Question

好的，我能够编写完整的代码，所以我现在可以将数据添加到我的数据库中。但我唯一的问题是，我希望我在html字段中输入的文本显示在数据库中。那么我如何将我输入html文件的数据转到php文件然后将数据发送到数据库呢？

代码：

<html>
    <head>
        <title>Tabelle</title>
        <form action="phpRICHTIG.php" action="post"/>
        Test1: <input type="text" name="Test1"/><br/>
        Test2: <input type="text" name="Test2"/><br/>
        Test3: <input type="text" name="Test3"/><br/>
        Test4:<input type="text" name="Test4"/><br/>
        <input type="submit" value="Absenden"/>
        </form>
      </body>
</html>

<?php

include "htmltest.php";

$link = mysqli_connect("localhost", "root", "", "vanille");


if($link === false){
die("Konnte nicht verbinden. " . mysqli_connect_error());
}


$sql = "INSERT INTO vanille (Test1, Test2, Test3, Test4) VALUES ('test', 'test', 'test', 'test')";
if(mysqli_query($link, $sql)){
echo "Hat geklappt.";
} else{
echo "Hat nicht geklappt und das hat nicht funktioniert: $sql. " .         mysqli_error($link);
}


mysqli_close($link);
?>

Answer 1

以下是您的代码更新：

<?php
//if the code on phpRICHTIG.php and htmltest.php is the same then you did need this line
//include "htmltest.php";

$link = mysqli_connect("localhost", "root", "", "vanille");


if($link === false){
die("Konnte nicht verbinden. " . mysqli_connect_error());
}


//First we check if we have some data posted .
if(isset($_POST)){

  //Then we set them to var
  //You will have to securize that by yourself later, insert directly $_POST is a huge security breach
  $test1 = $_POST['Test1'];
  $test2 = $_POST['Test2'];
  $test3 = $_POST['Test3'];
  $test4 = $_POST['Test4'];

  $sql = "INSERT INTO vanille (Test1, Test2, Test3, Test4) VALUES ('$test1', '$test2', '$test3', '$test4')";


  if(mysqli_query($link, $sql)){
    //It worked
    echo "Hat geklappt.";
  } else {
    //It failed
    echo "Hat nicht geklappt und das hat nicht funktioniert: $sql. " .         mysqli_error($link);
  }

}else{

  echo 'No form submitted';
}


mysqli_close($link);
?>

Answer 2

检查变量是否已设置，如果已设置，则定义它们。然后使用查询插入这些值。请参阅以下代码：

<?php

//include "htmltest.php"; you don't need to do this

$link = mysqli_connect("localhost", "root", "", "vanille");


if($link === false){
die("Konnte nicht verbinden. " . mysqli_connect_error());
}

if (isset($_POST['Test1']))//check if it's set
    $test1 = $_POST['Test1'];//define it

if (isset($_POST['Test2']))
    $test2 = $_POST['Test2'];

if (isset($_POST['Test3']))
    $test3 = $_POST['Test3'];

if (isset($_POST['Test4']))
    $test4 = $_POST['Test4'];
//make sure you escape values before inserting it to database 
$sql = "INSERT INTO vanille (Test1, Test2, Test3, Test4) VALUES ('".$test1."', '".$test2."', '".$test3."', '".$test4."')"; 

//then insert it into your database through the query

if(mysqli_query($link, $sql)){
echo "Hat geklappt.";
} else{
echo "Hat nicht geklappt und das hat nicht funktioniert: $sql. " .         mysqli_error($link);
}


mysqli_close($link);
?>

您最好在表单提交之前进行数据验证，以确保字段不为空。否则您将收到错误，或者您将空值插入数据库。

如何将html数据发送到我的php文件？

2 个答案: