从现在起不起作用,没有任何表现

问题描述 投票:0回答:1

我想向用户显示从规定日期到所述日期。

以下是查询。如果我不包括查询之间它显示所有,但如果我包括它,它不会显示任何东西

    <?php
      $iq=mysqli_query($conn,"select * from inventory left join product on product.productid=inventory.productid order by inventory_date desc ");
      while($iqrow=mysqli_fetch_array($iq)){


      ?>
        <tr>
          <td class="hidden"></td>
          <td><?php echo date('M d, Y h:i A',strtotime($iqrow['inventory_date'])); ?></td>  
          <td>
          <?php 
            $u=mysqli_query($conn,"select * from `user` left join customer on customer.userid=user.userid left join supplier on supplier.userid=user.userid where user.userid='".$iqrow['userid']."'");
            $urow=mysqli_fetch_array($u);
            if($urow['access']==1){
              echo "Admin";
            }
            elseif($urow['access']==2){
              echo $urow['customer_name'];
            }
            else{
              echo $urow['company_name'];
            }
          ?>
          </td>
          <td align="right"><?php echo $iqrow['action']; ?></td>
          <td align="right"><?php echo $iqrow['product_name']; ?></td>
          <td align="right"><?php echo $iqrow['quantity']; ?></td>
        </tr>
      <?php

} } ?>

php jquery
1个回答
0
投票

试试这个

$u = mysqli_query($conn, "select * from `user` left join customer on customer.userid=user.userid left join supplier on supplier.userid=user.userid where user.userid='".$iqrow['userid'].
      "AND (SALES_DATE BETWEEN '$from' AND '$to')");
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