如何根据选择在下拉列表旁边显示信息?

问题描述 投票:0回答:1

我是新手,我想从数据库中打印汽车品牌,具体取决于我已经放入数据库中的下拉列表所选择的ID。

我使用的是带有HTML的php,我知道这很简单,也许是if语句,但我无法得到想要的东西。我的意思是,我可以带上汽车品牌,但选择不会改变

<?php
    $database = new mysqli('localhost', 'root', 'password', 'viajes') 
    or die ('Unable to connect to the db');
        $result = $database->query("SELECT id,Placas,Marca FROM vehiculos GROUP BY Placas ASC");
        echo "<select name='id'>";
        while ($row = $result->fetch_assoc()) {
                      unset($id, $name);
                      $id = $row['id'];
                      $Placas = $row['Placas'];
                      $Marca = $row['Marca']; 
                      echo '<option value="'.$id.'">'.$Placas.'</option>';
    }
        echo "</select>";

//Marca means brand for those who doesn't know.

        if($Placas){
            echo "$Marca";
        };
    ?>

假设“ Placa”是下拉列表,我希望在我的表中有一个类似这样的输出(我正在创建列表,因为我无法在此处创建下拉列表):

Placa马卡(品牌)ABC-123雪佛兰DEF-456丰田GHI-789现代但是我得到以下信息:Placa马尔卡ABC-123丰田DEF-456丰田GHI-789丰田我只想匹配值Placa-Marca

php html mysql
1个回答
0
投票

如果可以使用'jquery',请这样做。

<?php
    $database = new mysqli('localhost', 'root', 'password', 'viajes') 
    or die ('Unable to connect to the db');
        $result = $database->query("SELECT id,Placas,Marca FROM vehiculos GROUP BY Placas ASC");
        echo "<select name='id'>";
        while ($row = $result->fetch_assoc()) {
                      unset($id, $name);
                      $id = $row['id'];
                      $Placas = $row['Placas'];
                      $Marca = $row['Marca']; 
                      echo '<option value="'.$id.'" data-marca="'.$Marca.'">'.$Placas.'</option>';
    }
        echo "</select>";
?>

<div id="marca"></div>

<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.4.1/jquery.min.js"></script>

<script>
jQuery(function($){
    $('select[name="id"]').change(function(){
        $("#marca").text($('select[name="id"] option:selected').attr("data-marca"));
    });
});
</script>
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