Question

我试图在php文件中设置更新功能从表单中获取数据但是它没有在phpmysql中更新它，这里是查询，可能是我遗漏了一些东西。

 $query="UPDATE controlpanel1 SET ftitle_p1_1 = '$_POST[ftitle_p1_1]'";

Answer 1

好吧，正如每个人都想说的那样，你不应该这样做，因为它很危险。让我试着用MySQLi（你想使用MySQLi或PDO）和准备好的声明给你一个或多或少更可接受的程序的基本样本：

 $query= $MysqliConnection->prepare("UPDATE controlpanel1 SET ftitle_p1_1 = ? WHERE id = ?"); //reason why it's not updating, probably. You have to tell the system where to update. Which row.
 $query->bind_param("si", $title, $id); //string, integral - title and id(?). Just guessing.
 $title = $_POST["title_p1_1"];
 $id = $_GET["I_have_no_idea"]; // or $_POST["I_have_no_idea"];
 $query->execute();
 $query->close();
 $MysqliConnection->close();

或者，请参考，this page。请清理数据...即使使用预处理语句，我也会检查字符串是否有效。我太担心了，你不是吗？

Answer 2

确保在php.ini配置文件中启用了mysqli扩展和pdo扩展。
那么你需要选择是否要采用程序或OO方式以及mysqli或PDO方式这是程序性的mysqli：

$DBConnect = new mysqli("localhost", "root", "myCoolPW", "myDBname");
$FTitle = '_';
$FTitle = $_POST['ftitle_p1_1'];
echo " Test posted FTitle: " . $FTitle . " ";
$query = "
  UPDATE controlpanel1
  SET ftitle_p1_1 = '$FTitle'
  WHERE id != '2'
",
echo " Test query: " . $query . " ";  
mysqli_query($DBConnect, $query);  
if (mysqli_affected_rows($DBConnect) == -1)
    echo "NOT successfull  :-(";
else
    echo "SUCCESS!!! :-)";

将数据更新到mysql无法正常工作

2 个答案: