正如标题所说,我想根据用户对年份的选择显示特定的数据行。
如果用户选择了2017,则我只想显示在2017年创建的数据行。
我知道查询有问题,我试图了解如何编写查询。谁能帮帮我吗?预先谢谢你!
<form action="Report1.php" method="POST">
Select Year:<select name="sortyear" id="sortyear">
<option>2017</option>
<option>2018</option>
</select>
<?php
$sortyear = $_POST['sortyear'];
$db = mysqli_connect('localhost','root','','customercaremodule');
date_default_timezone_set('Asia/Kuala_Lumpur');
if ($db->connect_error) {
die("Connection failed: " . $db->connect_error);
}
$sql = "SELECT first_name, last_name, email, message, FC_DateTime FROM fbcomplain WHERE (YEAR(FC_DateTime) = $sortyear)";
$result = mysqli_query($db,$sql);
?>
</form>
答案 0 :(得分:0)
忽略安全漏洞,您可以执行以下操作:
<?php $getYears = sqlquery() ?>
<form action="Report1.php" method="POST">
<select name="sortyear" id="sortyear">
<?php for ($i = 0; $i < $getYears.length; $i++): ?>
<option value="<?php echo $getYears[$i];?>"><?php echo $getYears[$i];?></option>
<?php endif; ?>
</select>
你可以
<?php for () {
}?>
但是for:允许您返回html代码并突出显示语法;)
答案 1 :(得分:0)
我设法通过添加按钮
解决了该问题<input type ="submit" name="search" class="btn btn-sm btn-primary" value="Filter" style="background-color: #474747;padding: 5px 5px !important;margin-bottom:5px;">
并以此方式编码php
if(isset($_POST['search']))
{
$db = mysqli_connect('localhost','root','','customercaremodule');
date_default_timezone_set('Asia/Kuala_Lumpur');
if ($db->connect_error) {
die("Connection failed: " . $db->connect_error);
}
$sortyear = $_POST['sortyear'];
$sql = "SELECT first_name, last_name, email, message, FC_DateTime FROM fbcomplain WHERE (YEAR(FC_DateTime) = $sortyear)";
$result = mysqli_query($db,$sql);
}
答案 2 :(得分:0)
您可以尝试以下方法:
<form action="Report1.php" method="POST">
Select Year:<select name="sortyear" id="sortyear">
<option>2017</option>
<option>2018</option>
</select>
<input type="submit" name="submit" value="Get Annual Data" />
<?php
if(isset($_POST['submit'])){
$sortyear = $_POST['sortyear'];
$db = mysqli_connect('localhost','root','','customercaremodule');
date_default_timezone_set('Asia/Kuala_Lumpur');
if ($db->connect_error) {
die("Connection failed: " . $db->connect_error);
}
$sql = "SELECT first_name, last_name, email, message, FC_DateTime FROM fbcomplain WHERE (YEAR(FC_DateTime) = " .$sortyear .")";
$result = mysqli_query($db,$sql);
}
?>
</form>
请注意,这主要是您的代码,我只是对此artcile进行了相应的修改。关键是我添加了按钮,使用户可以提交表单。