如何在复选框的每个选择上插入访问相同参考 ID 的多个选择

Question

同时在两个表中插入值时，我有名为 user 和 category 的表。 但是在类别表中，我使用 6 个不同选择的复选框插入不同的类别，这些复选框正在访问购买一个名称。但问题是我插入的选项我希望每个选项都能从用户表中获得相同的引用 ID，这样我就可以轻松跟踪该用户的选择。现在，当我插入它时，它会为每个选项提供一个不同的参考 ID，并且它只在第一个选项上使用原始 ID 请帮助我解决这个问题。 下面是代码，但我删除了一些，以便我们只关注这个问题。

  <?php



?>

<div class="form">

    <h1>Client informatinon <?php  echo $_SESSION['username']."   ";?></h1>
    <form action ="form.php" method = "post" id="postform">
<table><tr><td>

<tr><td>
<input type="checkbox" name="category_name[]" id="inlineCheckbox1" value="Architecture"> Architecture
</td></td><td>
<input type="checkbox" name="category_name[]" id="inlineCheckbox2" value="townplanning">Town Planning 
</td></tr><tr><td>
<input type="checkbox" name="category_name[]" id="inlineCheckbox3" value="civilengineering">Civil Engineering 
                                </td><td>
<input type="checkbox" name="category_name[]" id="inlineCheckbox3" value="buildingandrenovation"> Building & Renovation
                                 </td></tr><tr><td>
                                 <input type="checkbox" name="category_name []" id="inlineCheckbox3" value="other"> Other 
        </td><td>                   
<input type="checkbox" name="category_name[]" id="inlineCheckbox3" value="interiorgaphicdesign"> Interior graphic design 

                                </td></tr>



    </form>
    </table>
</div>

<?php


if(isset($_POST['category_name'])){
foreach($_POST['category_name'] as  $value){

?>
<?php
try{
$query="INSERT INTO tish_user(username,Password,Previllage,date_created)
VALUES(:username,:Password,:Previllage,:date_created)";
$insert = $con->prepare($query);
$insert->execute(array(
':username'=>$username,
':Password'=>(md5($Password)),
':Previllage'=>$Previllage,
':date_created'=>$date_created));
#end of first table
################################################
#You select the first Id and put it in a variable then 
$id_last = ("SELECT LAST_INSERT_ID()");
$result =$con->prepare($id_last);
$result->execute();
$last_id = $result->fetchColumn();
############################## Last Id query Ends here
#insert into  clientinfo table 
$clientinfor="INSERT INTO tish_clientinfo
(title, firstname, lastname, nickname, idnumber, client_code, 
 company, country, city, province, address, cell, 
tel, webaddress, satifiedstatus, email, job_approval, cash_with_vat, 
cash_paid, date_registered,user_id)
VALUES(:title,:firstname,:lastname,:nickname,:idnumber,:client_code,
:company,:country,:city,:province,:address,
:cell,:tel,:webaddress,:satifiedstatus, :email, :job_approval,
:cash_with_vat,:cash_paid, :date_registered,$last_id)";
$clientinfor_insert = $con->prepare($clientinfor);
$clientinfor_insert->execute(array(
':title'=>$title,
':firstname'=>$firstname,
':lastname'=>$lastname,
':nickname'=>$nickname,
':idnumber'=>$idnumber,
':client_code'=>$client_code,
':company'=>$company,
':country'=>$country,
':city'=>$city,
':province'=>$province,
':address'=>$address,
':cell'=>$cell,
':tel'=>$tel,
':webaddress'=>$webaddress,
':satifiedstatus'=>$satifiedstatus,
':email'=>$email,
':job_approval'=>$job_approval,
':cash_with_vat'=>$cash_with_vat,
':cash_paid'=>$cash_paid,
':date_registered'=>$date_registered
));
#end of clien infor 
################################################
$security="INSERT INTO tish_security(ip_address,user_id)
VALUES(:ip_address,$last_id)";
$security_insert = $con->prepare($security);
$security_insert->execute(array(
':ip_address'=>$ip_address));
##########################end of security 
############ images 
$images ="INSERT INTO tish_images(user_id,image_name,date_registered)
VALUES($last_id,:image_name,:date_registered)";
$images_insert = $con->prepare($images);
$images_insert->execute(array(
':image_name'=>$rename,
':date_registered'=>$date_created));
##############################category 
$catigory="INSERT INTO tish_catigory(user_id,category_name)
VALUES($last_id,:category_name)";
$catigory_insert = $con->prepare($catigory);
$catigory_insert->execute(array(
':category_name'=>$value));
############# property table##########################################################
/*$property ="INSERT INTO tish_propertyinfo(user_id,date_registered)
VALUES($last_id,:date_registered)";
$property_insert = $con->prepare($images);
$property_insert->execute(array(':date_registered'=>$date_created));
*/}catch(PDOException $e){
echo $e->getMessage();
}
#3 fo the 
}
}
var_dump($value);



?>


</body>

Answer 1

我认为挑战就在这里，对吧？因为，只能插入一只猫吗？

$catigory="INSERT INTO tish_catigory(user_id,category_name)
    VALUES($last_id,:category_name)";

好吧，也许这不是一个真正的答案，但假设您尝试插入多只猫但具有相同的 last_id；

$cats = $vals = array();
foreach ((array) $_POST['category_name'] as $cat) {
    if ('' !== ($cat = trim($cat))) {
        $cats[] = $cat;
        $vals[] = "({$last_id}, ?)";
    }
}

if (!empty($cats)) {
    $sql = 'INSERT INTO tish_catigory (user_id, category_name) VALUES'. join(',', $vals);
    print($sql); // INSERT INTO tish_catigory (user_id, category_name) VALUES(111, ?),(111, ?)

    $sth = $con->prepare($sql);
    foreach ($cats as $i => $cat) {
        $sth->bindValue($i+1, $cat, PDO::PARAM_STR);
    }
    $sth->execute();
    ...
}

在此处查看更多详细信息：http://php.net/manual/en/pdostatement.bindvalue.php