如何将表单数据传递给另一个页面和数据库？

debugcn 发表于 Dev

曼扎尔·古尔扎尔

实际上，我正在尝试按用户建议的名称创建一个表，并将数据插入该表中，也是根据用户的建议。

我有两个 php 文件：CreateTable.php和EnterData.php

这是我的CreateTable.php代码：

<?php 

    $conn = new mysqli("localhost","root","","mywebsite");

    if (isset($_POST['tbButton'])) {
        $qry = "Create Table ".$_POST['tableName']."(firstname varchar(25),lastname varchar(25));";
        $res = mysqli_query($conn,$qry);
        if ($res) {
            echo "Table Created!";
        }
        else{
            die("query failed!");
        }
    }

 ?>
 <!DOCTYPE html>
 <html>
 <head>
    <title>Create Table</title>
 </head>
 <body>
    <form action="EnterData.php" method="post">
        <p><input type="text" name="tableName" placeholder="Enter Table Name..."></p>
        <p><input type="submit" name="tbButton"></p>
    </form>
 </body>
 </html>

这是我的EnterData.php代码：

<?php 
    $tbname = $_POST['tableName']; 
    $conn = new mysqli("localhost","root","","mywebsite");

    if (isset($_POST['dataButton'])) {

        $qry = "Insert into ".$tbname."(firstname,lastname) values('".$_POST['firstname']."','".$_POST['lastname']."');";
        $res = mysqli_query($conn,$qry);
        if ($res) {
            echo "Data Inserted!";
        }
        else{
            die("query failed!");
        }
    }

 ?>
 <!DOCTYPE html>
 <html>
 <head>
    <title>Create Table</title>
 </head>
 <body>
    <form action="" method="post">
        <p><input type="text" name="firstname" placeholder="Enter First Name..."></p>
        <p><input type="text" name="lastname" placeholder="Enter Last Name..."></p>
        <p><input type="submit" name="dataButton"></p>
    </form>
 </body>
 </html>

问题是，当我写action="EnterData.php"表时，不会在数据库中创建，而是将表单值传递给'EnterData'文件。当我action="CreateTable.php"在数据库中创建表时，但值不会传递到 'EnterData'文件。我也想将值传递给EnterData文件和数据库。

这是我第一次尝试 stackoverflow，希望我能很好地解释我的问题

千人

您可以通过 get 方法传递您的表名

创建表.php

<?php 

$conn = new mysqli("localhost","root","","mywebsite");
$tableName = $_POST['tableName'];

if (isset($_POST['tbButton'])) {
    $qry = "Create Table ".$tableName ."(firstname varchar(25),lastname varchar(25));";
    $res = mysqli_query($conn,$qry);
    if ($res) {
        header("Location: EnterData.php?tableName=".$tableName);
    }
    else{
        die("query failed!");
    }
}

?>
<!DOCTYPE html>
<html>
<head>
  <title>Create Table</title>
</head>
<body>
  <form action="CreateTable.php" method="post">
    <p><input type="text" name="tableName" placeholder="Enter Table Name..."></p>
    <p><input type="submit" name="tbButton"></p>
  </form>
</body>
</html>

输入数据.php

<?php 
   $tbname = $_GET['tableName']; 
   $conn = new mysqli("localhost","root","","mywebsite");

   if (isset($_POST['dataButton'])) {

     $qry = "Insert into ".$tbname."(firstname,lastname) values('".$_POST['firstname']."','".$_POST['lastname']."');";
     $res = mysqli_query($conn,$qry);
     if ($res) {
        echo "Data Inserted!";
     }
     else{
        die("query failed!");
     }
  }

?>
 <!DOCTYPE html>
 <html>
 <head>
    <title>Create Table</title>
 </head>
 <body>
   <form action="EnterData.php?tableName=<?php echo $tbname;?>" method="post">
     <p><input type="text" name="firstname" placeholder="Enter First Name..."></p>
     <p><input type="text" name="lastname" placeholder="Enter Last Name..."></p>
     <p><input type="submit" name="dataButton"></p>
    </form>
  </body>
 </html>

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