第 19 行是 $Image_path = $row->Image_path;
嘿,大家好,我不知道为什么会收到此错误...因为这是我第一次调用此变量,而且 $row->Image_path 是表中的实际字段。有任何想法吗?打败了我
这是代码
<?php
error_reporting(E_ALL);
ini_set('display_errors', '1');
$search_output = "";
if(isset($_POST['searchquery']) && $_POST['searchquery'] != ""){
$searchquery = preg_replace('#[^a-z 0-9?!]#i', '', $_POST['searchquery']);
if($_POST['filter1'] == "Companies"){
$sqlCommand = "SELECT Company_ID, Company_Name, Image_path AS company FROM Company WHERE Company_Name LIKE '%$searchquery%'";
}
include_once("database_connect.php");
$query = mysqli_query($connection, $sqlCommand) or die(mysql_error());
$count = mysqli_num_rows($query);
if($count >= 1){
$search_output .= "<hr />$count results for <strong>$searchquery</strong><hr />$sqlCommand<hr />";
while($row = mysqli_fetch_array($query)){
$id = $row->Company_ID;
$Company_name = $row->Company_Name;
$Image_path = $row->Image_path;
$search_output .= "Item ID: $ $Image_path.$Company_name<br />";
} // close while
} else {
$search_output = "<hr />0 results for <strong>$searchquery</strong><hr />$sqlCommand";
}
}
?>
最佳答案
您需要将mysqli_fetch_array
更改为mysqli_fetch_object
。
关于php - 注意:尝试获取第 19 行/home/.../Welcome.php 中非对象的属性,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20307222/