我是php的新手。我试图将变量的值插入mariadb表，并试图使用mysqli\u real\u escape\u string来转义“$value”。我是从这里想到的。它向表中插入了一个空字符串（我确实向数据库添加了一个连接链接）。
所以，我从php手册中复制并粘贴了下面的代码，它仍然不起作用。我得到的输出仅仅是一个错误代码：error:42000。我错过了什么？
我使用的是virtualbox，os:centos7

<?php
$link = mysqli_connect("localhost", "my_user", "my_password", "world");

/* check connection */
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

mysqli_query($link, "CREATE TEMPORARY TABLE myCity LIKE City");

$city = "'s Hertogenbosch";

/* this query will fail, cause we didn't escape $city */
if (!mysqli_query($link, "INSERT into myCity (Name) VALUES ('$city')")) {
    printf("Error: %s\n", mysqli_sqlstate($link));
}

$city = mysqli_real_escape_string($link, $city);

/* this query with escaped $city will work */
if (mysqli_query($link, "INSERT into myCity (Name) VALUES ('$city')")) {
    printf("%d Row inserted.\n", mysqli_affected_rows($link));
}

mysqli_close($link);
?>

更新：感谢您的及时回复！我尝试了@pilan的代码，但没能插入一行。我在数据库中创建了一个名为“city”的表。我检查了代码中是否有数据库连接，它确实返回“connected”。以下是更新的代码：

<?php
$link = mysqli_connect("localhost", "my_user", "my_password", "world");

/* check connection */
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

else {

    echo "Connected";

$city = "'s Hertogenbosch";    

// Connect to db, returns mysqli-connection
$mysqli = new mysqli("localhost", "my_user", "my_password", "world");

// Prepare, "?" for placeholders, returns mysqli-statement
$stmt = $mysqli->prepare("INSERT INTO City (Name) VALUES (?)");

// Bin param to statement, with type "s" for string
$stmt->bind_param("s", $city);

//Execute
/* this query with escaped $city will work */
if ($stmt->execute()) {
    printf("%d Row inserted.\n", mysqli_affected_rows($link));
}

}
mysqli_close($link);
?>

更新：谢谢大家，代码成功了，它确实插入到了表中，但是“row inserted”没有出现：结果是，我忘了从if条件语句中的“execute（）”中去掉分号。