如何将动态生成的 table 数据保存到 php codeigniter 中的 mysql 数据库

Question

我在 PHP Code igniter 中创建了一个 Web 应用程序，它生成 table 包含计算员工收入和扣除额。因为我是 CI 的新手。我试图将动态生成的 table 数据保存到数据库，但失败了。我不知道如何 POST table 数据到控制器。欢迎任何帮助。
This is My Controller To generate dynamic table

function index()
{

    if($this->input->post("submit"))
    {
         $vals=$this->cpanel_model->esic();
     if($vals){
        $data['w_share']=$vals->w_share;
        $data['e_share']=$vals->e_share;
     }

    $data["wshare"]="";
        //echo $data['w_share'];
        $city=$_POST["city"];
        $v1= $data['w_share'];
        $v2= '100';
        $v3= $data['e_share'];
        $arr=$_POST["wid"];
        $arr1=$_POST["days"];
        $arr2=$_POST["payment"];
        $arr3=$_POST["wdays"];
        for ($i=0; $i <count($arr1) ; $i++)
         { 
            $net[$i]=($arr2[$i]/$arr3[$i])*$arr1[$i];

            $wshare[$i]=($net[$i]/$v2)*$v1;
            $eshare[$i]=($net[$i]/$v2)*$v3;
            $total[$i]=$eshare[$i]+$wshare[$i];
            $wdays[$i]=$arr3;
            $days[$i]=$arr1;
         }
         $data["net"]=$net;
         $data["wshare"]=$wshare;
         $data["total"]=$total;
         $data["eshare"]=$eshare;
         $data["wdays"]=$arr3;
         $data["days"]=$arr1;
         $data["list"]=$this->payment_model->search($city);
         $this->load->view("secure/php-version/esic",$data);

    }

这是我的观点以回应 table 值

  <?php echo form_open('esic/insert',array("role"=>"form")); ?>
             <div class="table-responsive">
               <table border="1" id="t01"  class="table table-striped table-bordered table-hover table-condensed">
<tr>
  <th>Worker_id</th>
  <th>Worker Name</th>
  <th>Worker City</th>
  <th>Designation</th>
  <th>Working Days</th>
   <th>Basic Payment</th>
   <th>Net Payment</th>
  <th>Present Days</th>
  <th>Worker Share (1.75%)</th>
  <th>Employee Share (4.75%)</th>
  <th>Total</th>
</tr>
<?php $i=0; $sum2=0; ?>
<?php  if(count($list)>0) { foreach ($list as $std) {?>
  <tr>
  <?php echo form_hidden("wid[]",$std["wid"]); ?>
  <?php echo form_hidden("wdays[]",$wdays); ?>
  <?php  echo form_hidden("payment[]",$std["payment"]); ?>
  <td><?php echo $std["wid"]; ?></td>
  <td><?php echo $std["wname"]; ?></td>
  <td><?php echo $std["city"]; ?></td>
   <td><?php echo $std["designation"]; ?></td>
   <td><?php echo $wdays[$i]; ?></td>
  <td><?php echo $std["payment"]; ?></td>
   <td><?php echo $net[$i]; ?></td>
<td><?php echo $days[$i]; ?></td>  
   <td><?php echo $wshare[$i];  ?></td>
  <td><?php echo $eshare[$i];  ?></td>
  <td><?php echo $total[$i];  ?></td>
  <?php $sum2 += $std["payment"]; ?>
    </tr>
<?php $i++;}}?>
<tr>
<?php
$sum=0;
$sum1=0;
$sum3=0;
$sum4=0;
$sum5=0;
$i=0;
foreach ($total as $value) {
  $sum = $sum + $value;
  $i++;
}
foreach ($days as $value) {
  $sum1 = $sum1 + $value;
  $i++;
}
foreach ($eshare as $value) {
  $sum3 = $sum3 + $value;
  $i++;
}
foreach ($wshare as $value) {
  $sum4 = $sum4 + $value;
  $i++;
}
foreach ($net as $value) {
  $sum5 = $sum5 + $value;
  $i++;
}
 ?>
<td>
  &nbsp;
</td>
<td>
  &nbsp;
</td>
<td>
  &nbsp;
</td>
<td>
  &nbsp;
</td>
<td>
  &nbsp;
</td>
<td style="font-weight:bold; font-size:10pt;">
  <?php echo $sum2; ?>
</td>
<td style="font-weight:bold; font-size:10pt;">
  <?php echo $sum5; ?>
</td>
<td style="font-weight:bold; font-size:10pt;">
 <?php echo $sum1; ?>
</td>
<td style="font-weight:bold; font-size:10pt;">
<?php echo $sum4; ?>
</td>
<td style="font-weight:bold; font-size:10pt;">
<?php echo $sum3; ?>
</td>
<td style="font-weight:bold; font-size:10pt;">
 <?php echo $sum; ?>
</td>
</tr>
</table>

</div>
         <?php echo form_submit(array("name"=>"submit","type"=>"submit","value"=>"Create ESIC")); ?>
<?php echo form_close(); ?>

Answer 1

目前还不太清楚您是如何生成 table 的，但是无论如何，当您生成 table 时，您必须将它放在一个表单中，并为输入分配键。试试这个方法：

HTML（这些是您生成 table 时应该得到的结果）

<form action="<?= base_url() ?>table_data" method="post">
    <table>
        <thead>
            <tr>
                <th>Name</th>
                <th>Earnings</th>
                <th>Deduction</th>
            </tr>
        </thead>
        <tbody>
            <tr>
                <td><input type="hidden" name="table[0][name]" value="name1">name1</td>
                <td><input type="hidden" name="table[0][earning]" value="earning1">earning1</td>
                <td><input type="hidden" name="table[0][deduc]" value="deduc1">deduc1</td>
            </tr>
            <tr>
                <td><input type="hidden" name="table[1][name]" value="name1">name1</td>
                <td><input type="hidden" name="table[1][earning]" value="earning1">earning1</td>
                <td><input type="hidden" name="table[1][deduc]" value="deduc1">deduc1</td>
            </tr>
        </tbody>
    </table>
    <input type="submit" value="Submit this table">
</form>

控制器：

function table_data() {
    if ($this->input->post()) {
        $table_data = $this->input->post('table');
        $this->load->model('model_name');
        $this->model_name->insert_table_data($table_data);
    }
}

型号：

function insert_table_data($table_data){
    $this->db->insert_batch('mytable', $table_data); 
}

如何将动态生成的 table 数据保存到 php codeigniter 中的 mysql 数据库

How to save dynamic generated table data to mysql database in php codeigniter

php

mysql

codeigniter