使用 nymber_format 使用 mysql 和 php 检索员工工资

Question

我正在使用 php 从 musql db 返回员工信息。我使用 number_format() 方法以格式化逗号返回工资，但是当返回数据时，所有员工的工资都相同。如何让php从employees表中返回个人工资？

PHP/MySQL

   <?php

    require_once("db.php");

    $sql = "SELECT `*` FROM `employees`";
    $results = mysqli_query($connect, $sql) or die(mysql_error());
    $row = mysqli_fetch_array($results, MYSQL_BOTH) or die(mysql_error());
    $salary = $row['salary']; 
    $rows_sal = number_format($salary); 

    echo("<table>");
    while($row = mysqli_fetch_array($results, MYSQL_BOTH))
    {
        echo("<tr>");
        echo "<td>" . $row['empid'] . '</td>' .
             "<td>" . $row['lastname'] . '</td>' .
             "<td>" . $row['firstname'] . '</td>' .
             "<td>" . $row['department'] . '</td>' .
             "<td>" . $row['position'] . '</td>' .
             "<td>" . $rows_sal . '</td>'; 


        echo '</br>';

        echo('</tr>');
    }


    echo("</table>");
    ?>

Answer 1

number_format 不适用于数组。这样做：

<?php

    require_once("db.php");

    $sql = "SELECT `*` FROM `employees`";
    $results = mysqli_query($connect, $sql) or die(mysql_error());
    $row = mysqli_fetch_array($results, MYSQL_BOTH) or die(mysql_error());
    //$salary = $row['salary'];
    //$rows_sal = number_format($salary);

    echo("<table>");
    while($row = mysqli_fetch_array($results, MYSQL_BOTH))
    {
        echo("<tr>");
        echo "<td>" . $row['empid'] . '</td>' .
             "<td>" . $row['lastname'] . '</td>' .
             "<td>" . $row['firstname'] . '</td>' .
             "<td>" . $row['department'] . '</td>' .
             "<td>" . $row['position'] . '</td>' .
             "<td>" . number_format($row['salary']) . '</td>';


        echo '</br>';

        echo('</tr>');
    }


    echo("</table>");
?>

Answer 2

那是因为您在 foreach 循环之外使用薪水。 （顺便说一句，您将第一行从表格中删除）。

在我建议你如何做之前，让我解释一下你的实际代码发生了什么：（跟随 cmets）

   <?php

    require_once("db.php");

    $sql = "SELECT `*` FROM `employees`";
    $results = mysqli_query($connect, $sql) or die(mysql_error());

    //here you're asking THE FIRST row:
    $row = mysqli_fetch_array($results, MYSQL_BOTH) or die(mysql_error()); 
    $salary = $row['salary']; 
    //so $rows_sal is the first salary. All this have nothing to do with the iteration below.
    $rows_sal = number_format($salary); 

    echo("<table>");

    //here you're asking for the next results (leaving 1st row outside the table!)
    while($row = mysqli_fetch_array($results, MYSQL_BOTH)) 
    {
        echo("<tr>");
        echo "<td>" . $row['empid'] . '</td>' .
             "<td>" . $row['lastname'] . '</td>' .
             "<td>" . $row['firstname'] . '</td>' .
             "<td>" . $row['department'] . '</td>' .
             "<td>" . $row['position'] . '</td>' .

             //and here you're referencing the value set before the while.
             "<td>" . $rows_sal . '</td>'; 


        echo '</br>';

        echo('</tr>');
    }


    echo("</table>");
    ?>

现在这是你应该做的：（跟随 cmets）

   <?php

    require_once("db.php");

    $sql = "SELECT `*` FROM `employees`";
    $results = mysqli_query($connect, $sql) or die(mysql_error());
    //don't ask for results yet! 

    echo("<table>");
    while($row = mysqli_fetch_array($results, MYSQL_BOTH)) //including 1st row
    {
        //now $row is EACH row, NOW you can format your salary:
        $salary = $row['salary']; 
        $rows_sal = number_format($salary); 
        echo("<tr>");
        echo "<td>" . $row['empid'] . '</td>' .
             "<td>" . $row['lastname'] . '</td>' .
             "<td>" . $row['firstname'] . '</td>' .
             "<td>" . $row['department'] . '</td>' .
             "<td>" . $row['position'] . '</td>' .
             "<td>" . $rows_sal . '</td>';
        echo '</br>';
        echo('</tr>');
    }


    echo("</table>");
    ?>

希望对你有帮助。