我如何可以调用使用JQuery切换和AJAX PHP脚本？ [英] How can I call a PHP script using JQuery toggle and AJAX?

问题描述

我想打电话给我的主PHP文件的PHP脚本。我想对正在运行的SQL查询显示从我的PHP脚本的结果。

I am trying to call a PHP script in my main PHP file . I want to display the results from my PHP script with the SQL queries that are being run.

我还想包括显示结果动态/通过不刷新页面的可能性。

I'd also like to include the possibility of showing the results dynamically/by not refreshing the page.

这是我试过到目前为止，即时通讯新jQuery和AJAX。在此先感谢！

this is what I tried so far, im new to Jquery and AJAX. thanks in advance!

工作拨弄： http://jsfiddle.net/52n861ee/ 这就是我想做的事情，但是当我点击它会告诉我的错误：第23行（在这里我使用json_en code

working fiddle: http://jsfiddle.net/52n861ee/ thats what I want to do but when I click on it will tell me error: line 23 (" where I am using json_encode

的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() {
                        $id = $(this).attr("data")
                    $("#station_info_"+$id).toggle();

                    $.ajax({
                        url:"display_stationinfo.php",
                        type: "GET",
                        success:function(result){
                    $("#station_info_"+$id).html(result);
                    }});//end ajax  
                    });//end click
                    });//end ready
    </script>
</div> <!-- end map_size -->

display_station.php（脚本，我要打电话）：

<?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>";
}//end while loop for station_result
    echo $_GET['callback'] . '(' .json_encode($html) . ')';             
mysqli_close($conn); // <-- DO I NEED TO INCLUDE IT HERE OR IN MY db_conn.php SINCE IM INCLUDING IT AT THE TOP?

？>

推荐答案

三个问题：

1. 在你的PHP脚本你治疗HTML字符串，就好像它是一个对象或数组--- 您不能输出一个字符串作为JSON

在你的JS脚本你治疗JSON[对象]，就好像它是一个字符串或HTML对象--- 您不能输出JSON这样，你需要使用DOT符号来访问它的数据

In your JS script you're treating the 'JSON' [object] as though it was a string literal or a html object --- you cannot output JSON that way, you need to use DOT notation to access it's data.

即使你的PHP脚本似乎发送 JSONP 您还没有指定一个 JSONP 的dataType 在 AJAX 电话。

Even though your PHP script seems to be sending JSONP you have not specified a jsonp dataType in your ajax call.

假设你所有的PHP脚本都在同一个域，将回声$ _GET ['回调']。 （.json_en code（$ HTML）。）; 来：

Assuming all your PHP scripts are on the same domain, change echo $_GET['callback'] . '(' .json_encode($html) . ')'; to:

echo $html; 

让你的PHP输出与您的JS所预期的一致。

So that your PHP output is consistent with what your JS expects.

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