Question

我是PHP的新手。我正在尝试将变量的值插入MariaDB表中，并试图使用mysqli_real_escape_string来转义'$ value'。我的想法来自here。它在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）。

因此，我从PHP Manual复制并粘贴了以下代码，但仍然无法使用。我得到的输出只是一个错误代码：错误：42000。我缺少什么？

我正在使用Virtualbox，操作系统：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，但无法插入行。我在数据库中创建了一个名为“城市”的表。我检查了代码中是否存在数据库连接，并且确实返回了“已连接”。这是更新的代码：

<?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，代码起作用了，确实插入了表格，但没有显示“ Row insert”：结果，我忘了从中取出分号 if 条件语句中的'execute（）'。

Answer 1

这里有一个准备好的声明的示例：

$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();

有关详细信息，请点击此处：prepare，bind

mysqli_real_escape_string似乎不起作用

1 个答案: