PDO在数据库中插入的是ID而不是名字。

问题描述 投票：0回答：1

所以我在这里有点困惑，我做了两个下拉菜单，它们都在工作。问题是，当我插入数据时，我得到的是下拉元素的id，而不是它们的名称。我很确定我需要用另一种方式来做，但我是个外行，所以任何解释都会很感激。这是我的代码。

<?php
session_start();
error_reporting(E_ALL);
ini_set('display_errors', 1);
include('includes/config.php');
if(strlen($_SESSION['alogin'])==0)
    {   
header('location:index.php');
}
else{
if(isset($_POST['submit']))
{

$id=null;
$name=$_POST['name'];
$email=$_POST['email'];
$user=$_POST['user'];
$password=md5($_POST['password']);
$mobileno=$_POST['mobileno'];
$bloc=$_POST['industry_name'];
$apartament=$_POST['sub_industry_name'];
$intretinere=$_POST['intretinere'];
$m2=$_POST['m2'];

$notitype='Create Account';
$reciver='Admin';
$sender=$email;

$sqlnoti="insert into notification (notiuser,notireciver,notitype) values (:notiuser,:notireciver,:notitype)";
$querynoti = $dbh->prepare($sqlnoti);
$querynoti-> bindParam(':notiuser', $sender, PDO::PARAM_STR);
$querynoti-> bindParam(':notireciver',$reciver, PDO::PARAM_STR);
$querynoti-> bindParam(':notitype', $notitype, PDO::PARAM_STR);
$querynoti->execute();    

$sql ="INSERT INTO users(id, name, email, user, password, mobile, bloc, apartament, intretinere, m2, status) VALUES(:id, :name, :email, :user, :password, :mobileno, :industry_name, :sub_industry_name, :intretinere, :m2, 1)";
$query= $dbh -> prepare($sql);
$query-> bindParam(':id', $id, PDO::PARAM_STR);
$query-> bindParam(':name', $name, PDO::PARAM_STR);
$query-> bindParam(':email', $email, PDO::PARAM_STR);
$query-> bindParam(':user', $user, PDO::PARAM_STR);
$query-> bindParam(':password', $password, PDO::PARAM_STR);
$query-> bindParam(':mobileno', $mobileno, PDO::PARAM_STR);
$query-> bindParam(':industry_name', $bloc, PDO::PARAM_STR);
$query-> bindParam(':sub_industry_name', $apartament, PDO::PARAM_STR);
$query-> bindParam(':intretinere', $intretinere, PDO::PARAM_STR);
$query-> bindParam(':m2', $m2, PDO::PARAM_STR);
$query->execute();
$lastInsertId = $dbh->lastInsertId();
if($lastInsertId)
{
echo "<script type='text/javascript'>alert('Registration Sucessfull!');</script>";
echo "<script type='text/javascript'> document.location = 'adduser.php'; </script>";
}
else 
{
$error="Something went wrong. Please try again";
}

}
?>

<script>
$(document).ready(function(){

  $('#category_item').selectpicker();

  $('#sub_category_item').selectpicker();

  load_data('category_data');

  function load_data(type, category_id = '')
  {
    $.ajax({
      url:"../admin/load_data.php",
      method:"POST",
      data:{type:type, category_id:category_id},
      dataType:"json",
      success:function(data)
      {
        var html = '';
        for(var count = 0; count < data.length; count++)
        {
          html += '<option value="'+data[count].id+'">'+data[count].name+'</option>';
        }
        if(type == 'category_data')
        {
          $('#category_item').html(html);
          $('#category_item').selectpicker('refresh');
        }
        else
        {
          $('#sub_category_item').html(html);
          $('#sub_category_item').selectpicker('refresh');
        }
      }
    });
  }

  $(document).on('change', '#category_item', function(){
    var category_id = $('#category_item').val();
    load_data('sub_category_data', category_id);
  });

});
</script>

<div class="form-group">
<label class="col-sm-2 control-label">Bloc<span style="color:red">*</span></label>
<div class="col-sm-4">
<select name="industry_name" id="category_item" class="form-control" required title="Selecteaza Bloc">
<option value="">Selecteaza Bloc</option>
</select>
</div>
<label class="col-sm-2 control-label">Apartament<span style="color:red">*</span></label>
<div class="col-sm-4">
<select name="sub_industry_name" id="sub_category_item" class="form-control" required data-live-search="true" title="Selecteaza Apartament">
<option value="">Selecteaza Apartament</option>
</select>
</div>
</div>

还有load_data.php

<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
include('includes/config.php');

if(isset($_POST["type"]))
{
 if($_POST["type"] == "category_data")
 {
  $query = "
  SELECT * FROM tbl_industry 
  ORDER BY industry_name ASC
  ";
  $statement = $dbh->prepare($query);
  $statement->execute();
  $data = $statement->fetchAll();
  foreach($data as $row)
  {
   $output[] = array(
    'id'  => $row["industry_id"],
    'name'  => $row["industry_name"]
   );
  }
  echo json_encode($output);
 }
 else
 {
  $query = "
  SELECT * FROM tbl_sub_industry 
  WHERE industry_id = '".$_POST["category_id"]."' 
  ORDER BY sub_industry_name ASC
  ";
  $statement = $dbh->prepare($query);
  $statement->execute();
  $data = $statement->fetchAll();
  foreach($data as $row)
  {
   $output[] = array(
    'id'  => $row["sub_industry_id"],
    'name'  => $row["sub_industry_name"]
   );
  }
  echo json_encode($output);
 }
}

?>

EDIT: 我已经更新了下面的代码，但现在我在下拉菜单中没有得到任何东西。

这里是js。

<script>
$(document).ready(function(){

  $('#category_item').selectpicker();

  $('#sub_category_item').selectpicker();

  load_data('category_data');

  function load_data(type, category_id = '')
  {
    $.ajax({
      url:"../admin/load_data.php",
      method:"POST",
      data:{type:type, category_id:category_id},
      dataType:"json",
      success:function(data)
      {
        var html = '';
        for(var count = 0; count < data.length; count++)
        {
    $.each( data, function( key, value ) {
    html += '<option value="'+value+'">'+value+'</option>';
    });

        }
        if(type == 'category_data')
        {
          $('#category_item').html(html);
          $('#category_item').selectpicker('refresh');
        }
        else
        {
          $('#sub_category_item').html(html);
          $('#sub_category_item').selectpicker('refresh');
        }
      }
    });
  }

  $(document).on('change', '#category_item', function(){
    var category_id = $('#category_item').val();
    load_data('sub_category_data', category_id);
  });

});
</script>

还有load_data.php

<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
include('includes/config.php');

if(isset($_POST["type"]))
{
 if($_POST["type"] == "category_data")
 {
  $query = "
  SELECT * FROM tbl_industry 
  ORDER BY industry_name ASC
  ";
  $statement = $dbh->prepare($query);
  $statement->execute();
  $data = $statement->fetchAll();
  foreach($data as $row){
  $output[$row["industry_id"]] = $row["industry_name"];
}
  echo json_encode($output);
 }
 else
 {
  $query = "
  SELECT * FROM tbl_sub_industry 
  WHERE industry_id = '".$_POST["category_id"]."' 
  ORDER BY sub_industry_name ASC
  ";
  $statement = $dbh->prepare($query);
  $statement->execute();
  $data = $statement->fetchAll();
  foreach($data as $row){
  $output[$row["sub_industry_id"]] = $row["sub_industry_name"];
}
  echo json_encode($output);
 }
}

?>
javascript php pdo
1个回答
0
投票

你说你在下拉菜单中插入了项目的ID，因为那是你在下面设置的。load_data().

而不是 

 html += '<option value="'+data[count].id+'">'+data[count].name+'</option>';

试试

 html += '<option value="'+data[count].name+'">'+data[count].name+'</option>';

除非你在AJAX调用前对数据进行操作。$_POST 将采取 value 来自 option 标签，而不是显示在开头和结尾标签之间的文本。

你在创建和读取数组时增加了不必要的步骤。在 load_data你可以根据你的结果创建一个带键的数组。

foreach($data as $row){
   $output[$row["industry_id"]] = $row["industry_name"];
}

然后使用jQuery的 $.each 函数来读取AJAX调用后的数据。

$.each( data, function( key, value ) {
   html += '<option value="'+value+'">'+value+'</option>';
});

参见 https:/api.jquery.comjquery.each。

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