我正在尝试制作一个页面,您可以在其中选择一个下拉选择框中的值。做出选择后,它应该使用所选值更新数据库。我使用了两个页面,一个包含html/javascript,另一个使用php。目前,当我选择某些东西时,什么都不会发生。
我做错了什么?
test.php
<html>
<head>
<script type="text/javascript" src="js/jquery.js"></script>
<script>
function updateDb() {
$.post("buh.php", $("#form").serialize());
}
</script>
</head>
<body>
<form id="form">
<?php
include 'Includes/database_connection.php';
$sql = "select * FROM sims ORDER BY phonenr asc" ;
$result = mysql_query($sql,$con);
echo "<select id='select' name='select' onChange='updateDb()'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['phonenr'] . "'>" . $row['phonenr'] . "</option>";
}
echo "</select>";
?>
</form>
</body>
和buh.php
<?php
include 'Includes/database_connection.php';
$sql = "select * FROM sims ORDER WHERE phonenr='".mysql_escape_string($_POST["select"])."'" ;
$result = mysql_query($sql,$con);
while ($row = mysql_fetch_array($result)) {
$id = $row['id'];
mysql_query("UPDATE pairings SET sim_id='$id' WHERE unit_id='1'")
or die(mysql_error());
}?>
发布于 2013-05-23 19:42:19
发布于 2013-05-23 19:42:06
select * FROM sims ORDER ..You在buh.php中错误地在此处添加了订单
$sql = "select * FROM sims ORDER WHERE phonenr='".mysql_escape_string($_POST["select"])."'" ;发布于 2013-05-23 19:45:04
$sql = "select * FROM sims ORDER WHERE phonenr='".mysql_escape_string($_POST["select"])."'" ;
删除订单。
https://stackoverflow.com/questions/16712929
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