Как я могу отобразить число на основе количества данных из базы данных.
Например, на приведенном выше рисунке показана извлечение данных из базы данных. В моей базе данных есть две данные. Поэтому я хотел бы отобразить номера 1 и 2 в столбце Нет.
И если в базе данных есть четыре данных, я бы хотел отобразить цифры 1, 2, 3 и 4 в столбце Нет.
Кодирование для отображения таблицы выше
<?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>";
?>
Благодарю.
Попробуйте с приведенным ниже кодом:
<?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>";
?>
@ Hardy Change. $ ++ i. до. ++ $ я.