php - 无法将FORM提交到Mysql

我有一个页面，其中有一个输入文本框，该数据需要插入或更新现有行。

我稍后会回应，那就是在控制之下。

<form method="post" name="tapahtuma">
Submit:
<input type="text" name="auts[]">
<input type="button" value="Ilmoita">
</form >

<?php

if(isset( $_POST["tapahtuma"]) ) {
$link = mysql_connect("localhost","root","","db");
mysqli_query($link, "INSERT INTO table (column) VALUES ('".$auts."')");
$auts = mysql_real_escape_string($_POST['auts']);
mysqli_close($link);
}

?>

最佳答案

根据提供的代码。

变化


将<input type="button" value="Ilmoita">更改为<input type="submit" name="Ilmoita" value="Ilmoita">
将if(isset( $_POST["tapahtuma"]) ) {更改为if(isset( $_POST["Ilmoita"]) ) {
将$link = mysql_connect("localhost","root","","db");更改为$link = mysqli_connect("localhost","root","","db");
将$auts = mysql_real_escape_string($_POST['auts']);放在mysqli_query之前


更新的代码

<form method="post" name="tapahtuma">
  <input type="text" name="auts">
  Submit: <input type="submit" name="Ilmoita" value="Ilmoita">
</form >
<?php

if(isset( $_POST["Ilmoita"]) ) {
  $link = mysqli_connect("localhost","root","","db");
  $auts = mysql_real_escape_string($_POST['auts']);
  mysqli_query($link, "INSERT INTO table (column) VALUES ('".$auts."')");
  mysqli_close($link);
}?>

我建议您使用prepared statements来避免SQL Injections

<form method="post" name="tapahtuma">
  <input type="text" name="auts">
  Submit: <input type="submit" name="Ilmoita" value="Ilmoita">
</form>

<?php
if(isset($_POST["Ilmoita"])) {
  $link = mysqli_connect("localhost","root","","db") or die("connection failed: " . mysqli_connect_error());
  $result = mysqli_prepare($link, "INSERT INTO `table` (`column`) VALUES (?)");
  mysqli_stmt_bind_param($result, "s", $_POST['auts']);
  mysqli_stmt_execute($result);
  mysqli_close($link);
}?>

快速链接


mysqli_prepare
How can I prevent SQL injection in PHP?