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
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>