Php 将特定值插入数据库，并将其显示在表上？

我试图将特定值（刀子和毯子）插入数据库，但根本没有插入DB/表。另外，我想在下表中显示插入的值，但这也不起作用。它取决于在表上显示的插入。我确定，因为我通过phpmyAdmin插入了一个值，它显示在表中。请，我需要修复插入方面

插入代码/错误处理程序

<?php
if (isset($_POST['Collect'])) {
if(($_POST['Object'])!= "knife" && ($_POST['Object'])!= "blanket")
{
  echo "This isn't among the room objects.";
}else {
// this makes sure that all the uses that sign up have their own names
$sql = "SELECT id FROM objects WHERE object='".mysql_real_escape_string($_POST['Object'])."'";
$query = mysql_query($sql) or die(mysql_error());
$m_count = mysql_num_rows($query);

 if($m_count >= "1"){
    echo 'This object has already been taken.!';
    } else{
   $sql="INSERT INTO objects (object)
VALUES
('$_POST[Object]')";

echo "".$_POST['object']." ADDED";
}
} 
} 

?>

---

您的SQL语法错误。您应该更改以下选项：

INSERT INTO objects SET id = '', object = '".$_POST['Object']."'

到

如果希望插入也替换可能存在的任何值，请使用“替换”而不是“插入”

if(($_POST['Object'])!= knife || ($_POST['Object'])!= blanket)

这些价值刀和毯子是细绳。因此，您可能需要使用引号将其定义为字符串，否则php将无法理解；）
 如果对象的主键是id并且设置为自动递增

$sql = "INSERT INTO objects SET id = '', object = '".$_POST['Object']."'";

试一试

您可能也应该在其中添加一个转义符
您插入的查询也不正确

$sql=“插入到对象（id，object）值（“”，“$\u POST['object']。”）”

这个密码呢

if(($_POST['Object'])!= "knife" || ($_POST['Object'])!= "blanket")
{
  echo "This isn't among the room objects.";
}

将始终执行对象的值为刀或橡皮布，因为变量可以有一个值。你必须使用

if(($_POST['Object'])!= "knife" && ($_POST['Object'])!= "blanket")
    {
      echo "This isn't among the room objects.";
    }

非常感谢。我已经修好了，但还是不行。如果问题来自于此，请检查错误处理
$sql= "INSERT INTO objects(object) VALUES ('".$_POST['Object'].")';

if(($_POST['Object'])!= "knife" || ($_POST['Object'])!= "blanket")
{
  echo "This isn't among the room objects.";
}

if(($_POST['Object'])!= "knife" && ($_POST['Object'])!= "blanket")
    {
      echo "This isn't among the room objects.";
    }