douqin231881 2016-02-13 09:05
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如何以安全的方式执行PHP Select查询?

How can I do this piece of code in a safe way to prevent SQL injections?

I tried to read the php manual of mysqli->prepared but I was not able to convert it since I'm new to PHP development.

NOTE: DAL::$conn is $msqli = new mysqli()

$objects = array();
        if($id != null)
        {
            $sql = "select * from Pages where id = ".$id;
        }
        else
        {
            $sql = "select * from Pages";
        }

        $result = mysqli_query(DAL::$conn, $sql);

        if (mysqli_num_rows($result) > 0) {
            // output data of each row
            $records = 0;

            while($row = mysqli_fetch_assoc($result)) {
                $records++;
                $data = new Pages();
                $data->id       =   $row["id"];
                $data->title    =   $row['title'];
                $data->content  =   $row["content"];
                $objects[$records] = $data;
            }
        } else {
            //No results
        }
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  • doujishan2247 2016-02-13 09:08
    关注

    Any query can be injected whether it's read or write, persistent or transient. Injections can be performed by ending one query and running a separate one (possible with mysqli), which renders the intended query irrelevant.

    Any input to a query from an external source whether it is from users or even internal should be considered an argument to the query, and a parameter in the context of the query. Any parameter in a query needs to be parameterized. This leads to a properly parameterized query that you can create a prepared statement from and execute with arguments. For example:

    SELECT col1 FROM t1 WHERE col2 = ?
    

    ? is a placeholder for a parameter. Using mysqli, you can create a prepared statement using prepare, bind a variable (argument) to a parameter using bind_param, and run the query with execute. You don't have to sanitize the argument at all (in fact it's detrimental to do so). mysqli does that for you. The full process would be:

    $stmt = mysqli->prepare("SELECT col1 FROM t1 WHERE col2 = ?");
    $stmt->bind_param("s", $col2_arg);
    $stmt->execute();
    

    There is also an important distinction between parameterized query and prepared statement. This statement, while prepared, is not parameterized and is thus vulnerable to injection:

    $stmt = mysqli->prepare("INSERT INTO t1 VALUES ($_POST[user_input])");
    

    To summarize:

    1. All Queries should be properly parameterized (unless they have no parameters)
    2. All arguments to a query should be treated as hostile as possible no matter their source
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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