php - MySQL如何从MySQL选择插入的变量

Question

我有这个问题：
当我想将变量插入MySQL数据库时，它正在工作：D
但是当我想选择它时，它将向我输出此错误：
可捕获的致命错误：无法将类mysqli_result的对象转换为字符串
我的代码在此文本下：D
谢谢你的每一个答案：D

<meta charset="utf-8">
<style type="text/css">body {text-align: center;}</style>
<form action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>" method="POST">
<table border="0" align="center">
    <tr>
        <th>
            <textarea  name="comment" placeholder="Okomentuj túto stránku...(Najviac 500 znakov)" rows="10" cols="100" maxlength="5000"></textarea>
        </th>
    </tr>
    <tr>
        <th>
            <input type="submit" name="submit" value="Odoslať">
        </th>
    </tr>
</table>
</form>
<?php
$submit = $_POST["submit"];
$name = $_SESSION['user_name'];
$comment = $_POST["comment"];
$location = 1;
$conn = mysqli_connect("localhost", "Kubajsk0", "", "login");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if ($location = "1") {
if($submit)
    {if($name&&$comment){
        $insert=mysqli_query($conn,"INSERT INTO serverlist_comments (name, carriage,   location) VALUES ('$name','$comment','$location') ");
    }else {
    echo "<script>alert('Please fill out all fields.');</script>";/*echo "please fill  out all fields";*/
}
}
} else {
echo "Nepodarilo sa tvoj koment uložiť do databázi :(<br/>
      Skús to neskôr :D";
}
$show_name = "SELECT name FROM `serverlist_comments` WHERE 1";
$result = $conn->query($show_name);
echo $result;
?>

Answer 1

试试这个

$show_name = "SELECT name FROM `serverlist_comments` WHERE 1";
$result = mysqli_query($conn,$show_name);
print_r(mysqli_fetch_assoc($result));
echo $result['name'];

您也可以这样使用。

$show_name = "SELECT name FROM `serverlist_comments` WHERE 1";
$result = mysqli_query($conn,$show_name);
print_r(mysqli_fetch_array($result));
echo $result[0];

如果您有多个未加工的原料，请像这样使用。

$show_name = "SELECT name FROM `serverlist_comments` WHERE 1";
$result = mysqli_query($conn,$show_name);
while($row = mysqli_fetch_assoc($result)){
    print_r($row);
    echo "<hr />";
}