Question

我有一个表格，显示我网站页面上的用户ID和用户名。我试图使用表的最后一列中的删除按钮从SQL表中删除行。我确实遵循了here的一些指示，但我似乎遇到了问题。我收到这个错误：

mysqli_real_escape_string（）需要2个参数，1个给定

我似乎无法解决这个问题。谁能解释我做错了什么？

$sql = "SELECT id, username FROM users";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
  echo "<table classs><tr><th>ID</th><th>Name</th></tr>";
    while($row = $result->fetch_assoc()) {
      echo "<tr>";
      echo "<td>" . $row["id"]. "</td>";
      echo "<td>" . $row["username"]. "</td>";
      echo "<td align='right'><a href='delete.php?did=".$row['id']."'>Delete</a></td>";
      echo "</tr>";
    }
  echo "</table>";
} else {
    echo "0 results";
}

这是我的delete.php

require_once'config.php';

    $conn = new mysqli($servername, $username, $password, $dbname);

    if ($conn->connect_error) {
      die("Connection failed: " . $conn->connect_error);
    }

if(isset($_GET['did'])) {
    $delete_id = mysqli_real_escape_string($_GET['did']);
    $sql = ("DELETE FROM users WHERE id = '".$delete_id."'");
    if($sql) {
        echo "<br/><br/><span>deleted successfully...!!</span>";
    } else {
        echo "ERROR";
    }
}

Answer 1

尝试通过此

更改您的delete.php文件

require_once'config.php';

    $conn = new mysqli($servername, $username, $password, $dbname);

    if ($conn->connect_error) {
      die("Connection failed: " . $conn->connect_error);
}

if(isset($_GET['did'])) {

    //Escape the id using the function intval (Only if is a integer)
    $delete_id = intval($_GET['did']);

    //Use the method query of $conn
    $sql = $conn->query("DELETE FROM users WHERE id = '".$delete_id."'");
    if($sql) {
        echo "<br/><br/><span>deleted successfully...!!</span>";
    } else {
        echo "ERROR";
    }
}

使用PHP从SQL表中删除行

1 个答案: