我是php / mysql的新手,我学到了很多东西。我的复选框遇到了麻烦。 我看了,我遇到的一切对我来说都没有意义。
我正在做一个管理员可以添加用户的网站,并且有一些复选框可以说出用户感兴趣的内容。(可以选择多个。
实施例 对什么运动感兴趣。 (这些是用户可以选择的复选框)
等等
如何将选项存储在数据库中?
这是我到目前为止所拥有的。
HTML
<div class="col-md-8">
<label style="margin-right:10px; width:130px"><input name="interested[]" type="checkbox" value="a6"><span class="cats">Baseball</span></label>
<label style="margin-right:10px; width:130px"><input name="interested[]" type="checkbox" value="a6"><span class="cats">Football</span></label>
<label style="margin-right:10px; width:130px"><input name="interested[]" type="checkbox" value="a6"><span class="cats">Hockey</span></label>
</div>
PHP
<?php
//error_reporting(E_ALL);
//ini_set("display_errors", 1);
// Pick up the form data and assign it to variables
$id = @$_POST['id'];
$fname = $_POST['fname'];
$lname = $_POST['lname'];
$email = $_POST['email'];
$altemail = $_POST['altemail'];
$notes = $_POST['notes'];
$company = $_POST['company'];
$address = $_POST['address'];
$home = $_POST['home'];
$cell = $_POST['cell'];
$telephone = $_POST['telephone'];
$category = $_POST['category'];
$usertype = $_POST['usertype'];
$assigned = $_POST['assigned'];
$othercat = $_POST['othercat'];
$interested=$_POST['interested'];
//Get data in local variable
$id = @$_POST['id'];
$fname = $_POST['fname'];
$lname = $_POST['lname'];
$email = $_POST['email'];
$altemail = $_POST['altemail'];
$notes = $_POST['notes'];
$company = $_POST['company'];
$address = $_POST['address'];
$home = $_POST['home'];
$cell = $_POST['cell'];
$telephone = $_POST['telephone'];
$category = $_POST['category'];
$usertype = $_POST['usertype'];
$assigned = $_POST['assigned'];
$othercat = $_POST['othercat'];
$interested=$_POST['interested'];
// You have to loop through the array of checked box values ...
$interested="";
foreach($interested as $entry){
$interested .= $entry.",";
}
if ($fname=="" || $email=="")
{
echo "All fields must be entered, hit back button and re-enter information";
}else{
$query="INSERT INTO users(`id`, `fname`, `lname`, `email`, `notes`,`company`,`address`,`cell`,`home`,`telephone`,`category`,`usertype`,`assigned`,`altemail`,`othercat`,`interested`) VALUES('$id','$fname','$lname','$email','$notes','$company','$address','$telephone','$category','$usertype','$assigned','$altemail','$othercat','$cell','$home','$interested')";
就像我提到的,我是PHP / Mysql的新手。我现在只做了大约8天。我已经采取了很多方法,但这让我很难过。
任何帮助将不胜感激。不是试图让自己失望,而是与我所看到的所有其他人都无法理解,请为我愚蠢。
提前致谢。