我有3个选择选项搜索数据表中的记录与ajax jQuery php。
我已经写了这段代码,但选择选项数据不去其他页面&没有从ajax请求返回。
PHP代码: -
<form name="search_form" id="search_form" method="POST">
<div class="col-md-3">
<div class="formrow">
<select class="form-control" name="job_title" class="select_filter">
<option value ='' disabled selected>Job Title</option>
<option>PHP Developer</option>
<option>Andorid Developer</option>
</select>
</div>
</div>
<div class="col-md-3">
<div class="formrow">
<select class="form-control" name="emptype" class="select_filter">
<option value ='' disabled selected>Employment Status</option>
<option>Permanent</option>
<option>Contract</option>
<option>Freelance</option>
</select>
</div>
</div>
<div class="col-md-3">
<div class="formrow">
<select class="form-control" name="experience" class="select_filter">
<option value ='' disabled selected>Experience</option>
<option>Fresher</option>
<option>1 Year</option>
<option>2 Years</option>
<option>3 Years</option>
<option>4 Years</option>
<option>5 Years</option>
<option>6 Years</option>
<option>7 Years</option>
<option>8 Years</option>
<option>9 Years</option>
<option>10 Years</option>
</select>
</div>
</div>
</form>
jQuery/Ajax代码:-<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.5.1/jquery.min.js"></script>
<script type="text/javascript" rel="stylesheet">
$(document).ready(function(){
$('.select_filter').on('change',function(){
$.ajax({
type: "POST",
url: "ajaxCompany_search.php",
data: $('#search_form').serialize(),
success:function(data){
console.log(data);
alert(data);
$("#projects").html(data);
}
});
});
});
</script>
ajaxCompany_search.php
<?php
include('../../config.php');
print_r($_POST);
?>
您必须将类分组在一个唯一的标签中:
改变class="form-control" ... class="select_filter"
class="form-control select_filter"
您已经在select字段中两次给出class,因此没有显示数据。
<select class="form-control" name="job_title" class="select_filter">
将代码写成
<select class="form-control select_filter" name="job_title">
就像这样写在这里