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大家好，抱歉，如果这是一个双重帖子。
我创建了一个数据库，其中包含 4 个值 FLID、DEPID、ARRID、Distance。
我设法使用 Ajax 方法来显示数据库中某一行的数据:
<?php
if( isset($_POST['DEPID']) === true && empty($_POST['DEPID']) ===false){
require'../db/connect.php';
$query = mysql_query("
SELECT `Flights`.`FLID`,`Flights`.`DEPID`,`Flights`.`ARRID`,`Flights`.`Distance`
FROM `Flights`
WHERE `Flights`.`DEPID` ='".mysql_real_escape_string(trim($_POST['DEPID'])) ."'");
echo(mysql_num_rows($query)!== 0) ? mysql_result($query, 0, 'FLID') : 'Departure Airport not found ';
echo(mysql_num_rows($query)!== 0) ? mysql_result($query, 0, 'DEPID') : 'Departure Airport not found ';
echo(mysql_num_rows($query)!== 0) ? mysql_result($query, 0, 'ARRID') : 'Departure Airport not found ';
echo(mysql_num_rows($query)!== 0) ? mysql_result($query, 0, 'Distance') : 'Departure Airport not found ';
}
?>
我的问题是如何让这段代码检索数据库中具有相同 DEPID 的所有行，以及如何将结果添加到表中。
我已经创建了以下代码来尝试解决我的问题并且我已经达到了这一点:
<?php
if( isset($_POST['DEPID']) === true && empty($_POST['DEPID']) ===false){
require'../db/connect.php';
$query = mysql_query("SELECT * FROM Flights WHERE DEPID ='DEPID'");
$result = mysql_query($mysql_connect,$query) or die ("Error");
echo "<table><tr><th>Flight ID</th><th>Departure Airport</th><th>Arrival Airport</th><th>Distance</th></tr>";
while($row = mysql_fetch_array($result)) {
echo "<tr><td>" . $row['FLID'] . "</td><td>" . $row['DEPID'] . "</td><td>" . $row['ARRID'] . "</td><td>" . $row['Distance'] . "</td></tr>";
}
echo "</table>";
}
现在我遇到了代码失败并显示此消息的问题:
Warning: mysql_query() expects parameter 2 to be resource, boolean given in /home/ak118043/public_html/ajax/name.php on line 9 Thanks in advance.