PHP代碼從SQL數據庫的查詢結果

Question

我試圖打開從數據庫中包含的URL，但得到的錯誤採取了網址像

Catchable fatal error: Object of class mysqli_result could not be converted to string in D:\xampp\htdocs\randomizer\index.php on line 23

我的PHP程序

<?php 

$dbhost = 'localhost:3306'; 
$dbuser = 'root'; 
$dbpass = '#####'; 

$dbname = 'random'; 
$conn= mysqli_connect($dbhost,$dbuser,$dbpass,$dbname); 
if (mysqli_connect_errno($conn)) 
{ 
echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
} 
mysqli_select_db($conn,'random') or die ("no database"); 
$query=mysqli_query($conn,"SELECT url FROM ufc 
ORDER BY RAND() 
LIMIT 1"); 

echo '<a href="javascript: void(0)" onclick="window.open(' . '\'' . $query .    '\'' . ');">YYAAY</a>'; 
} 
?> 

Answer 1

您是打開網址執行選擇查詢，但不收集任何結果。 你需要獲取數據，嘗試這樣的事情：

if ($result = mysqli_fetch_array($query)) 
{ 
echo '<a href="javascript: void(0)" onclick="window.open(\'' . $result['url'] . '\');">YYAAY</a>'; 
} 

更多信息：http://php.net/manual/en/mysqli-result.fetch-array.php

Answer 2

除了艾倫的回答，您可以重定向到URL自動像這樣... header("location:".$result['url']);