php-在mousemove上调用jQuery AJAX
作者:互联网
我有一个用gdimage创建的图像,它有40000个5×5块链接到不同的用户配置文件,我希望当您将鼠标悬停在这些块之一上时,AJAX会通过检测x和y co-来从数据库中获取该配置文件.将其移到图像上方时会发出命令.
然后,单击该链接,并获得该信息,并获得指向该用户配置文件的链接.
这是到目前为止我得到的:
Javascript / jQuery:
<script type="text/javascript">
jQuery.fn.elementlocation = function() {
var curleft = 0;
var curtop = 0;
var obj = this;
do {
curleft += obj.attr('offsetLeft');
curtop += obj.attr('offsetTop');
obj = obj.offsetParent();
} while ( obj.attr('tagName') != 'BODY' );
return ( {x:curleft, y:curtop} );
};
$(document).ready( function() {
$("#gdimage").mousemove( function( eventObj ) {
var location = $("#gdimage").elementlocation();
var x = eventObj.pageX - location.x;
var x_org = eventObj.pageX - location.x;
var y = eventObj.pageY - location.y;
var y_org = eventObj.pageY - location.y;
x = x / 5;
y = y / 5;
x = (Math.floor( x ) + 1);
y = (Math.floor( y ) + 1);
if (y > 1) {
block = (y * 200) - 200;
block = block + x;
} else {
block = x;
}
$("#block").text( block );
$("#x_coords").text( x );
$("#y_coords").text( y );
$.ajax({
type: "GET",
url: "fetch.php",
data: "x=" + x + "&y=" + y + "",
dataType: "json",
async: false,
success: function(data) {
$("#user_name_area").html(data.username);
}
});
});
});
</script>
PHP:
<?
require('connect.php');
$mouse_x = $_GET['x'];
$mouse_y = $_GET['y'];
$grid_search = mysql_query("SELECT * FROM project WHERE project_x_cood = '$mouse_x' AND project_y_cood = '$mouse_y'") or die(mysql_error());
$user_exists = mysql_num_rows($grid_search);
if ($user_exists == 1) {
$row_grid_search = mysql_fetch_array($grid_search);
$user_id = $row_grid_search['project_user_id'];
$get_user = mysql_query("SELECT * FROM users WHERE user_id = '$user_id'") or die(mysql_error());
$row_get_user = mysql_fetch_array($get_user);
$user_name = $row_get_user['user_name'];
$user_online = $row_get_user['user_online'];
$json['username'] = $user_name;
echo json_encode($json);
} else {
$json['username'] = $blank;
echo json_encode($json);
}
?>
的HTML
<div class="tip_trigger" style="cursor: pointer;">
<img src="gd_image.php" width="1000" height="1000" id="gdimage" />
<div id="hover" class="tip" style="text-align: left;">
Block No. <span id="block"></span><br />
X Co-ords: <span id="x_coords"></span><br />
Y Co-ords: <span id="y_coords"></span><br />
User: <span id="user_name_area"> </span>
</div>
</div>
现在,mousemove位置的’block’,’x_coords’和’y_coords’变量可以正常工作并显示在span标记中,但是它没有从AJAX函数获取PHP变量,我不明白为什么.
我也不知道该怎么做,所以当单击鼠标时,它将获取从fetch.php中获取的变量,并将用户定向到诸如“ / user / view /?id = VAR_ID_NUMBER”的页面
我是用错误的方式来处理还是做错了?有人可以帮忙吗?
标签:ajax,mousemove,html,php,jquery 来源: https://codeday.me/bug/20191208/2093878.html