显示从php程序到数据库的插入信息,然后显示插入的infor

时间:2013-02-08 19:29:29

标签: php html mysql

我正在尝试在提交数据并插入数据库后显示用户从php程序输入的信息。当我运行它时,我看不到任何结果。我只是得到一个空白页面。我在代码中找不到错误,希望你们能帮助我找到它,所以我可以修复它并继续前进。这只是我的代码的一部分,我想把一些信息插入到我的数据库中,然后显示我插入的内容......

if ( $valid ) { 
    $lines = file('/home/user/Documents/file.txt');
    $uid = trim($lines[0]);
    $pw = trim($lines[1]);
    $dbserver = trim($lines[2]);
    $dbname = trim($lines[3]);

    //Connecting to mysql
    $link = mysqli_connect($dbserver, $uid, $pw, $dbname);
    or die('Could not connect: ' . mysql_error());

    //Our SQL Query
    $firstname = $_POST['firstname'];
    $lastname = $_POST['lastname'];
    $organization = $_POST['organization'];
    $emailaddress = $_POST['emailaddress'];
    $phonenumber = $_POST['phonenumber'];
    $sql_query = "INSERT INTO table VALUES ('$firstname','$lastname','$organization','$emailaddress','$phonenumber')";

        //Run our sql query
    $result = mysqli_query($link, $sql_query) or die('query failed'. mysql_error());

    // Get all records now in DB
    $sql_query = "SELECT * FROM table";
    //Run our sql query
    $result = mysqli_query($link, $sql_query) or die('query failed'. mysql_error());

    //iterate through result printing each record
    echo "<br>Names in DB: <br>";
    while($row = mysqli_fetch_assoc($result)) {
    echo $row['firstname'];
    echo $row['lastname'];
    echo $row['organization']; 
    echo $row['emailaddress'];
    echo $row['phonenumber'];
    echo "<br>";
    }

    // Free resultset (optional)
    mysqli_free_result($result);

    //Close the MySQL Link
    mysqli_close($link);
    }
}

2 个答案:

答案 0 :(得分:0)

更改$ sql_query = "INSERT INTO table VALUES $firstname','$lastname','$organization','$emailaddress','$phonenumber')";

$sql_query = "INSERT INTO table VALUES ($firstname','$lastname','$organization','$emailaddress','$phonenumber')";

您编写的代码也非常不安全,很容易被SQL注入攻击。

花点时间学习MySQLiPDO一系列Mysql_ *函数。以下是使用mysqli参数化查询的基本查询:

$ query = $ mysqli-&gt; prepare(“SELECT COLUMN_LIST FROM TABLE WHERE COLUMN =?”); $查询 - &GT; bind_param( 'S',$ UNSAFE_VARIABLE); $查询 - &GT;执行();

答案 1 :(得分:0)

如果您坚持使用过时的mysql函数,那么您正在寻找的是mysqli_insertid我相信它有类似的功能。