Question

我想通过搜索功能（搜索栏和＆＃34; Ok＆＃34;按钮）来使用下拉列表（带过滤器/选项）。

当我在下拉列表中选择过滤器/选项时，例如＆＃34; Team＆＃34;。下拉然后列表应该知道该列表已更改为过滤器/选项＆＃34; Team＆＃34;。在此之后，我使用搜索功能输入搜索词，如＆＃34; USA＆＃34;然后按好的按钮。按下确定按钮后，会出现一个sql-query：SELECT * FROM mytable WHERE Team =＆＃39; USA＆＃39 ;; 生成以获取db＆＃34; mytable＆＃34;。

中的所有匹配行

选择过滤器＆＃34; Team＆＃34;。
选择一个团队，＆＃34; USA＆＃34;。
使用db mytable中的相关数据获取匹配的行。

希望我让自己更好地理解，谢谢大家的时间！

这是我的代码：

<html>
    <head> 
        <title>AiFind</title>
        <link rel="stylesheet" href="Style.css">
        <script src="logic.js"></script>

    </head>
    <body>
        <h1>AiFind</h1>
    </body>

</html>


<?php

include "connection.php";

$sql = "SELECT * FROM mytable";

if (isset($_POST['search'])) {

    $search_term = mysql_real_escape_string($_POST['search_box']);

    $sql .= " WHERE F_ar = '$search_term' ";
    $sql .= " OR Postnr = '$search_term' ";
    $sql .= " OR Postort = '$search_term' ";
    $sql .= " OR Vardgivare = '$search_term' ";
    $sql .= " OR Team = '$search_term' ";
    $sql .= " OR Orsak = '$search_term' ";
    $sql .= " OR Planerat_datum = '$search_term' ";
    $sql .= " OR fran = '$search_term' ";
    $sql .= " OR AAA_diam = '$search_term'; ";
}

$query = mysql_query($sql) or die(mysql_error());

?>

<form name="Select_filter" method="POST" action="VGR_data_display.php">
<select id="dropdown" name="filter">
<option value=""></option>
<option value="1">ID</option>
<option value="2">Alder</option>
<option value="3">Postnummer</option>
<option value="5">Postort</option>
<option value="6">Vårdgivare</option>
<option value="7">Planerat Datum</option>
<option value="8">Status</option>
<option value="9">AAA_diameter</option>
</select>
</form>


<!--search bar for search term input -->
<form name ="search_form" method="POST" action="VGR_data_display.php">

    <input id="search_box" type="text" name="search_box" value="" />
    <input id="submit" type ="submit" name ="search" value ="Ok">

</form>

<table style="margin:auto;" id="table" border='1'>
<tr>
<th>ID</th>
<th>F_ar</th>
<th>Postnr</th>
<th>Postort</th>
<th>Vardgivare</th>
<th>Team</th>
<th>Orsak</th>
<th>Planerat_datum</th>
<th>fran</th>
<th>AAA_diam</th>
</tr>


<?php while($row = mysql_fetch_array($query)) { ?>
  <tr>
  <td><?php echo $row['id']; ?></td>
  <td><?php echo $row['F_ar']; ?></td>
  <td><?php echo $row['Postnr']; ?></td>
  <td><?php echo $row['Postort']; ?></td>
  <td><?php echo $row['Vardgivare']; ?></td>
  <td><?php echo $row['Team']; ?></td>
  <td><?php echo $row['Orsak']; ?></td>
  <td><?php echo $row['Planerat_datum']; ?></td>
  <td><?php echo $row['fran']; ?></td>
  <td><?php echo $row['AAA_diam']; ?></td>
  </tr>

<?php } ?>

Answer 1

更改下拉列表，以便选择值是数据库列名。即

<select id="dropdown" name="filter">
  <option value=""></option>
  <option value="ID">ID</option>
</select>

然后在PHP中：

if (isset($_POST['search'])) {

    $search_term = mysql_real_escape_string($_POST['search_box']);
    $column = mysql_real_escape_string($_POST['filter']);
}

SQL

SELECT * FROM mytable where $column = '$search_term'

这意味着添加新选项很容易，并且保存了一个冗长的if / switch语句。它还允许您在线下构建多个选项。

注意：如果您担心在表单中显示列名，可以在构建下拉列表时加密它们，然后在需要使用它们时解密

编辑 - 基于评论。

在SQL之前：

$allowed[] = array("ID") //Add all allowed columns.
if (in_array($column, $allowed))
{
  //Run SQL
}

如果希望自动化，可以在构建DDL时构建此数组。

Answer 2

如果我已正确理解，那么您只想搜索下拉列表中选择的列。然后你想用这个：

if (isset($_POST['search'])) {

    $search_term = mysql_real_escape_string($_POST['search_box']);

    switch($_POST['filter']) {

        Default: 
            $sql .= " WHERE F_ar = '$search_term' ";
            break;
        case 1:
            $sql .= " WHERE Postnr = '$search_term' ";
            break;
        case 2:
            $sql .= " WHERE Postort = '$search_term' ";
            break;
        case 3:
            $sql .= " WHERE Vardgivare = '$search_term' ";
            break;
        case 4:
            $sql .= " WHERE Team = '$search_term' ";
            break;
        case 5:
            $sql .= " WHERE Orsak = '$search_term' ";
            break;
        case 6:
            $sql .= " WHERE Planerat_datum = '$search_term' ";
            break;
        case 7:
            $sql .= " WHERE fran = '$search_term' ";
            break;
        case 8:
            $sql .= " WHERE AAA_diam = '$search_term'; ";
            break;

    }
}

<小时/> 至于你的错误，你有两种形式，你需要一切形式如下：

<form name="Select_filter" method="POST" action="VGR_data_display.php">
    <select id="dropdown" name="filter">
        <option value=""></option>
        <option value="1">ID</option>
        <option value="2">Alder</option>
        <option value="3">Postnummer</option>
        <option value="5">Postort</option>
        <option value="6">Vårdgivare</option>
        <option value="7">Planerat Datum</option>
        <option value="8">Status</option>
        <option value="9">AAA_diameter</option>
    </select>


    <!--search bar for search term input -->

    <input id="search_box" type="text" name="search_box" value="" />
    <input id="submit" type ="submit" name ="search" value ="Ok">

</form>

使用下拉列表从mysql数据库中获取数据，然后使用搜索功能从选定的下拉列表选项中过滤数据

2 个答案: