如何将选项字段中的两个值发布到表

Question

job_category{
category_id PRIMARY KEY,
category_name
}

job{
job_id PRIMARY KEY,
role
category_id FOREIGN KEY
}

exam{
exam_id AUTO INCREMENT PRIMARY KEY,
exam_name,
job_cat
}

// job_cat是con-cat值as job_id-category_id

这是我的选择选项

<select name="job_cat" id="select_item" required="true">
<option value="">Select the Job Role</option>
<?php 

$sql1="SELECT Cat.category_id category_id, Cat.category_name category_name, Job.job_id job_id, Job.role role FROM job Job LEFT JOIN job_category Cat ON Job.category_id=cat.category_id";

$result1= mysqli_query($dbcon, $sql1);
while($row1=mysqli_fetch_array($result1)){?>
<option value="<?php echo $row1["job_id"] . "-" . $row1["catergory_id"]; ?>"><?php echo $row1["role"] . " - " . $row1["catergory_name"] ;?></option>
<?php }?>
</select>

//插入值

$exam_name=$_POST['exam_name'];
$job_cat=$_POST['job_cat'];

$sql2= "INSERT INTO exam_paper (exam_name,job_cat) VALUES ('$exam_name','$job_cat')";

这不能正常工作。 只将job_id插入'job_cat'列。 category_id未添加。  我该怎么办？

Answer 1

在表格中

将job category替换为job_category 以及

的插入查询

INSERT INTO exam_paper (`exam_name`,`job_cat`) VALUES ('".$exam_name."','".$job_cat."')

Answer 2

这将插入您的数据只是将您的列名称而不是column_name

$job_cat=$_POST['job_cat'];
$ins=mysqli_query("insert into column_name values ('".$job_cat."')");
if($ins)
{echo "data inserted";}
else
{echo "data not inserted";}