我是新手，我有一个问题，无法调用第二个选择的数据，下面是我的数据库结构和代码的表和列。请帮帮我 TQVM!
表1名称 - 电影
id、电影 ID、电影名称
表2名称-电影 field
id、movieid、venueid、mvenue
index.php
<html>
<head>
<script type="text/javascript" src="js/jquery-1.7.1.min.js"></script>
<script type="text/javascript" src="js/jquery-ui-1.8.17.custom.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$("#movie").change(function(){
var movie=$("#movie").val();
$.ajax({
type:"post",
url:"venue.php",
data:"movie="+movie,
success:function(data){
$("#venue").html(data);
}
});
});
});
</script>
</head>
<body>
Movie :
<select name="movie" id="movie">
<option>-select your movie-</option>
<?php
include "db.php";
$result=mysql_query("SELECT movieid, moviename from movie order by moviename");
while($movie=mysql_fetch_array($result)){
echo "<option value=$movie[moviename]>$movie[moviename]</option>";
} ?>
</select>
Venue :
<select name="venue" id="venue">
<option>-select your venue-</option>
</select>
</body>
</html>
venue.php(我希望在第二个选择框中显示“mvenue”，它取决于第一个选择 ID，即“movieid”。)
<?php
include "db.php";
$movie=$_POST["movie"];
$result=mysql_query("select movieid,mvenue FROM movievenue where mvenue='$movie' ");
while($venue=mysql_fetch_array($result)){
echo"<option value=$venue[mvenue]>$venue[mvenue]</option>";
}
?>
最佳答案
<?php
include "db.php";
$movie=$_POST["movie"];
$result=mysql_query("select movieid,mvenue FROM movievenue where mvenue='{$movie['movieid']}' ");
while($venue=mysql_fetch_array($result)){
echo"<option value=$venue[mvenue]>$venue[mvenue]</option>";
}
?>
此外，在您的第一个文件中，您可能希望选择值是电影的 id
echo "<option value=$movie[movieid]>$movie[moviename]</option>";
关于php - 依赖选择框 php mysql，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/25608265/