【发布时间】:2016-06-17 15:16:59
【问题描述】:
我的代码很乱,但我想在我的网站上做一个论坛。
<?php
$conn = mysqli_connect("localhost", "root", "");
if($conn->connect_error){
die("Connection Failed");
}
mysqli_select_db($conn, 'forum');
$get= "select * from forum";
$runq = mysqli_query($conn, $get);
while($fetch_value=mysqli_fetch_array($runq)){
$get_usr_name=$fetch_value['username'];
$get_body = $fetch_value['info'];
}
$result = $conn->query($get);
if(is_null($get_body)){
}
else{
echo "<div id='inputform1'>";
for($row = 1; $row < $result->num_rows; $row++){
$dbhost = "localhost";
$dbuser = "root";
$dbpass = "";
$dbname = "forum";
$connection = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname);
if(mysqli_connect_errno()){
die("database connection failed: ") .
mysqli_connect_errno() .
" (" . mysqli_connect_errno() . ")";
}
$result1 = mysqli_query($connection, "select info from forum where id='$row';");
$result = mysqli_query($connection, "select username from forum where id='$row';");
echo "<strong class=\"after_username\">by: {$result}</strong><br><br><p id=\"bodyforum\">{$result1}<br><br></p>";
}
echo "</div>";
}
?>
我似乎无法发布用户名和密码;它一直在说:
mysqli_result 类的对象无法转换为字符串
我尝试了许多不同的“策略”,但似乎都没有奏效。有人可以帮我吗?
【问题讨论】:
-
哪一行报错了?为什么要运行两次相同的查询?
-
chris85 是正确的:您可以使用
select info, username from forum where id='$id'"来避免两次查询。但是,$id应该取自查询结果;您假设查询返回的所有行的 ID 将从 1 开始并定期递增。不太可能是这种情况。