INSERT维护清单表表格PHP HTML MYSQL

时间:2019-03-22 09:44:40

标签: php html mysql sql

我需要创建一个表来注册对机器安全设备进行的维护。

我有一个数据库表,其中包含与自己的安全设备关联的所有机器的列表。

每个安全设备可以为“正常”,“不正常”,“正常”或“不适用”

我用HTML / PHP创建了一个网页,维护人员在该网页上选择要进行维护的机器,这将显示他必须控制和注册设备状态的安全设备列表(好吧,不好...)。

<html>
    <? include 'home.php' ?>    
        <body>
            <div>
                <form action="insert_controllo.php" method="POST">
                <table>
                    <th>Macchina</th>
                    <th>Dispositivo</th>
                    <th>Descrizione</th>
                    <th>OK</th> 
                    <th>NOK</th>    
                    <th>PARZ</th>   
                    <th>NA</th> 
                    <th>Note</th>   
<?                      while($row = mysql_fetch_array($risultato)){
                            echo "<tr><td>  <input class='hide' type= 'text' name='macc' value ='".$row['macc']."' readonly> 
                                  </td><td> <input class='hide' type= 'text' name='iddispssic' value ='".$row['iddispsic']."' readonly> 
                                  </td><td> <input class='hide' type= 'text' name='descgen' value ='".$row['descgen']."' readonly>
                                  </td><td> <input type= 'checkbox' name='chk1[]' value ='OK'>  
                                  </td><td> <input type= 'checkbox' name='chk1[]' value ='NOK'>
                                  </td><td> <input type= 'checkbox' name='chk1[]' value ='PARZ'>    
                                  </td><td> <input type= 'checkbox' name='chk1[]' value ='NA'> 
                                  </td><td> <input type='text' name='note' placeholder = 'Note'> 
                                 </td></tr>";   
                        }?>
                </table>
                <button class="bottone" name='submit' type='submit'>Concludi</button>
                </form>
            </div>
        </body>
</html>

我的问题是在POST VAR DUMP中,我仅收到表的第一行

        <!-- POST: Array
(
    [macc] => 09
    [iddispssic] => 2
    [descgen] => Ripari Fissi
    [chk1] => Array
        (
            [0] => NOK
        )

    [note] => 
    [submit] => 
)
 -->

我不知道如何在选中的复选框中插入数据库

有人可以帮我吗?

我想在数据库中插入类似的内容

机器安全设备状态(确定,不确定,...)-说明(有输入类型的文本在每行上添加注释)-日期时间

1 个答案:

答案 0 :(得分:1)

您必须更改输入的名称才能创建元素数组(可以为另一个元素更改“ elements”名称)。

$count = 0;
while($row = mysql_fetch_array($risultato)){
  echo '<tr><td>  <input class="hide" type= "text" name="elements['.$count.'][macc]" value ="'.$row["macc"].'" readonly> 
    </td><td> <input class="hide" type= "text" name="elements['.$count.'][iddispssic]" value ="'.$row["iddispsic"].'" readonly> 
    </td><td> <input class="hide" type= "text" name="elements['.$count.'][descgen]" value ="'.$row["descgen"].'" readonly>
    </td><td> <input type= "checkbox" name="elements['.$count.'][chk1][]" value ="OK">  
    </td><td> <input type= "checkbox" name="elements['.$count.'][chk1][]" value ="NOK">
    </td><td> <input type= "checkbox" name="elements['.$count.'][chk1][]" value ="PARZ">    
    </td><td> <input type= "checkbox" name="elements['.$count.'][chk1][]" value ="NA"> 
    </td><td> <input type="text" name="elements['.$count.'][note]" placeholder = "Note"> 
    </td></tr>';   

  $count ++;
}

如果同时只允许一个状态,则必须更改复选框以放置单选按钮,如下所示:

$count = 0;
while($row = mysql_fetch_array($risultato)){
  echo '<tr><td>  <input class="hide" type= "text" name="element['.$count.'][macc]" value ="'.$row["macc"].'" readonly> 
    </td><td> <input class="hide" type= "text" name="element['.$count.'][iddispssic]" value ="'.$row["iddispsic"].'" readonly> 
    </td><td> <input class="hide" type= "text" name="element['.$count.'][descgen]" value ="'.$row["descgen"].'" readonly>
    </td><td> <input type= "radio" name="element['.$count.'][chk1]" value ="OK">  
    </td><td> <input type= "radio" name="element['.$count.'][chk1]" value ="NOK">
    </td><td> <input type= "radio" name="element['.$count.'][chk1]" value ="PARZ">    
    </td><td> <input type= "radio" name="element['.$count.'][chk1]" value ="NA" checked> 
    </td><td> <input type="text" name="element['.$count.'][note]" placeholder = "Note"> 
    </td></tr>';   

  $count ++;
}

在“ insert_controllo.php”中,您可以通过以下方式读取数据:

  if(isset($_POST['elements'])){
    for($i=0; $i<count($_POST['elements']); $i++){
      // the query. For example 'UPDATE TABLENAME SET chk1 = "'.$_POST['elements']['chk1'].'" WHERE macc = "'.$_POST['elements']['macc'].'" ';
    }
  }

}