如何在jquery中选择多个复选框时仅显示一次消息或数据?

Question

code：

<script>
$(document).ready(function(){
$(".choose").click(function(){
job_type = $(':checked').map(function() { 
return this.value; 
}).get().join(',');
category = $("#cat_gory").val();
$.ajax({
type:"POST",
data:{"job_type":job_type,"category":category},
url:"type-jobs.php",
success:function(data){
$(".success").html(data);
}
});
});
});
</script>
<input type="hidden" name="cat_gory" id="cat_gory" value="<?php echo $job_category; ?>" />  
<p class="flchek">
<input type="checkbox" name="choosetype" value="freelance" class="choose" id="33r">
<label for="33r">Freelance</label>
</p>
<p class="ftchek">
<input type="checkbox" name="choosetype" value="full time" class="choose" id="dsf">
<label for="dsf">Full Time</label>
</p>
<p class="ischek">
<input type="checkbox" name="choosetype" value="internship" class="choose" id="sdd">
<label for="sdd">Internship</label>
</p>
<div class="success"></div>

键入作业.php

<?php
include("config.php");
$job_category = mysqli_real_escape_string($con,$_POST['category']);
$job_type = mysqli_real_escape_string($con,$_POST['job_type']);
$tag = explode(",",$job_type);
foreach($tag as $type)
{
$sql_job = mysqli_query($con,"select p.job_title,c.state,c.city from company c inner join jobs p on p.company_name = c.company_name where p.category = '".$job_category."' and job_type='".$type."'");
$result_job = mysqli_num_rows($sql_job);
if($result_job > 0)
{
while($row_jobs = mysqli_fetch_array($sql_job))
{
echo '<h3>'.$row_jobs['job_title'].'</h3>';
}
}
else
{
echo '<p id="c_mm">record not found</p>';
}
}
?>

在此代码中，我有多个复选框。现在，当我单击复选框通过jquery/ajax获取数据时，它会将数据返回给我，但问题是如果我单击复选框freelance当我单击full time时它会再次显示record not found，然后它显示record not found以及jobs。现在，我希望如果我的数据库中没有找到数据，它将向我显示一条消息，如果找到数据，则隐藏消息并显示数据库记录。那么，我该怎么做呢？请帮助我。

谢谢

Answer 1

无需炸毁$jobtype使用'jobtype IN('.$jobtype.') '

<?php
include("config.php");
$job_category = mysqli_real_escape_string($con,$_POST['category']);
$job_type = mysqli_real_escape_string($con,$_POST['job_type']);
$sql_job = mysqli_query($con,"select p.job_title,c.state,c.city from company c inner join jobs p on p.company_name = c.company_name where p.category = '".$job_category."' and job_type IN('".$jobtype."') ');
$count = mysqli_num_rows($sql_job);
$row_data = mysqli_fetch_array($chek_result);
if($count > 0){
foreach($row_data as $data){
$message = '<h3>'.$data['job_title'].'</h3>';
}
}else{
$message = "No Data found";
}
echo $message;exit;
?>

Answer 2

如果你只想在找不到任何记录时才显示record not found消息，那么你可以在php中使用一个标志

，如下所示：

<?php
include("config.php");
$job_category = mysqli_real_escape_string($con,$_POST['category']);
$job_type = mysqli_real_escape_string($con,$_POST['job_type']);
$found = 0;
$message = '';
$tag = explode(",",$job_type);
foreach($tag as $type)
{
$sql_job = mysqli_query($con,"select p.job_title,c.state,c.city from company c inner join jobs p on p.company_name = c.company_name where p.category = '".$job_category."' and job_type='".$type."'");
$result_job = mysqli_num_rows($sql_job);
if($result_job > 0)
{
$found = 1;
while($row_jobs = mysqli_fetch_array($sql_job))
{
$message .= '<h3>'.$row_jobs['job_title'].'</h3>';
}
}
}
if(found == 0) {
$message = '<p id="c_mm">record not found</p>';
}
echo message;
?>

请注意代码中的$found标志。呵呵。