Question

我有一个数据表，用户可以使用各种选项进行过滤。

我可以使用表单提交来处理数据，但表单接受$_POST个变量，在某些情况下我希望用户只使用一个。

MP | Conservative | Year Elected

例如：

案例＃1 - 用户搜索“保守派”派对，但并不关心当选的年份。

案例＃2 - 用户搜索＆＃39; 2005＆＃39;选举年，但并不关心政党。

案例＃3 - 用户搜索“对话”＆＃39;派对和2005年＆＃39;。

这是我的代码：

全局变量

if(empty($_POST['year_elec'])) {
    $year = "show_all";
} 
else {
    $year = $_POST['year_elec'];
}

if(empty($_POST['party'])) {
    $party = "show_all";
}
else {
    $party = $_POST['party'];
}

主表生成：

        if($year == "show_all" && $party == "show_all") {
                        $query = "SELECT * FROM mps";
                    }
                    else if(!empty($year) && $party ="show_all"){
                        $query = "SELECT * FROM mps WHERE year_elec = $year";
                    }
                    else if($year == "show_all" && !empty($party)) {
                        $query = "SELECT * FROM mps WHERE party = $party";

                    }
                    else {
                        $query = "SELECT * FROM mps WHERE year_elec = $year AND party = $party";
                    }

                        $res = mysqli_query($conn, $query);
                        while ($a = mysqli_fetch_assoc($res)) {
                            $name = $a['name'];
                            $party = $a['party'];
                            $consti = $a['cont'];
                            $year_elec = $a['year_elec'];

                            echo "<tr>";
                            echo "<td>" .$name. "</td>";
                            echo "<td>" .$party. "</td>";                       
                            echo "<td>" .$consti. "</td>";
                            echo "<td>" .$year_elec. "</td>";       
                            echo "</tr>";           
                        }

表格

<form method="post" action="mps.php">
            <span class="css-select-moz">
                <select name="year_elec" id="selectYear">
                    <option <?php if (isset($year) && $year=="2010") echo "selected";?> name="2010" value="2010">First elected 2010</option>
                    <option <?php if (isset($year) && $year=="2005") echo "selected";?> name="2005" value="2005">First elected 2005</option>
                    <option <?php if (isset($year) && $year=="2001") echo "selected";?> name="2001" value="2001">First elected 2001</option>
                </select>
            </span> <br />

            <span class="css-select-moz">
                <select name="party" id="selectParty">                  
                    <?php
                        $query = "SELECT DISTINCT party FROM mps";
                        $res = mysqli_query($conn, $query);
                        while($a = mysqli_fetch_assoc($res)) {
                            $party = $a['party'];
                            echo "<option name='$party' value='$party'>$party</option>";
                        }
                    ?>
                </select>
            </span>
            <input class="submit" type="submit" value="Load" />
        </form>

Answer 1

除了第一个=中的一个else if之外，我的代码中确实没有问题。

但是，您可以使用三元运算符进行操作，以便为$year和$party指定值。并且可以使用not equal运算符而不是not empty：

$year = $_POST['year_elec']?:'show_all';
$party = $_POST['party']?:'show_all';

if($year == "show_all" && $party == "show_all") {
    $query = "SELECT * FROM mps";
}
else if($year != "show_all" && $party == "show_all"){
    $query = "SELECT * FROM mps WHERE year_elec = $year";
}
else if($year == "show_all" && $party != "show_all") {
    $query = "SELECT * FROM mps WHERE party = $party";
}
else {
    $query = "SELECT * FROM mps WHERE year_elec = $year AND party = $party";
}

另请注意，您对SQL注入攻击持开放态度，理想情况下使用预准备语句和使用者可以选择的值列表（一系列政党和一系列日期）

使用PHP中的多个表单字段提交来处理HTML表数据

1 个答案: