Php 根据选中的复选框选择查询

我有一个表单，用户可以在其中选择可用的技能：

<form name="myForm" action="jssearch.php" method="post">

 <input type="checkbox" name="chk1[]" value="1">Helpdesk Support
 <input type="checkbox" name="chk1[]" value="2">DB Admin<br>
 <input type="checkbox" name="chk1[]" value="3">C++ Programming
 <input type="checkbox" name="chk1[]" value="5">HTML<br>
 <input type="checkbox" name="chk1[]" value="6">PHP<br>
 <input type="checkbox" name="chk1[]" value="7">Memory Dump Analysis<br>
 <input type="checkbox" name="chk1[]" value="8">SQL<br><br>

 <input type="submit" name="Update" value="Search">

</form> 

帮助台支持
数据库管理

C++程序设计
HTML

PHP

内存转储分析

SQL

基于这些选择，我希望对多对多表运行查询，并显示包含所选技能的可用作业

这是我目前的疑问：

<?php

session_start();
mysql_connect("localhost", "root", "root") or die(mysql_error());
mysql_select_db("jobsearch") or die(mysql_error());

$variable=$_POST['chk1'];
foreach ($variable as $variablename)
{
    $query = mysql_query(
       "SELECT jobs.jobid AS job_id, jobs.jobtitle AS
        job_title,jobs.salary AS salary_desc, GROUP_CONCAT(skills.Desc) AS skills_desc    
        FROM jobskillsjoin
        INNER JOIN jobs ON jobs.jobid = jobskillsjoin.JobID
        INNER JOIN skills ON skills.skill_id = jobskillsjoin.SkillID
        WHERE skills.skill_id = '".$variablename."'
        GROUP BY jobs.jobid
        ")
    or die(mysql_error());
}

echo "<table border='1'>
<tr>
<th>Job ID</th>
<th>Job Title</th>
<th>Skills required</th>
<th>Salary Offered</th> 
</tr>";

while($row = mysql_fetch_array($query))
{
    echo "<tr>";
    echo "<td>" . $row['job_id'] . "</td>";
    echo "<td>" . $row['job_title'] . "</td>";
    echo "<td>" . $row['skills_desc'] . "</td>";
    echo "<td>" . $row['salary_desc'] . "</td>"; 
    echo "</tr>";
}
echo "</table>";


?>

---

当您像以前一样提交一个复选框时，它将成为PHP端的一个数组。您需要在where子句中使用此数组。使用函数infrade将数组转换为字符串，并使用运算符“in”。因此，您的where子句将是：

$query = mysql_query("SELECT jobs.jobid AS job_id, jobs.jobtitle AS
        job_title,jobs.salary AS salary_desc, GROUP_CONCAT(skills.Desc) AS skills_desc    
    FROM jobskillsjoin
    INNER JOIN jobs ON jobs.jobid = jobskillsjoin.JobID
    INNER JOIN skills ON skills.skill_id = jobskillsjoin.SkillID
    WHERE skills.skill_id in (". implode(",",$_POST['chk1']) .")
    GROUP BY jobs.jobid
    ")

这样，查询将返回所有已检查的技能

要返回至少选择了一项技能的作业的所有技能，您需要将查询逻辑更改为：

SELECT j.jobid AS job_id, j.jobtitle AS
            job_title, GROUP_CONCAT(skills_Desc) AS skills_desc    
        FROM jobskillsjoin
        INNER JOIN jobs j ON j.jobid = jobskillsjoin.JobID
        INNER JOIN skills ON skills.skill_id = jobskillsjoin.SkillID
where exists(select 1 from jobskillsjoin where jobid = j.jobid and SkillID in (1,2))
        GROUP BY j.jobid;

请注意，我更改了查询。不要复制并粘贴到代码中。调整它以确保您不会错过任何东西。
Hmmm。。。虽然它可以工作，但它只显示选定的技能。我的意思是，如果一份工作有与之相关的技能12和3，我搜索技能1，它就会找到这份工作。但这是唯一的打印技能1。。。我需要它来打印特定工作的所有技能。有什么想法吗？