2013-12-12 48 views
1

在製作網頁方面我很新穎。但我正在做一個表格插入到我的數據庫的主頁。這沒問題,我的問題是,我想顯示最後一行的特定列。而且,我已經這麼遠的代碼是這樣的:在html中顯示數據庫中的特定值

<html> 
<body> 

<form action="insert.php" method="post"> 
Publiceringsdag (OBS! En dag tidigare an foregaende):<br> 
<?php 
$con=mysqli_connect("localhost","rss","Habb0","kalender"); 
if (mysqli_connect_errno()) 
{ 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
} 
$lastPub = mysql_query("SELECT DISTINCT pub FROM event ORDER BY `id` DESC LIMIT 1") 
or die(mysql_error()); 
echo $lastPub 
?> 
<br> 
<input type="text" name="pub"><br> 
<input type="submit"> 
</form> 

</body> 
</html> 
+2

你應該真的考慮使用mysqli –

回答

0

必須先獲取結果:

$lastPub = mysql_query("SELECT DISTINCT pub FROM event ORDER BY `id` DESC LIMIT 1") 
or die(mysql_error()); 
$result = mysql_fetch_array($lastPub); 
echo $result['pub']; 
3

其實,這是不使用過時mysql_功能一個非常好的主意。改用PDOMysqli

同時,在當前的實現,你只需要在查詢執行後,獲取你的數據:

$con = mysql_connect("localhost", "rss", "Habb0", "kalender"); 

if (mysql_connect_errno())  
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 

$lastPub = mysql_query("SELECT DISTINCT pub FROM event ORDER BY `id` DESC LIMIT 1") 
    or die(mysql_error()); 

if($row = mysql_fetch_assoc($lastPub))) 
    $result = $lastPub['pub']; 

現在的結果應該是在你的$result變量。

編輯:我只注意到在你的代碼中使用mysqli_connectmysqli_connect_errnomysql_querymysql_error在同一時間。但它們屬於不同的PHP擴展。

0

試試這個。

<html> 
<body> 

<form action="insert.php" method="post"> 
Publiceringsdag (OBS! En dag tidigare an foregaende):<br> 
<?php 
$con=mysql_connect("localhost","rss","Habb0") or die("Failed to connect to MySQL: " . mysql_error()); 
$db=mysql_select_db("kalender",$con) or die("Failed to connect to MySQL: " . mysql_error()); 
$result = mysql_query("SELECT DISTINCT pub FROM event ORDER BY `id` DESC LIMIT 1"); 
$data = mysql_fetch_array($result); 

echo $data['pub']; 
?> 
<br> 
<input type="text" name="pub"><br> 
<input type="submit"> 
</form> 

</body> 
</html> 
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