Question

我有5个来自数据库的php变量，使用这些变量我想计算一个付款计划表。这是我的loan_calc.php代码。此变量成功回显其值，但未构建表。非常感谢你。

<p>
<?php require_once('include_header.php'); ?>
<?php 

    $sql='SELECT * FROM tbl_loan_application where  application_id= 1563256222';
    $result = mysql_query($sql, $conn) or die(mysql_error());
    $row = mysql_fetch_assoc($result);

    // echo $row['application_id'];
    // echo "<br />";
    // echo $row['number_of_rentel'];
    // echo "<br />";
    // echo $row['loan_amount'];
    // echo "<br />";
    // echo $row['effective_rate'];
    // echo "<br />";
    // echo $row['accepted_amount'];
    // echo "<br />";
    // echo $row['repayment'];

    $npy=$row['number_of_rentel'];
    $apr=$row['effective_rate'];
    $amount=$row['accepted_amount'];
    $Repayment=$row['repayment']; 
    $payment=$row['payment']; 

    echo "<table class='table table-striped table-hover' border='1' cellspacing='1' cellpadding='5'>";

    echo "<tr class='success'><td align='center'>Payment<br>Number</td><td align='center'>Payment<br>Amount</td><td align='center'>Interest</td><td align='center'>Principle</td><td align='center'>Balance</td></tr>";

    //$amount= eval($amount);
    for($i=1; $i<=$Repayment*$npy;$i++){
        $tbldata='<td>';
        $interest=$amount*$apr/$npy/100;
        $amount+=$interest;
        $principle=$payment-$interest;
        $amount-= $payment;
        echo'<tr>'+$tbldata+$i+':</td>'+$tbldata+round($payment)+'</td>'+$tbldata+round($interest)+'</td>'+$tbldata+round($principle)+'</td>'+$tbldata+round($amount)+'</td></tr>';
    }                  
    echo'</table>';                     

?>

Answer 1

尝试在echo中使用 . （点）代替 + ，如以下代码

  echo'<tr>'.$tbldata.$i.':</td>'.$tbldata.round($payment).'</td>'.$tbldata.round($interest).'</td>'.$tbldata.round($principle).'</td>'.$tbldata.round($amount).'</td></tr>';

如何使用PHP生成贷款支付计划

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