Question

我在SQL中创建了一个数据库，我正在尝试回显HTML表中的数据。我创建了一个单独的PHP文件，它将引用特定条件（在本例中为行业），用于确定表中显示的记录。

我的代码提取所有匹配的条件，但只显示表格中的第一个结果。所有其他结果只是在没有空格的情况下编写。知道如何解决这个问题吗？谢谢！

mysql_select_db("project") or die(mysql_error());
print ("<h1>Database</h1>");
print("</br>");
print("</br>");


if($industry == "Media")
    {
        $strSQL = "SELECT * FROM db WHERE industry = 'Media' ";
        $data = count($id);             

        } elseif ($industry == "Finance")
    {
        $strSQL = "SELECT * FROM db WHERE industry = 'Finance' ";               

    } elseif ($industry == "Technology")
    {
        $strSQL = "SELECT * FROM db WHERE industry = 'Technology' ";                

    } elseif ($industry == "Health")
    {
        $strSQL = "SELECT * FROM db WHERE industry = 'Health care/biotechnology' ";             

    } elseif ($industry == "RealEstate")
    {
        $strSQL = "SELECT * FROM db WHERE industry = 'Real Estate' ";               

    } elseif ($industry == "Retail")
    {
        $strSQL = "SELECT * FROM db WHERE industry = 'Retail' ";                

    } elseif ($industry == "Other")
    {
        $strSQL = "SELECT * FROM db WHERE industry = 'Other' ";             

    } elseif ($industry == "Technology")
    {
        $strSQL = "SELECT * FROM db WHERE industry = 'Technology' ";                

    } 
    ?>

<table border="1">
<tr>
    <th>First Name</th>
    <th>Last Name</th>
    <th>Column 3</th>
    <th>Column 4</th>
    <th>Column 5</th>
    <th>Column 6</th>
</tr>
<?php
$id = $row['id'];
$firstName = $row['firstName'];
$lastName = $row['lastName'];
$company = $row['company'];
$totaldollar = $row['totaldollar'];
$formatted_dollar = number_format($totaldollar);

$dollar2 = $row['dollar2'];
$formatted_dollar2 = number_format($dollar2);

$year = $row['year'];
$industry = $row['industry'];


?>


<?php 
for($i = 0; $i < count($id); $i++)
{
 echo "<tr><td>" . $firstName . "</td><td>" . $lastName . "</td><td>" . $company . "</td><td>" . "$" . $formatted_dollar . "</td><td>" . "$" . $formatted_dollar2 . "</td><td>" . $industry . "</td></tr>" ;

}
?>    



</table>

  <?php

}

mysql_close($link);
?>


</body>
</html>

Answer 1

不确定你在哪里煽动$row。但是，既然您将$id视为数组 - $i < count($id)，那么您可能希望将它们设置为数组 -

<?php
    $id[] = $row['id'];
    $firstName[] = $row['firstName'];
    $lastName[] = $row['lastName'];
    $company[] = $row['company'];
    $totaldollar[] = $row['totaldollar'];
    $formatted_dollar[] = number_format($totaldollar);

    $dollar2[] = $row['dollar2'];
    $formatted_dollar2[] = number_format($dollar2);

    $year[] = $row['year'];
    $industry[] = $row['industry'];  
?>

然后在循环中，您需要使用$i来设置数组键

<?php 
    for($i = 0; $i < count($id); $i++)
    {
        echo "<tr><td>" . $firstName[$i] . "</td><td>" . $lastName[$i] . "</td><td>" . $company[$i] . "</td><td>" . "$" . $formatted_dollar[$i] . "</td><td>" . "$" . $formatted_dollar2[$i] . "</td><td>" . $industry[$i] . "</td></tr>" ;

}
?>

在php表中回显出sql数据

1 个答案: