试图获取非对象的属性"num_rows" [英] Trying to get property 'num_rows' of non-object

问题描述

有人可以帮助我更正此代码吗?

Could someone help me to correct this code?

我不断收到此错误:-

注意:尝试获取非对象的属性'num_rows'

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>";
    }
  } 

?>

推荐答案

更改

if ($result->num_rows > 0) {

到

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

然后它将起作用.

编辑:之所以不会崩溃，是因为首先您要查看是否有任何结果.如果没有，它将不会尝试显示它们(这是发生错误的地方).

The reason it won't blow up is that first you are seeing if there are any results, first. If there are not, then it will not attempt to show them (this is where the error was occuring).

您真的应该替换

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

与

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

，然后使用 准备好的声明 .但这是另一个问题.

and then use prepared statements. But that is another question.

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