我想改變一個PHP表的tr的bgcolor,它從循環中獲取其數據,我認爲。我是一個體面的noob在所有的PHP HTML的東西。一直在嘗試這一整天,無法通過查找數據來改變bgcolor。PHP的MySQL查詢循環,如果語句
從mysql數據庫獲取所有數據的循環工作正常,現在我不知道如何進一步輸出更改bgcolor如果超過30天,並具有OUT的狀態。
這裏是我目前有和嘗試的代碼。任何幫助,將不勝感激。
$con=mysqli_connect($host, $db_user, $db_pass, $db);
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM data");
$sql77 = "SELECT * FROM data WHERE dt_in < date_sub(curdate(), INTERVAL 30 DAY) AND status != OUT";
$result77 = mysql_query($con,$sql77);
echo "<table border='1'>
<tr bgcolor='lightgrey'>
<th>Signed in by</th>
<th>Reference Number</th>
<th>Asset Number</th>
<th>Make Model</th>
<th>Operating System</th>
<th>Office</th>
<th>Profile</th>
<th>Extra Apps</th>
<th>Time IN</th>
<th>Status</th>
<th>Time OUT</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
if (!$result77) {
echo "<tr>";
} else {
echo "<tr bgcolor='red'>";
}
echo "<td>" . $row['who'] . "</td>";
echo "<td>" . $row['ref'] . "</td>";
echo "<td>" . $row['asset'] . "</td>";
echo "<td>" . $row['make_model'] . "</td>";
echo "<td>" . $row['os'] . "</td>";
echo "<td>" . $row['office'] . "</td>";
echo "<td>" . $row['swp'] . "</td>";
echo "<td>" . $row['ea'] . "</td>";
echo "<td>" . $row['dt_in'] . "</td>";
echo "<td>" . $row['status'] . "</td>";
echo "<td>" . $row['dt_out'] . "</td>";
echo "</tr>";
}
echo "</table>";
echo $result77;
mysqli_close($con);
?>
這是終於解決了我的問題。
while($row = mysqli_fetch_array($result))
{
if(time() - strtotime($row['dt_in']) > 2592000 && $row['status'] != 'OUT') //2592000 sec == 30 days
{
echo "<tr class='MyOut'>";
} else {
echo "<tr>";
}
echo "<td>" . $row['who'] . "</td>";
echo "<td>" . $row['ref'] . "</td>";
所以...到底是'$ result77'定義在哪裏?它不在代碼中的任何地方,所以它會'如果(!$ result77)'永遠是真的。 –
不是$ result77這裏定義的:$ sql77 =「選擇*從數據WHERE dt_in
Fishy
html5中不支持bgcolor屬性,並且已在html 4.01中棄用。改用css。 – elitechief21