mysqli_real_escape_string似乎不起作用

Question

I am a novice in PHP. I am trying to insert a variable's value into a MariaDB table and was trying to use mysqli_real_escape_string to escape '$value'. I got the idea from here . It inserted an empty string to the table(I did add a connection link to the database).

So, I copied and pasted the following code from PHP Manual , it still didn't work. The output I got was an error code alone: Error: 42000. What am I missing?

I am using a 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);
?>

Update: Thank you for your prompt response! I tried @Pilan's code but I was not able to insert a row. I created a table in the database called 'City'. I checked whether there was a database connection in the code and it did return "Connected". Here is the updated code:

<?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);
?>

Update : Thanks guys, The code worked, It did insert into the table but 'Row inserted' didn't show up: turns out, I forgot to take out the semicolon from 'execute()' inside if conditional statement.

Answer 1

Here you got an example of a prepared statement :

$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 myCity (Name) VALUES (?)");

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

// Well execute :D
$stmt->execute();

For details have a look here: prepare , bind