Question

我为我的一个项目创建了这个。我们有一个网上商店，用户可以在其中输入其凭证并订购产品。当前的解决方案将所有数据放入一个.csv文件中，我的任务是创建一个mysql数据库作为新解决方案。

我为用户输入了一个简单的HTML插入符来输入他的名字，但是如果我尝试运行该脚本，则会得到第19行的语法错误。我是编程新手，因此不确定该错误在哪里。

<!DOCTYPE html>
<html>
<body>


<?php

$servername = "localhost";
$username = "localhost";
$password = "";
$dbname = "test"

    // create a variable
$Vorname=$_POST['Vorname'];
$Nachname=$_POST['Nachname'];

    //Execute the query

mysqli_query($connect "INSERT INTO tbl_bestellungen(Vorname,Nachname)
				VALUES('$Vorname','$Nachname')");
    <?php include 'database.php';>


if(mysqli_affected_rows($connect) > 0){
    echo "<p>Bestellung erfasst</p>";
   } else {
    echo "Bestellvorgang fehlgeschlagen<br />";
    echo mysqli_error ($connect);

<h2>Text Input</h2>

<form>
  Vorname:<br>
  <input type="text" name="Vorname">
  <br>
  Nachname:<br>
  <input type="text" name="Nachname">
  
  <input type="submit" name="button1"  value="Senden">
</form>

</body>
</html>

谢谢。

Answer 1

好吧，你应该这样：

$servername = "localhost";
$username = "localhost";
$password = "";
$dbname = "test"

$dbConn = mysqli_connect($servername, $username, $password, $dbname);
if(!$dbConn){
   echo "No Db connected"; 
}

//above connection code should be in a separate file and include in all files or header



$Vorname=$_POST['Vorname'];
$Nachname=$_POST['Nachname'];

$query = "INSERT INTO tbl_bestellungen (Vorname,Nachname)
            VALUES('$Vorname','$Nachname')";

or you can set query like

$query = "INSERT INTO tbl_bestellungen (Vorname,Nachname)
            VALUES('".$Vorname."','".$Nachname."')";

if($dbConn->query($query)){
   echo "Record inserted !";
}else{
   echo "Record cannot be inserted !";
}

使用php / html中的数据填充mysql数据库时出错

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