小编jog*_*_pi的帖子

$stmt = "insert into {CI}payment_logs (customer_id, invoice_id, invoice_total_amount, invoice_paid_amount, customer_paid_amount, invoice_payment_on ) "
                . "values( ?, ?, ?, ?, ?, NOW() )";
$this->db->query( $stmt, array( $Customer_Id, $Invoice_Id, $Invoice_Total_Amount, $Deducted_Amount, $Paid_By_Cust ) );        
$Log_Id = $this->db->insert_id(); // Returns Last Insert ID

根据我的需要,这是正常工作,但我必须在特定条件下更新同一个表中每一行的状态.看看图片,

这是上述查询的插入数据.如您所见,invoice_total_amount是总金额,但这个金额在第三行数据中完成.我必须更新该invoice_status字段,因为paid在高度点亮量SUM等于invoice_total_amount和之后,我必须将状态更新为partial.我尝试更新状态以添加触发器但不成功.

然后我尝试使用此命令的PHP.

update {CI}payment_logs 
    set invoice_status = IF( invoice_total_amount = sum(invoices_paid_amount), 'paid', 'partial' ) 
where log_id = ?

但我没有得到invoices_paid_amountinvoice_id = 的SUM 5,我尝试在IF条件下使用子查询,但它无效我的查询.

请帮我解决单查询或触发解决方案..

我有一个简单的HTML表.我想获得我点击的行的第一列的值.如果我点击第一行的任何地方它应该获得第一行的第一列的值,如果我点击第二行任何地方,它应该给第二列的第一列的值..这是我的表:

<table style="border:1px solid black" id="tab">
    <tr>
        <td style="border-right:1px solid black">Sachin</td>
        <td>Yuvraj</td> 
    </tr>
    <tr>
        <td style="border-right:1px solid black">Virat</td>
        <td>Gautam</td>              
    </tr>
</table>

我的多重上传与codeigniter有问题,

1. 使用图像
2. 另一个使用pdf

当我上传两次上传的文件以及如何调用上传到数据库的不同路径时.这是我的代码

调节器

public function upload(){

    $catalog='catalog';
    $userfile='userfile';

    //for cover
    $config['upload_path'] = './file/book/'; //Use relative or absolute path
    $config['allowed_types'] = 'gif|jpg|png|'; 
    $config['max_size'] = '100000';
    $config['max_width'] = '1024';
    $config['max_height'] = '768';
    $this->load->library('upload', $config);
    $this->upload->initialize($config); 

    //$c=$this->upload->do_upload($userfile);

    //for catalog
    $config['upload_path'] = './file/book/pdf/'; //Use relative or absolute path
    $config['allowed_types'] = 'pdf'; 
    $config['max_size'] = '1000000';
    $this->load->library('upload', $config);
    $this->upload->initialize($config);

    //$cat=$this->upload->do_upload($catalog);


    if(!$this->upload->do_upload($catalog) &&         $this->upload->do_upload($userfile)){


        $error = array('error' => $this->upload->display_errors());
        $this->load->view('uploadds', $error);

    }else{ 



        $this->load->model("book_model"); 
        $this->book_model->addnewbook();
        redirect('book/book');      
    }

}

这个模型

function addnewbook(){
     $fcat=array('upload_data' => $this->upload->data($userfile));
     $fcatalog=array('upload_dataa' => …

我想加密url参数值,如

http://www.sitename.com/index.php?userid=12546

成

http://www.sitename.com/index.php?userid=SADFFHGFE

防止机器人破解自动递增到数据库的用户ID,我不确定base64_encode和的安全性base64_decode.有没有办法做到这一点？

我正在尝试user_status用条件更新提交的文件:

update pm_users 
    set user_status = if ( 
      (select u.user_status from pm_users u where u.user_id = 3
      ) = '1', '0', '1' )
    where user_id = 3

表示如果user_status = 1然后用0更新状态,如果用户状态为0,则为1.

我收到错误: You can't specify target table 'pm_users' for update in FROM clause

我认为这意味着我不能像上面那样使用这个查询同一个表吗？我不确定.
请帮助我以正确的方式前进,并让我纠正.

我试图用VueRouter实现VueJs.Home组件显示日志消息但不显示模板部分.我收到以下错误:

Home.vue

<template>
    <div class="wrap" id="app">
        <h1 class="inline">Hello There</h1>
    </div>
</template>

<script>
export default {
    name: 'app',
    data () {
        return {
            maps: []
        }
    },
    mounted() {
        console.log( 'Mounted Homepage' );
    }
}
</script>

<style lang="scss">
#app {
   font-family: 'Avenir', Helvetica, Arial, sans-serif;
   -webkit-font-smoothing: antialiased;
   -moz-osx-font-smoothing: grayscale;
}
</style>

routes.js

import Vue from 'vue';
import Home from './components/Home.vue';
import NotFound from './components/NotFound.vue';

const routes = [
    { name: 'home', path: '/', components: Home },
    { path: '*', …

我已经构建了自定义电子邮件模板.并指定了一些变量{#paid_amount}等等.

所有变量都被替换但paid_amount不是预期的.我已经取代了这样的东西:

// Text file with HTML markups
$template = file_get_contents($template_url);

$paid_amount = '$1.00';
$pattern = array( 
              '/\{\#user_name\}/i', 
              '/\{\#paid_amount\}/i', 
              '/\{\#duration\}/i'  );
$replacement = array( 
              $user_name, 
              $paid_amount, 
              $duration );

$new_template = preg_replace($pattern, $replacement, $template);

它.00在电子邮件中打印金额,如果我$从打印金额中删除标志1.00.我在Gmail中测试了它.以前有人遇到过这个吗？

即使我尝试$但不工作.任何人都可以告诉我我错过了什么或为什么它不工作？