根據數據庫中的數據數顯示數字序列

enter image description here 根據數據庫中的數據數顯示數字序列

如何根據數據庫中的數據顯示數量。

例如，上面的圖片顯示了數據庫中的數據摘錄。在我的數據庫中有兩個數據。因此，我想在No欄中顯示編號1和2。

如果數據庫中有四個數據，我想在No列顯示數字1,2,3和4。

編碼來顯示上述表

<?php 
$sql= mysql_query ("SELECT * FROM employee INNER JOIN cash ON employee.emp_id = cash.emp_id WHERE cash_status='Pending'"); 
echo "<table id='dataTable' width='850' border='1' align='center'>"; 

echo "<tr> 
     <th height='50'>No</th> 
     <th height='50'>Employee Number</th> 
     <th height='50'>Name</th> 
     <th height='50'>Department</th> 
     <th height='50'>Date Apply</th> 
     <th height='50'>Date Cash To Be Use</th> 
     <th height='50'>Amount</th> 
     <th height='50'>Status</th> 
     <th height='50'>Cash Id</th> 
     <th height='50'>View</th> 
    </tr>"; 


    while ($row = mysql_fetch_array($sql)) 
    { 
     echo "<tr>"; 
     echo "<td align='center' height='30'></td>"; 
     echo "<td align='center' height='30'>" .$row['emp_id']. "</td>"; 
     echo "<td align='center' height='30'>" .$row['emp_name']. "</td>"; 
     echo "<td align='center' height='30'>" .$row['emp_department']. "</td>"; 
     echo "<td align='center' height='30'>" .$row['cash_dapply']. "</td>"; 
     echo "<td align='center' height='30'>" .$row['cash_duse']. "</td>"; 
     echo "<td align='center' height='30'>RM" .$row['cash_amount']. "</td>"; 
     echo "<td align='center' height='30'>" .$row['cash_status']. "</td>"; 
     echo "<td align='center' height='30'>" .$row['cash_id']. "</td>"; 
     echo"<td height='30'><a href= cadvance_approval.php?id=".$row['emp_id']."&cash_id=".$row['cash_id']."><img src='../img/view_user.png' width='20' height='20'></a></td>"; 
     echo "</tr>"; 
    } 
echo "</table>";  
?>

感謝。

來源

2015-05-04 amln_ndh

你能告訴我們打印這張表的代碼嗎？ – Halcyon

據我所知，**沒有**列沒有值... – sitilge

它很簡單，你可以採取'$計數器'，並實現這一點 –

嘗試用下面的代碼：

<?php 
$i = 0; 
$sql= mysql_query ("SELECT * FROM employee INNER JOIN cash ON employee.emp_id = cash.emp_id WHERE cash_status='Pending'"); 
echo "<table id='dataTable' width='850' border='1' align='center'>"; 

echo "<tr> 
     <th height='50'>No</th> 
     <th height='50'>Employee Number</th> 
     <th height='50'>Name</th> 
     <th height='50'>Department</th> 
     <th height='50'>Date Apply</th> 
     <th height='50'>Date Cash To Be Use</th> 
     <th height='50'>Amount</th> 
     <th height='50'>Status</th> 
     <th height='50'>Cash Id</th> 
     <th height='50'>View</th> 
    </tr>"; 


    while ($row = mysql_fetch_array($sql)) 
    { 
     echo "<tr>"; 
     echo "<td align='center' height='30'>".$++i."</td>"; 
     echo "<td align='center' height='30'>" .$row['emp_id']. "</td>"; 
     echo "<td align='center' height='30'>" .$row['emp_name']. "</td>"; 
     echo "<td align='center' height='30'>" .$row['emp_department']. "</td>"; 
     echo "<td align='center' height='30'>" .$row['cash_dapply']. "</td>"; 
     echo "<td align='center' height='30'>" .$row['cash_duse']. "</td>"; 
     echo "<td align='center' height='30'>RM" .$row['cash_amount']. "</td>"; 
     echo "<td align='center' height='30'>" .$row['cash_status']. "</td>"; 
     echo "<td align='center' height='30'>" .$row['cash_id']. "</td>"; 
     echo"<td height='30'><a href= cadvance_approval.php?id=".$row['emp_id']."&cash_id=".$row['cash_id']."><img src='../img/view_user.png' width='20' height='20'></a></td>"; 
     echo "</tr>"; 
    } 
echo "</table>";  
?>

來源

2015-05-04 14:49:13

你的_No_列從'0'開始。既可以做'$ i = 1'或'++ $ i'。 – Halcyon

@Halcyon ..謝謝男人.. !!代碼已更新 –

@Hardy更改$ ++我。到。++ $ i。

來源

2016-08-27 08:03:56

根據數據庫中的數據數顯示數字序列

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