我一直在尝试解决这个问题,但到目前为止我还无法让它发挥作用。不会引发任何错误,页面会在提交时刷新。我很茫然,但我不完全是专家,对此还很陌生。
这是代码(为了发布而简化):
<?php if (!isset($_POST['submit'])) {
echo "<!-- Form starts here -->
<form id=\"billing\" action=\"\" method=\"post\">
<!-- Name -->
<div class=\"control-group\">
<label class=\"control-label\"><b>Name</b></label>
<div class=\"controls\">
<input type=\"text\" id=\"name\" name=\"name\" placeholder=\"your name\" class=\"input-large\">
</div>
</div>
<!-- Zip -->
<div class=\"control-group\">
<label class=\"control-label\"><b>Zip Code</b></label>
<div class=\"controls\">
<input type=\"text\" id=\"billingzip\" name=\"billingzip\" placeholder=\"5 digit zip\" class=\"input-large\">
</div>
</div>
<!-- Submit -->
<div class=\"control-group\">
<div class=\"controls\">
<button class=\"button save small_green_button\" type=\"submit\">
Save
</button>
</div>
</div>
</form>";
}
else
{
$host="localhost";
$user_name="user";
$pwd="password";
$database_name="database";
$db=mysql_connect($host, $user_name, $pwd) or die(mysql_error());
$dbsel=mysql_select_db($database_name, $db);
if (mysql_error() > "") print mysql_error() . "<br>";
if (mysql_error() > "") print mysql_error() . "<br>";
$account_id = users::getAttr('Account', 'account_id');
$zip = mysql_real_escape_string($_POST['billingzip']);
$name = mysql_real_escape_string($_POST['name']);
$sql = "INSERT INTO `billing`
SET `account_id` = '{$account_id}',
`zip` = '{$billingzip}',
`name` = '{$name}',
`updated_at` = NOW()";
$result = mysql_query($sql, $dbsel)
or die(mysql_error().$sql);
mysql_close($db);
}
?>
最佳答案
首先我看到的是
$result = mysql_query($sql, $dbsel) or die(mysql_error().$sql);
我认为你应该写:
$result = mysql_query($sql, $db) or die(mysql_error().$sql);
关于php - 表单和PHP函数将表单数据保存到mysql在同一个文件中,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20394680/