Question

我试图在PHP中动态显示选择选项的值。基本上用户创建存储在数据库中的那些选项，然后我试图在不同的地方获取它。为此我编写了以下函数：

function fetch_acad_yr($conn) { 
    $query = "SELECT a.fk_acadyear_id as acadyearid,b.acad_year as acadyear FROM tbl_admparam a inner join list_acad_years b on a.fk_acadyear_id = b.pk_acad_year_id";
    $stmt = $conn->prepare($query);
    if ($stmt->execute()) {
        foreach ($stmt as $row) {
            ?>
            <option value="<?php echo $row['acadyearid'] ?>"><?php echo $row['acadyear'] ?></option>
            <?php
        }
    }
}

然后在html中我调用了这个函数

<select class="form-control" id="acad_period" name="acad_period" required>
<option value="">Please select...</option>
 <?php fetch_acad_yr($conn); ?>
 </select>

但是选项会显示任何值，它只是空白而不是显示存储在数据库中的值。我尝试单独运行查询，查询返回所需的结果

Answer 1

XpsDocument xpsDocument = new XpsDocument(@"C:\pathRoot\fileName.xps", FileAccess.ReadWrite);

XpsDocumentWriter xpsDocumentWriter = XpsDocument.CreateXpsDocumentWriter(xpsDocument);

xpsDocumentWriter.Write(@"C:\pathRoot\fileName.png");

Answer 2

确保您拥有正确的连接以及查询结果中的数据。我使用的基本php mysql代码是下面的，它的工作原理。您将其更改为pdo或推进并获得结果。但是使用与while或for循环相同。

$conn = mysql_connect($host, $username, $pass);
mysql_select_db($dbname, $conn);

echo "<pre>";
echo "<select>";
fetch_acad_yr($conn);
echo "<\select>";

function fetch_acad_yr($conn) { 
    $query = "SELECT a.fk_acadyear_id as acadyearid,b.acad_year as acadyear FROM tbl_admparam a inner join list_acad_years b on a.fk_acadyear_id = b.pk_acad_year_id";
    $res = mysql_query($query);
    while($row = mysql_fetch_assoc($res)){
            ?>
            <option value="<?php echo $row['acadyearid'] ?>"><?php echo $row['acadyear'] ?></option>
            <?php
    }
}
?>

适合我。

选择不使用功能显示的选项

2 个答案: