Question

在两个表中同时插入值时，我有一个名为user和category的表。但是在类别表I中，使用6个不同选项的复选框插入不同的类别，这些选项正被访问购买一个名称。但问题是我插入的选项我希望每个人都从用户的表中获得相同的引用ID，这样我就可以很容易地跟踪该用户的选择。到现在，当我插入它给每个选择一个不同的参考ID，它只使用第一个选择的原始ID请帮我解决这个问题。下面是代码，但我删除了一些，所以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

我认为挑战在这里，对吗？是的，它只能插入一只猫吗？

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

好吧，也许这不是一个真正的答案，但假设你试图插入多只猫，但使用相同的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();
    ...
}

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

如何插入多个选项，在每个复选框选项上访问相同的引用ID

1 个答案: