function getDepartmentAndCondition($dep, $userid, $cond) {
$result = mysql_query("SELECT * FROM department WHERE ID='$dep'");
while($row = mysql_fetch_array($result))
{
$DepConInfo['Department'] = $row['Department'];
}
$userName = mysql_query("SELECT * FROM users WHERE FacebookID = '$userid'") or die ("<hr>error in SQL query: " . mysql_error() . "<hr>");
while($row = mysql_fetch_array($username)) {
$DepConInfo['Name'] = $row['name'];
}
$result2 = mysql_query("SELECT * FROM condition WHERE ID= '$cond' ")
or die("<hr>error in SQL query: " . mysql_error() . "<hr>");
while($row2 = mysql_fetch_array($result2))
{
$DepConInfo['Condition'] = $row2['Condition'];
}
return $DepConInfo;
}
$dep
、$userid
、および$cond
はすべてint
秒です。最初の1 $DepConInfo['Department']
は、右の文字列を返しているが、他の2つは[OK]を、私は機能 PHP、MySQLのクエリの異なるテーブルは
function getCondition($cond) {
$query = "SELECT * FROM condition WHERE ID = '$cond' ";
$sql = mysql_query($query);
if (!$sql) {
$message = 'Invalid query: ' . mysql_error() . "\n";
$message .= 'Whole query: ' . $query;
die($message);
}
while($row2 = mysql_fetch_array($sql))
{
$condition = $row2['Name'];
}
return $condition;
}
を書き直し
Warning:
mysql_fetch_array()
: supplied argument is not a valid MySQL result resource in ...
エラーで失敗が、私はまだエラーを取得しています:
Invalid query: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'condition WHERE id = '1'' at line 1 Whole query: SELECT * FROM condition WHERE ID = '1'
表 "条件"には、2つの列 "ID"と "名前"があります。
データベースのテーブルに条件用の行があることを確認してください。 – footy