javascript TypeError: 'stepUp' 在未在 jquery 中实现接口 HTMLInputElement 的对象上调用
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TypeError: 'stepUp' called on an object that does not implement interface HTMLInputElement in jquery
提问by Manju
I am calling the php file using jquery $.post method.At the time of the retriving the data I am getting an error.
我正在使用 jquery $.post 方法调用 php 文件。在检索数据时出现错误。
I did search on google,found 4 to 5 solutions but those are not working.
我确实在谷歌上搜索过,找到了 4 到 5 个解决方案,但这些都不起作用。
Please help me.
请帮我。
this is my code,
这是我的代码
Jquery
查询
$(document).ready(function(){
$("[name='side_door_on_size']").click(function (){
var side_door_on_size=$(this).val();
$.post('<?php echo get_home_url(); ?>/wp-content/plugins/thermax/filter.php',
{side_door_on_size:side_door_on_size},
function(data){
alert(data);
});
})
})
Html
html
<input type="radio" name="side_door_on_size" class="side_door_on_size" id="side_door_on_size" value="25m"/>25m
回答by Girish
Please add element name in jQuery selector $("input[name=side_door_on_size]")
,
请在 jQuery 选择器中添加元素名称$("input[name=side_door_on_size]")
,
try this code
试试这个代码
$(document).ready(function(){
$("input[name=side_door_on_size]").click(function (){
var side_door_on_size=$(this).val();
$.post('<?php echo get_home_url(); ?>/wp-content/plugins/thermax/filter.php',
{side_door_on_size:side_door_on_size},
function(data){
alert(data);
});
})
})
回答by Laukik Patel
please try with
请尝试
<script>
$(document).ready(function() {
$("[name=side_door_on_size]").click(function() {
var side_door_on_size = $(this).val();
$.post('<?php echo get_home_url(); ?>/wp-content/plugins/thermax/filter.php', {side_door_on_size: side_door_on_size}, function(data) {
alert(data);
});
});
});
</script>