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Php 数据不保存且不显示任何错误-Codeigniter_Php_Mysql_Codeigniter - Fatal编程技术网

Php 数据不保存且不显示任何错误-Codeigniter

Php 数据不保存且不显示任何错误-Codeigniter,php,mysql,codeigniter,Php,Mysql,Codeigniter,如果不提示代码中的任何错误,则无法保存数据 这是我控制器中的代码 function saveDebit(){ $jevseries=$this->input->post('series'); $account='debit'; $count='1'; $this->form_validation->set_error_delimiters('<div class="error">', '</div>'); $this->form_valid

如果不提示代码中的任何错误,则无法保存数据

这是我控制器中的代码

function saveDebit(){
$jevseries=$this->input->post('series');
$account='debit';
$count='1';
$this->form_validation->set_error_delimiters('<div class="error">', '</div>'); 
$this->form_validation->set_rules('accountName', 'Account Name', 'required');
$this->form_validation->set_rules('amountDebit', 'Amount', 'required|numeric');

$this->formValidation();
    if ($this->form_validation->run() == FALSE) {
        $this->entryAccount($jevseries);
        } else {
          $debitData = array(
            'Series' => $jevseries,
            'AccountCode' => $this->input->post('accountName'),
            'Account' => $account,
            'Amount' => $this->input->post('amountDebit'),
            'Count' => $count);

      $this->Jev_model->entryDebit($debitData);
    }
这是我的看法

<?php echo form_open('Jev/saveDebit'); ?>
        <table class="jev_entry" id="jev_entry">
            <tr>
                <td>Debit</td>
                <?php echo form_hidden('series', $series);?>
                <td><?php
                 $account_array = array();
                 foreach($accountNames as $account ){
                     $account_array []= $account->accountName;
                 }

                 echo form_dropdown('accountName',  $account_array,'', 'class="form-control"');

                 ?></td>
            </tr>
            <tr>
                <td>Amount</td>
                <td><?php  echo form_input('amountDebit','', 'class="form-control"');?><div class="error"><?php echo form_error('amountDebit'); ?></div></td>
            </tr>
            <tr>
                <td></td>
                <td align="center"><?php echo '<center>'. form_submit('submit','Save').'</center>'; ?> </td>
            </tr>
        </table>

借记

好的,我已经发现了问题所在,codeigniter的表单_输入只给出带有标签的表单。它没有名字,因此我们无法发布,因为它没有名字。这样做:)

就是这样,我们把名字贴上去。当然,您可以删除一些参数。更多信息请参见codeigniter:)中的form helper用户指南


Godluck meyt

你能给我们表格吗?我已经更新了我的问题:)很高兴帮忙,谢谢你的检查和投票!!:)@我已经解决了我的问题。这是我的错误,我复制并粘贴了其他函数中的代码,然后粘贴到了新函数中$此->formValidation();很高兴知道你已经解决了。祝你在编码方面好运!!
<?php echo form_open('Jev/saveDebit'); ?>
        <table class="jev_entry" id="jev_entry">
            <tr>
                <td>Debit</td>
                <?php echo form_hidden('series', $series);?>
                <td><?php
                 $account_array = array();
                 foreach($accountNames as $account ){
                     $account_array []= $account->accountName;
                 }

                 echo form_dropdown('accountName',  $account_array,'', 'class="form-control"');

                 ?></td>
            </tr>
            <tr>
                <td>Amount</td>
                <td><?php  echo form_input('amountDebit','', 'class="form-control"');?><div class="error"><?php echo form_error('amountDebit'); ?></div></td>
            </tr>
            <tr>
                <td></td>
                <td align="center"><?php echo '<center>'. form_submit('submit','Save').'</center>'; ?> </td>
            </tr>
        </table>
$data = array(
        'name'          => 'username',
        'id'            => 'username',
        'value'         => 'johndoe',
        'maxlength'     => '100',
        'size'          => '50',
        'style'         => 'width:50%'
);

echo form_input($data);