尝试使用带有for循环的php更新我的数据库

This is what I did so far but my code is not updating the database, no error showing, nothing, but it's not working. Please help me, I really need it and I'm new to php.

这是我到目前为止所做的,但我的代码没有更新数据库,没有错误显示,没有,但它不起作用。请帮助我,我真的需要它,我是php的新手。

Basically what I am trying to do is to show the id name and firstname of each one on the database, and the teacher has to put the notes for each one.

基本上我想要做的是在数据库上显示每个人的id名称和名字,老师必须为每个人写下笔记。

    <?php
            include('config.php');
        ?>
    <html>
        <head>
            <title>crypto </title>
        </head>
        <body>
            <h1>crypto !</h1>
            <h2>Entrez les données demandées :</h2>
            <table>
                <div class="content">
                    <form action="crypto.php" method="post">
                    <div class="center">


            <td class="left"> <?php mysql_query ( ' select id  from       etudient_l2'); ?></td>
            <td class="left"> <?php  mysql_query ( ' select nom  from etudient_l2'); ?> </td>
            <td class="left"> <?php  mysql_query ( ' select prenom  from etudient_l2'); ?> </td>

            <label for="test1">test1</label><input type="text" name="test1" value="<?php if(isset($_POST['test1'])); ?>" /><br />
            <label for="test2">test2</label><input type="text" name="test2" value="<?php if(isset($_POST['test2'])); ?>" /><br />
            <label for="participation">participation</label><input type="text" name="participation" value="<?php if(isset($_POST['participation'])); ?>" /><br />
            <label for="examen">Examen</label><input type="text" name="examen" value="<?php if(isset($_POST['examen'])); ?>" /><br />
            <input type="submit" value="Envoyer" />
        </div>
        </form>
        </div>
       </table>


      <?php
    if(isset($_POST['test1'], $_POST['test2'], $_POST['participation'],    $_POST['examen']) )
{




    //On récupère les valeurs entrées par l'utilisateur :

     $test1=$_POST['test1'];
    $test2=$_POST['test2'];
    $participation=$_POST['participation'];
    $examen=$_POST['examen'];



    $result=mysql_query('select * from etudient_l2');
    $num_rows=mysql_num_rows($result);
    echo "$num_rows";
    for($i=1; $i<= $num_rows ; $i++)
    {
    $sql = "UPDATE etudient_l2 SET                  test1_res='$test1',test2_res='$test2',participation_res='$participation',examen_         res='$examen' where id='$i'"; 


        }


        // on ferme la connexion
       mysql_close();
       >


      <?php
      }
       ?>
      </body>
      </html>

1 个解决方案

#1

just correcting your code first change your HTML like below

只需更正您的代码,首先更改您的HTML,如下所示

<label for="test1">test1</label>
<input type="text" name="test1" value="<?php if(!empty($_POST['test1'])){echo $_POST['test1'];} ?>" />
<br />
<label for="test2">test2</label>
<input type="text" name="test2" value="<?php if(!empty($_POST['test2'])){ echo $_POST['test2'];} ?>" />
<br />
<label for="participation">participation</label>
<input type="text" name="participation" value="<?php if(!empty($_POST['participation'])){ echo $_POST['participation'];} ?>" />
<br />
<label for="examen">Examen</label>
<input type="text" name="examen" value="<?php if(!empty($_POST['examen'])){ echo $_POST['examen'];} ?>" />
<br />
<input type="submit" value="Envoyer" name="submit" />

than change some php code like below

比改变一些像下面的PHP代码

<?php if(isset($_POST['submit']))
{
$test1=$_POST['test1'];
$test2=$_POST['test2'];
$participation=$_POST['participation'];
$examen=$_POST['examen'];

$result=mysql_query("select * from etudient_l2");
$num_rows=mysql_num_rows($result);
echo $num_rows;
for($i=1; $i<=$num_rows;$i++)
{
$sql =mysql_query("UPDATE `etudient_l2` SET `test1_res`='".$test1."',`test2_res`='".$test2."',`participation_res`='".$participation."',`examen_res`='".$examen."' where `id`='".$i."'"); 
}


    // on ferme la connexion
   mysql_close();

  }
   ?>

Rest is in your hand i don't take the guarantee based on your code again i am just correcting you.

休息是在你手中我不再根据你的代码采取保证我只是纠正你。

To help you during testing:

在测试期间帮助您:

Add error reporting to the top of your file(s) which will help find errors.

将错误报告添加到文件的顶部,这将有助于查找错误。

<?php 
error_reporting(E_ALL);
ini_set('display_errors', 1);

// rest of your code

Sidenote: Error reporting should only be done in staging, and never production.

旁注:错误报告应该只在暂存中完成,而不是生产。

Also add or die(mysql_error()) to mysql_query().

还要添加或死(mysql_error())到mysql_query()。

#1