php - 尝试获取非对象的属性 'num_rows'

Question

这个问题已经有答案了:

What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such (1 个回答)

已关闭 3 年前。

有人可以帮我纠正这个代码吗？

我不断收到此错误:-

Notice: Trying to get property 'num_rows' of non-object

<?php

  $test = $_GET['param'];
  $sql =" SELECT * FROM img WHERE id = $test ";
  $result = $conn->query($sql);

  if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
      echo " <div class='col-lg-9'> ";

      echo " <div class=card mt-4> ";
      echo " <img src=http://placehold.it/900x400 class=img-responsive alt=Responsive image> ";
      echo " <div class='card-body'> ";
      echo " <a class=pull-right> <button type=button class='btn btn-primary'>Prezzo " .$row["prz"]. " €</button> </a> ";

      echo " <h3 class=card-title>" .$row["nome"]. "</h3> "  ;
      echo "<br>";
      echo "<br>";
      echo " <p class=card-text>" .$row["ldscr"]. "</p> ";
      echo "</div>";
      echo "</div>";
      echo "<br>";
    }
  } 

?>

Answer 1

改变

if ($result->num_rows > 0) {

至

if (!empty($result) && $result->num_rows > 0) {

然后它就会起作用。

编辑:它不会爆炸的原因是，首先您要查看是否有任何结果。如果没有，那么它不会尝试显示它们(这是发生错误的地方)。

你真的应该更换

$sql =" SELECT * FROM img WHERE id = $test ";

与

$sql =" SELECT * FROM img WHERE id = ?";

然后使用 prepared statements 。但这是另一个问题。