php - sql + php - 致命错误：不能使用 mysqli_result 类型的对象作为数组为什么

Question

这是我的代码：

<?php

$con = mysqli_connect("127.0.0.1:3306","root","root","photoshare");
$query = "SELECT ID,nickname,photoLikes FROM tbl_photo";
$result = mysqli_query($con,$query);
$query_dislike = "SELECT nickname, idGivenLike FROM tbl_check_like";
$resultDislike = mysqli_query($con,$query_dislike);
$photoIDdislike;
$photoID;
while($photoID = mysqli_fetch_assoc ($result) || $photoIDdislike = mysqli_fetch_assoc($resultDislike)   )
{
//checks if both of results
if($result["ID"] != $resultDislike["idGivenLike"])
{
echo "true";
}
}
?>

当我运行它时，它说下一行

if($result["ID"] != $resultDislike["idGivenLike"])

带有 h 致命错误： Fatal error: Cannot use object of type mysqli_result as array

为什么它不起作用，我该如何解决？

Answer 1

if($result["ID"] != $resultDislike["idGivenLike"])

应该：

if($photoID["ID"] != $photoIDdislike["idGivenLike"])

$resultDislike/$result是您的 MySQL 资源，$photoID/$photoIDdislike是获取的行（数组）。

Answer 2

它应该是

$photoID["ID"] != $photoIDdislike["idGivenLike"]
// In script 
if($photoID["ID"] != $photoIDdislike["idGivenLike"])
{
   echo "true";
}