Question

这是我将数据插入数据库表的代码，但我想抓取数据并将其显示在下面的div中：

<?php
// only proccss the form if $_POST isn't empty
if ( ! empty( $_POST ) ) {

// connect to MySQL
$mysqli = new mysqli( 'localhost', 'root', '', 'mydb' );

// Check our connection
if ( $mysqli->connect_error ) {
    die( 'Connect Error: ' . $mysqli->connect_errno . ': ' . $mysqli->connect_error );
}

// insert our data
$sql = "INSERT INTO table1 ( name, email ) VALUES ('{$mysqli->real_escape_string($_POST['name'])}', '{$mysqli->real_escape_string($_POST['email'])}' )";
$insert = $mysqli->query($sql);

// print response from MySQL
if ( $insert ) {
    echo "Success! Row ID: {$mysqli->insert_id}";
} else {
    die("Error: {$mysqli->errno} : {$mysqli->error}");
}

// close our connection
$mysqli->close();
}

?> 

<form action="<?php echo $_SERVER['PHP_SELF'];?>" method="post">
    <input name="name" type="text">
    <input name="email" type="email">
    <input type="submit" value="Add">
</form>

Display each name and email here like:

<div class="name">Get Name Here from Database</div><div class="email">Get Email Here from Database</div>

上面的代码适用于发布但我希望这样做，每当我转到同一页面时，它必须在表单下方显示数据rom数据库。

Answer 1

尝试使用绑定变量和预处理语句：

<?php
$mysqli = new mysqli("localhost", "user", "pass", "db");

/* check connection */
if (mysqli_connect_errno()) {
  printf("Connect failed: %s\n", mysqli_connect_error());
  exit();
}

if (isset($_POST['scriptname'])) {
  $scriptname = $_POST['scriptname'];

  $stmt = $mysqli->prepare("INSERT INTO appslist(listall) VALUES (?)");
  $stmt->bind_param('s', $scriptname);

  /* execute prepared statement */
  $stmt->execute();

  printf("%d Row inserted.\n", $stmt->affected_rows);

  /* close statement and connection */
  $stmt->close();
 }
?>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
   Script Name: <input type="text" name="scriptname">
   <input type="submit">
</form>

输出继电器：

1 Row inserted.

http://php.net/manual/en/mysqli.quickstart.prepared-statements.php

试图只将一个列发布到数据库中

1 个答案: