[英]PHP MYSQL Get data from database and echo message if no data retrieved
我試圖編寫一個代碼,以顯示數據庫中“ clanwars”設置為1(int)的所有名稱,如果所有名稱均為0,則回顯一條消息,例如:“尚未注冊!”
這是我的一些代碼:
$result = mysqli_query($con,"SELECT * FROM users WHERE clanwars = 1 ORDER BY mantra");
while($row = mysqli_fetch_array($result))
{
if ($row['clanwars'] != '0') {
echo $row['mantra']."<br>";
} else {
echo 'Noone has signed up yet!';
}
}
mysqli_close($con);
首先,在示例row['clanwars']
永遠不會等於0
因為您已經在查詢中指定了WHERE clanwars = 1
,因此MySQL將僅返回那些clanwars = 1的查詢。 如果我理解得很好,則需要執行以下操作:
<?
$result = mysqli_query($con,"SELECT * FROM users WHERE clanwars = 1 ORDER BY mantra");
if (mysqli_num_rows($result)==0) echo 'Noone has signed up yet';
else {
while ($row = mysqli_fetch_array($result)) {
//do what you need
}
}
?>
因此,基本上,您檢索數據庫中將CLANWARS設置為1的所有人。 如果有記錄,請對其進行處理,如果沒有記錄,則表示沒有人簽名。
這是您需要的嗎?
嘗試這個:
$result = mysqli_query($con,"SELECT * FROM users WHERE clanwars = 1 ORDER BY mantra");
if (mysqli_num_rows($result) == 0)
{
echo 'Noone has signed up yet!';
}
else
{
while ($row = mysqli_fetch_array($result))
{
echo $row['mantra']."<br>";
}
}
mysqli_close($con);
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