2011-03-17 41 views
0

我試圖在phpMyAdmin中顯示數據庫中的所有記錄。由於最近安裝了統一服務器軟件,該數據庫位於我的計算機上,該計算機當前是服務器。我是新建構的PHP和HTML代碼。我使用了這個鏈接http://www.siteground.com/tutorials/php-mysql/display_table_data.htm的最後一個競爭documenet代碼,這是我的網頁上的輸出。在html網頁上顯示數據庫內容

<HTML> 

<HEAD> 

<TITLE> LOG INFORMATION FOR ALL customers </TITLE> 

</HEAD> 


<BODY> 

<H1> LOG INFORMATION FOR ENTIRE DATABASE </H1> 

<?php 


$database="mobile_app_tracking_log"; 

mysql_connect(localhost); 
@mysql_select_db($database) or die("Unable to select database"); 
$query="SELECT * FROM tablename"; 
$result=mysql_query($query); 



mysql_close(); 
?> 

<table border="2" cellspacing="7" cellpadding="7"> 
<tr> 
<th><font face="Arial, Helvetica, sans-serif">Index</font></th> 
<th><font face="Arial, Helvetica, sans-serif">Identification Number</font></th> 
<th><font face="Arial, Helvetica, sans-serif">Date</font></th> 
<th><font face="Arial, Helvetica, sans-serif">Time</font></th> 
<th><font face="Arial, Helvetica, sans-serif">Application</font></th> 
<th><font face="Arial, Helvetica, sans-serif">Usage</font></th> 
<th><font face="Arial, Helvetica, sans-serif">Latitude</font></th> 
<th><font face="Arial, Helvetica, sans-serif">Longitude</font></th> 
<th><font face="Arial, Helvetica, sans-serif">Address</font></th> 
</tr> 

<?php 
$i=0; 
while ($i < $num) { 


$f1=mysql_result($result,$i,"Index"); 
$f2-name=mysql_result($result,$i,"Identification Number"); 
$f3-name=mysql_result($result,$i,"Date"); 
$f4-name=mysql_result($result,$i,"Time"); 
$f5-name=mysql_result($result,$i,"Application"); 
$f6-name=mysql_result($result,$i,"Usage"); 
$f7-name=mysql_result($result,$i,"Latitude"); 
$f8-name=mysql_result($result,$i,"Longitude"); 
$f9-name=mysql_result($result,$i,"Address"); 
?> 

<tr> 
<td><font face="Arial, Helvetica, sans-serif"><?php echo $f1; ?></font></td> 
<td><font face="Arial, Helvetica, sans-serif"><?php echo $f2; ?></font></td> 
<td><font face="Arial, Helvetica, sans-serif"><?php echo $f3; ?></font></td> 
<td><font face="Arial, Helvetica, sans-serif"><?php echo $f4; ?></font></td> 
<td><font face="Arial, Helvetica, sans-serif"><?php echo $f5; ?></font></td> 
<td><font face="Arial, Helvetica, sans-serif"><?php echo $f6; ?></font></td> 
<td><font face="Arial, Helvetica, sans-serif"><?php echo $f7; ?></font></td> 
<td><font face="Arial, Helvetica, sans-serif"><?php echo $f8; ?></font></td> 
<td><font face="Arial, Helvetica, sans-serif"><?php echo $f9; ?></font></td> 
</tr> 


<?php 

$i++; 
} 

?> 

有人可以告訴我爲什麼數據庫中的記錄沒有顯示嗎?任何幫助將非常感激。

回答

0

它看起來像要保存的變量$f2-name但是你echo'ing $f2,同樣有$f3$f4

如果你想讓你的命名模式,你必須要這麼做:

echo $f2-name; 
echo $f3-name; 
// etc 
+0

謝謝,但我認爲我爲每個像這樣的所有行做了這樣的事情:​​<?php echo $ f1; ?>是不是代碼段<?php echo $ f1; ?>應該打印該值。我不太明白。 – 2011-03-17 03:38:31

+0

根據您的更改編輯帖子。 – 2011-03-17 03:41:04

+0

感謝邁克,我做了你的建議,但我仍然得到空白表。還有什麼建議?如何嘗試連接到統一網絡服務器上的數據庫有問題嗎? – 2011-03-17 03:53:08

0

與您提供的鏈接比較,查看您的代碼。你缺少$ num變量初始化

這一個: $ num = mysql_numrows($ result);

所以我想你的循環不會迭代,增加這可能會解決你的問題。在代碼中可能會丟失更多關於您所介紹的代碼的更多內容。

+0

感謝Ace,我很感激。我會試試看,並告訴你它是如何發生的。 – 2011-03-17 05:36:43

0

檢查這些代碼。這可能會幫助你..

<HTML> 
<HEAD> 
<TITLE> LOG INFORMATION FOR ALL customers </TITLE> 
</HEAD> 
<BODY> 
<H1> LOG INFORMATION FOR ENTIRE DATABASE </H1> 

<?php 
$user = 'root'; 
$password = ''; 
$database="test"; 

mysql_connect(localhost,$user, $password); 
@mysql_select_db($database) or die("Unable to select database"); 
echo $query = "SELECT * FROM customer"; 
$result = mysql_query($query); 

mysql_close(); 
?> 

<table border="2" cellspacing="7" cellpadding="7"> 
<tr> 
<th><font face="Arial, Helvetica, sans-serif">Index</font></th> 
<th><font face="Arial, Helvetica, sans-serif">Customer Name</font></th> 
<th><font face="Arial, Helvetica, sans-serif">Sex</font></th> 
</tr> 

<?php 

while ($row = mysql_fetch_assoc($result)) { 

?> 

<tr> 
<td><font face="Arial, Helvetica, sans-serif"><?php echo $row[id]; ?></font></td> 
<td><font face="Arial, Helvetica, sans-serif"><?php echo $row[customer_name]; ?></font></td> 
<td><font face="Arial, Helvetica, sans-serif"><?php echo $row[sex]; ?></font></td> 
</tr> 


<?php 
} 
?> 
0

我測試過此代碼,我發現有必要把引號localhostmysql_connect

mysql_connect('localhost',$user, $password); 

以及$row結果行以避免變得"Notice: Use of undefined constant" errors.

<td><font face="Arial, Helvetica, sans-serif"><?php echo $row['id']; ?></font></td>