."'/>";
        echo "<a href='javascript:void(0)' id='".$rows['id']."' class='apply'>click me</a>";
    }
?>

<?php
    $query = "select * from login where id = '".$ids."'";
    $result = mysqli_query($con,$query);
    while($row = mysqli_fetch_array($result))
    {
?>
    <input type="hidden" name="fname" id="fname" value="<?php echo $row['fname']; ?>"/>
    <input type="hidden" name="phone" id="phone" value="<?php echo $row['phone']; ?>"/>
<?php
    }
?>

Ajax回调两次，即它提醒值两次，但我想要一次。我不知道为什么？那么，我该如何解决这个问题？

谢谢