Question

我的用户提交的数据不会进入我的数据库，我有一个表单，在其中插入他们的信息，然后它应该只插入数据库。我可以很好地连接到我的数据库并获取信息，它只是不插入它？

表格：

<div class="form">
    <form id="addsteps" action="add.php" method="post" autocomplete="off">
         <div class="field-wrap">
             <input type="date" required autocomplete="off" name="date" style="color: black; text-align: center;"/>
          </div>
          <div class="field-wrap">
              <input type="number" required autocomplete="off" name="steps" style="color: black;"/>
          </div>
          <button type="submit" class="button button-block" name="submit" />Submit</button>
    </form>
</div>

代码：

<?php   include '../db_connection.php';

session_start();

$unique_id = $_SESSION['unique_id'];

$_SESSION['date'] = $_POST['date'];
$_SESSION['steps'] = $_POST['steps'];

$_POST['date'];
$_POST['steps'];

$date = $link->escape_string($_POST['date']);
$steps = $link->escape_string($_POST['steps']);

$sql = "INSERT INTO step_count (unique_id, date, steps) "
  . "VALUES ('$unique_id','$date','$steps')";

 ?>

它可以很好地获取数据，当使用var_dump时我得到了

string(81) "INSERT INTO step_count (unique_id, date, steps) VALUES ('1','2017-12-16','12345')"

所以我只是想知道我哪里出错了，因为它不会进入数据库？

这是db_connection文件：

<?php
$link = mysqli_connect("localhost", "root", "", "repsandsteps");

 if (!$link) {
  echo "Error: Unable to connect to MySQL." . PHP_EOL;
  echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
  echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
  exit;
  } ?>

Answer 1

你应该像这样执行查询。

<?php   include '../db_connection.php';
session_start();

$unique_id = $_SESSION['unique_id'];
$_SESSION['date'] = $_POST['date'];
$_SESSION['steps'] = $_POST['steps'];

$date = mysqli_real_escape_string($link, $_POST['date']);
$steps = mysqli_real_escape_string($link, $_POST['steps']);

$sql = "INSERT INTO step_count (unique_id, date, steps) "
  . "VALUES ('$unique_id','$date','$steps')"; 

mysqli_query($link,$sql);

PHP用户提交的数据是不是要进入数据库？

1 个答案: