我刚刚开始使用SQL,PHP和HTML,我打算学习一些课程,但在继续前进之前,我需要一个基本的工作模型。我有一个SQL数据库,有几个字段。我可以毫无问题地显示这些数据。我想用下拉框过滤这些数据。目前,我在下拉框中填充了Select Distinct命令。我似乎无法弄清楚修改我的SQL Select命令的语法,只显示与我的下拉框匹配的记录。哈,我是第一个承认我不知道我在做什么的人,但我的问题是我可能不知道我需要查找如何解决我的问题。谷歌和我都很紧张,但我们无法理解这一点。你能为我指出正确的方向吗?谢谢!
<!DOCTYPE html>
<head>
<style>
table {
font-family: arial, sans-serif;
border-collapse: collapse;
width: 100%;
}
td, th {
border: 1px solid #dddddd;
text-align: left;
padding: 8px;
}
tr:nth-child(even){
background-color: #dddddd;
}
</style>
<body>
<form action='new1.php' method="get" name='fPIC'>
<select class="form-dropdown validate[required]" style="width:150px" id="input_PIC" name="sPIC">
<?php
$link = mysqli_connect("localhost", "root", "**********", "pd4poc");
mysqli_select_db($link,'pd4poc');
$query = "SELECT DISTINCT PIC FROM dummydata order by PIC asc";
$result = mysqli_query($link,$query);
$menu=" ";
while($row = mysqli_fetch_array($result))
{
$menu .= "<option value=".$row['PIC'].">" .$row['PIC']. "</option>";
}
echo $menu;
echo "<table>";
while($row = mysqli_fetch_array($result))
{
echo "<tr><td>" . $row['PIC'] . "<td><td>" , $row['Reason'] . "<td><td>", $row['Status'] . "</td></tr>";
}
echo "</table>";
mysqli_close($link);
?>
</select>
<input type="submit" name="nPIC" value="Filter">
</form>
<?php
$link = mysqli_connect("localhost", "root", "**********", "pd4poc");
mysqli_select_db($link,'pd4poc');
$query = "SELECT * FROM dummydata"; //need to be picked from the pull down
$result = mysqli_query($link,$query);
echo "<table>";
echo "<tr>
<th>PIC</th>
<th>Reason</th>
<th>PlanApply</th>
<th>HOT</th>
<th>Status</th>
<th>Comments</th>
</tr>\n";
while($row = mysqli_fetch_array($result))
{
echo "<tr><td>" .
$row['PIC'] . "<td>",
$row['Reason'] . "<td>",
$row['PlanApply'] . "<td>",
$row['HOT'] . "<td>",
$row['Status'] . "<td>",
$row['Comments'] . "</td>\n</tr>";
}
echo "</table>";
mysqli_close($link);
?>
</body>
</head>