我正试图在我的主PHP文件中调用一个PHP脚本。下面是主php文件的Jquery/Ajax部分。display_stationinfo.php应该在main中创建DIV,但事实并非如此。
这是我迄今为止所尝试的,我对Jquery和AJAX是新手。提前感谢!
工作小提琴:http://jsfiddle.net/52n861ee/这就是我想做的,但当我点击desk_boxDIV时,我的display_stationinfo.php脚本并没有创建toggle-station_infoDIV。
当我查看源代码时,两个DIV都应该已经创建,但只有desk_box是..我做错了什么?
JQuery/AAJAX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
</script>
</div> <!-- end map_size -->
display_station.php(我想调用的脚本):
<?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?
?>
"SELECT coordinate_id, x_coord, y_coord, section_name FROM coordinates";
从表坐标中提取每一行,这就是你想要做的吗?还是只想返回带有用户单击的id的行?
jQuery
$.ajax({
url: 'display_stationinfo.php',
data: { 'id': id },
type: 'POST',
success: function(result) {}
});
php
$id = $_POST['id']
"SELECT coordinate_id, x_coord, y_coord, section_name FROM coordinates WHERE coordinate_id == " $id;
看看你的例子,我还猜测问题可能是你返回了一个字符串,并将其放入目标div中,这样完成的div看起来像这样:
<div class="station_info_" id="station_info_84" style="position: absolute; left: 20px; top: 90px; display: block;">
<div class="station_info_" id="station_info_84" style="position:absolute;left:20px;top:90px;">
Hello the id is:84<br>
Section:Section B<br>
</div>
</div>
您可以返回json对象并仅将数据附加到目标div ,而不是返回字符串
php
while ($row = mysqli_fetch_assoc($station_result)) {
$id = $row['coordinate_id'];
$x_pos = $row['x_coord'];
$y_pos = $row['y_coord'];
$sec_name = $row['section_name'];
$result = array('id' => $id, 'x_pos' => $x_pos, 'y_pos' => $y_pos, 'sec_name' => $sec_name);
echo json_encode($array);
}
jQuery
$.ajax({
url: 'display_stationinfo.php',
data: { 'id': id },
type: 'POST',
dataType: "json",
success: function(json) {
$("#station_info_"+id)
.css({'left':json.x_pos ,'top': json.y_pos})
.append('<p>Hello the id is:'+ json.id +'</br>Section:'+ json.sec_name +'</p>');
}
});