Question

我正在尝试在我的主PHP文件中调用PHP脚本。我想用我的PHP脚本显示正在运行的SQL查询的结果。

这是我到目前为止所尝试的，我是Jquery和AJAX的新手。提前谢谢！

工作小提琴：http://jsfiddle.net/52n861ee/ 这就是我想要做的事情但是当我点击desk_box DIV时，我的display_stationinfo.php脚本没有创建切换station_info DIV。

当我查看源代码时，两个DIV应该已经创建但只有desk_box是..我做错了什么？

JQuery / AJAX部分：

<div id="map_size" align="center">
<script type="text/javascript">
//Display station information in a hidden DIV that is toggled
                    //And call the php script that queries and returns the results LIVE
                    $(document).ready(function() {
                    $(".desk_box").click(function() {
                        alert("before toggle");
                        var id = $(this).attr("data")
                        alert(id);
                        alert($(this));
                    $("#station_info_"+id).toggle();
                        alert("after toggle");
                    $.ajax({
                        url:'display_stationinfo.php',
                        type: 'GET',
                        success:function(result){
                            alert("before result");
                    $("#station_info_"+id).html(result);
                    alert("result: " + result); //it shoes every DIV being created and not the one that I clicked on
                    alert("after result");
                    }});//end ajax  
                    });//end click
                    });//end ready
</div> <!-- end map_size -->

display_station.php（我要调用的脚本）：

<?php
include 'db_conn.php';
//query to show workstation/desks information from DB for the DESKS
$station_sql = "SELECT coordinate_id, x_coord, y_coord, section_name FROM coordinates";
$station_result = mysqli_query($conn,$station_sql);

//see if query is good
if($station_result === false) {
    die(mysqli_error()); 
}


//Display workstations information in a hidden DIV that is toggled
while($row = mysqli_fetch_assoc($station_result)){
    //naming values
    $id       = $row['coordinate_id'];
    $x_pos    = $row['x_coord'];
    $y_pos    = $row['y_coord'];
    $sec_name = $row['section_name'];
    //display DIV with the content inside
    $html = "<div class='station_info_' id='station_info_".$id."' style='position:absolute;left:".$x_pos."px;top:".$y_pos."px;'>Hello the id is:".$id."</br>Section:".$sec_name."</br></div>";
    echo $html;
}//end while loop for station_result
mysqli_close($conn); // <-- DO I NEED TO INCLUDE IT HERE OR IN MY db_conn.php SINCE IM INCLUDING IT AT THE TOP?

＆GT;

如何在JQuery切换功能中使用AJAX调用PHP脚本？

0 个答案: