如何在复选框的每个选项中插入访问相同参考ID的多个选择

Question

当同时在两个表中插入值时，我有称为用户和类别的表。 但是在类别表格中，我使用6个不同选项的复选框插入不同的类别，这些选项被访问时会购买一个名称。但问题是我插入的选择我想每个都从用户的表中获取相同的引用标识，以便我可以轻松跟踪该用户的选择。现在，当我插入它给每个选择一个不同的参考ID，它只使用第一选择的原始ID请帮助我解决这个问题。下面是 是代码，但我删除了一些，以便我们只关注问题。

<?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

我觉得挑战在这里吧？ Cos，可以插入一只猫吗？

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

好吧，也许这不是一个真正的答案，但假设你试图插入多个猫，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(); 
    ... 
} 

在这里看到更多的细节：http://php.net/manual/en/pdostatement.bindvalue.php