在php中更新一行mysql [英] Update a row mysql in php

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本文介绍了在php中更新一行mysql的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我有一张如下所示的表格:


姓氏用户名密码角色



student student student@csd.auth.gr学生学生

student2 student2 student2@csd.auth.gr student2学生 p>

我希望能够在php中编辑一行。
这些值来自像这样的html文件:

用户名



student2@csd.auth.gr < - 将写入文本框



密码

strong>



student2 < - 将写入文本框



姓名



student2 < - 将写入文本框





student2 < - 将写入文本框



角色



学生< - This将被写入文本框中



我的php文件是:

 <?php 
$ hostname =localhost;
$ database =mydb;
$ username =myuser;
$ password =mypsw;
$ link = mysql_connect($ hostname,$ username,$ password)或
die(Prosoxi!Provlima stin sundesi me ton server:。mysql_error());
mysql_select_db($ database,$ link);
$ b $ mysql_query(UPDATE user
SET username ='.mysql_real_escape_string($ _ POST [nusername])。',
SET password ='.mysql_real_escape_string($ _ POST [ ($ _ POST [nname])。',
SET surname ='.mysql_real_escape_string($ _ POST [nsurname])。'',
SET name ='.mysql_real_escape_string ,
SET role ='.mysql_real_escape_string($ _ POST [nrole])。'
WHERE username ='。mysql_real_escape_string($ _ POST [us])。');

mysql_close($ link);
header(Location:users.php);
?>

1.更新不会发生,所以php文件有问题,找到。



2。如果我选择某个用户名,我该如何在html文件中实现已填充的框,并使用正确的值?

有人可以帮助我吗?先谢谢你。 :

解决方案

这里有很多事情要做。


  • 始终将整数作为主键,例如:id MEDIUMINT NOT NULL AUTO_INCREMENT。然后把它作为主键。

  • 您需要使用 mysql_real_escape_string()


  • 您需要连接您的查询,因此它应该看起来像这样:

    mysql_query 更新用户
    SET用户名='.mysql_real_escape_string($ _ POST [nusername])。'
    SET password ='.mysql_real_escape_string($ _ POST [npassword])。''
    SET name''.mysql_real_escape_string($ _ POST [nname])。''
    SET surname ='.mysql_real_escape_string($ _ POST [nsurname])。'
    SET role ='.mysql_real_escape_string($ _BOST [nrole])。''
    WHERE username ='。mysql_real_escape_string($ _ POST [us])。');


  • 这是更正的代码:

     <?php 
    $ hostname = localhost的;
    $ database =mydb;
    $ username =myuser;
    $ password =mypsw;
    $ link = mysql_connect($ hostname,$ username,$ password)或
    die(Prosoxi!Provlima stin sundesi me ton server:。mysql_error());
    mysql_select_db($ database,$ link);
    $ b mysql_query(UPDATE user
    SET username ='.mysql_real_escape_string($ _ POST ['nusername'])。',
    SET password ='.mysql_real_escape_string($ _POST ['npassword'])。'',
    SET name ='.mysql_real_escape_string($ _ POST ['nname'])。'',
    SET surname ='.mysql_real_escape_string($ _ POST [ 'nsurname'])。',
    SET role ='.mysql_real_escape_string($ _ POST ['nrole'])。''
    WHERE username ='。mysql_real_escape_string($ _ POST ['us' ]) ')。

    mysql_close($ link);
    header(Location:users.php);
    ?>

    请注意围绕_POST var,$ _POST ['nusername']的单引号您有$ _POST [ n用户名]。



    立即试用,看看它是否更新。

    I have a table that looks like this:

    name surname username password role

    student student student@csd.auth.gr student student

    student2 student2 student2@csd.auth.gr student2 student

    and I want to be able to edit a row in php. The values are taken from a html file like this:

    Username

    student2@csd.auth.gr <--This will be written in a text box

    Password

    student2 <--This will be written in a text box

    Name

    student2 <--This will be written in a text box

    Surname

    student2 <--This will be written in a text box

    Role

    student <--This will be written in a text box

    My php file is:

    <?php
    $hostname = "localhost"; 
    $database = "mydb"; 
    $username = "myuser"; 
    $password = "mypsw";
    $link = mysql_connect( $hostname , $username , $password ) or 
            die("Prosoxi!Provlima stin sundesi me ton server : " . mysql_error());
    mysql_select_db($database,$link);
    
    mysql_query("UPDATE user 
             SET username = '".mysql_real_escape_string($_POST[nusername])."', 
             SET password = '".mysql_real_escape_string($_POST[npassword])."', 
             SET name = '".mysql_real_escape_string($_POST[nname])."', 
             SET surname = '".mysql_real_escape_string($_POST[nsurname])."', 
             SET role = '".mysql_real_escape_string($_POST[nrole])."' 
             WHERE username='".mysql_real_escape_string($_POST[us])."'");
    
    mysql_close($link);
    header("Location: users.php");
    ?>
    

    1.The update does not happen, so there's something wrong in the php file, that I can't find.

    2. How can I achieve already filled boxes in the html file, with the right values, if I choose a certain username?

    Can someone help me? Thank you in advance. :)

    解决方案

    There is a lot going on here.

    1. Always us a integer as a primary key, example: id MEDIUMINT NOT NULL AUTO_INCREMENT. Then make it the primary key.

    2. you need to sanitize your input to the database using mysql_real_escape_string()

    3. you need to concatenate your query, so it should look like this:

      mysql_query("UPDATE user SET username = '".mysql_real_escape_string($_POST[nusername])."' SET password = '".mysql_real_escape_string($_POST[npassword])."' SET name = '".mysql_real_escape_string($_POST[nname])."' SET surname = '".mysql_real_escape_string($_POST[nsurname])."' SET role = '".mysql_real_escape_string($_POST[nrole])."' WHERE username='".mysql_real_escape_string($_POST[us])."'");

    Here is corrected code:

        <?php
    $hostname = "localhost"; 
    $database = "mydb"; 
    $username = "myuser"; 
    $password = "mypsw";
    $link = mysql_connect( $hostname , $username , $password ) or 
            die("Prosoxi!Provlima stin sundesi me ton server : " . mysql_error());
    mysql_select_db($database,$link);
    
    mysql_query("UPDATE user 
             SET username = '".mysql_real_escape_string($_POST['nusername'])."', 
             SET password = '".mysql_real_escape_string($_POST['npassword'])."', 
             SET name = '".mysql_real_escape_string($_POST['nname'])."', 
             SET surname = '".mysql_real_escape_string($_POST['nsurname'])."', 
             SET role = '".mysql_real_escape_string($_POST['nrole'])."' 
             WHERE username='".mysql_real_escape_string($_POST['us'])."'");
    
    mysql_close($link);
    header("Location: users.php");
    ?>
    

    notice the single quote surrounding the _POST var, $_POST['nusername'] you had $_POST[nusername].

    Try it now, and see if it updates.

    这篇关于在php中更新一行mysql的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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