我想从数据库中获取信息，并将其作为选项放入selectbox中。我试着去做，但我做不出来。我的错误是什么？（db可以连接，我只删除服务器名称）我不知道如何将数据库行作为选项放在selectbox中。因此，我认为我的代码有问题。
p.php
<?php
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT * FROM test" ;
$result = mysqli_query($conn, $sql) or die("Query: ($sql) [problem]");
$row = mysqli_fetch_assoc($result);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_row($result)) {
display("<option value=$row[seat_id]>",$row[seatnumber]."'n");
}
display ("</select>", "'n");
} else {
echo "0 results";
}
mysqli_close($conn);
function display($tag , $value) {
echo $tag . $value ;
}
?>
p.html
<html>
<head>
<meta charset="utf-8">
<link href="" rel="stylesheet" type="text/css" />
</head>
<script type="text/javascript">
function transfer(){
var pix = document.getElementById('pix').value;
document.abc.test.value =pix;
}
</script>
<script
src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.2/jquery.min.js">
</script>
<script>
function ajaxWay() {
// syntax: $.post(URL,data,callback);
$.get("p.php", function(dataFromtheServer) {
$("#result").html(dataFromtheServer);
});
}
</script>
<body>
<div id="a" style="text-align:center;">
<form name="abc" method="get" action="p.php">
<select id='pix' onchange='ajaxWay()'>
<input type="button" value="click" onclick="transfer();">
<input type="text" name="test" id="test">
</form>
</div>
</body>
</html>
如果您的主要问题是无法将选项嵌入HTML中，请尝试以下操作：
<html>
<head>
<meta charset="utf-8">
<link href="" rel="stylesheet" type="text/css" />
</head>
<script type="text/javascript">
function transfer(){
var pix = document.getElementById('pix').value;
document.abc.test.value =pix;
}
</script>
<script
src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.2/jquery.min.js">
</script>
<script>
function ajaxWay() {
// syntax: $.post(URL,data,callback);
$.get("prefinal.php", function(dataFromtheServer) {
$("#result").html(dataFromtheServer);
});
}
</script>
<body>
<div id="a" style="text-align:center;">
<form name="abc" method="get" action="p.php">
<?php
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT * FROM test" ;
$result = mysqli_query($conn, $sql) or die("Query: ($sql) [problem]");
?>
<select id='pix' onchange='ajaxWay()'>
<?php
$row = mysqli_fetch_assoc($result);
if (mysqli_num_rows($result) > 0)
{
while($row = mysqli_fetch_row($result))
{
echo '<option value="' . $row[0] . ">' . $row[0] . '</option>';
}
}
mysqli_close($conn);
?>
</select>
<input type="button" value="click" onclick="transfer();">
<input type="text" name="test" id="test">
</form>
</div>
</body>
</html>
请注意，在上面的文章中，我使用来自p.php的代码来生成select选项的实际内容。表单不应该提交给p.php，而应该提交给其他一些脚本，这些脚本将处理表单并执行所需的操作。如果我更了解你正在努力实现的目标，我会帮助你的！