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这是我第一次尝试使用PHP与数据库(MySQLi)。我终于得到了数据库连接(至少,我没有得到任何错误)。无法显示MySQLi SELECT查询的结果
现在我试图获得选择查询的结果并将其显示在我的表单中,但我无法获取任何要显示的内容。
我哪里错了?
这是为了从数据库中检索查询结果的代码:
/*Connect To DB*/
$conn = mysqli_connect($host, $user, $pwd)
or die("Could not connect: " . mysql_error()); //connect to server
mysqli_select_db($conn, $database)
or die("Error: Could not connect to the database: " . mysql_error());
/*Check for Connection*/
if(mysqli_connect_errno()){
/*Display Error message if fails*/
echo 'Error, could not connect to the database please try again later.';
exit();
}
/*Query for states*/
$query = "SELECT StateAbbreviation, StateName FROM USState ORDER BY StateName";
$result = mysqli_query($conn, $query);
$num_results = mysqli_num_rows($result);
?>
这是结果应显示的形式:
<form id="StateSelector" action="" method="post">
<select size="1" name="states" id="states">
<option value "">--Select State--</option>
<!--Loops through the states-->
<?
/*Loop through through each stat and display as an option as a drop-down field */
for($i=0; $i<$num_results; $i++) {
$row = mysqli_fetch_assoc($result);
echo 'option value="' .$row['StateAbbreviation'] . '">' . $row['StateName'] . '</option>' . "\n";
}
?>
</select>
Zip:
<input type="text" name="zip" size="5" /></p>
</form>
<p>Your email address:<br/>
<input type="text" name="email" size="20" /></p>
<p>Please let us know what you think:<br/>
<textarea name="feedback" rows="12" cols="40" wrap="virtual" /></textarea></p>
<p><input type="submit" value="Send feedback" /></p>
</form>