2011-08-10 13 views
1

我試圖顯示數據從MySQL數據庫使用PHP這樣試圖展示數據從MySQL使用PHP和HTML

  $query = "SELECT CONCAT(usrFirstname,'',usrSurname) As FullName,usrNickname AS Nickname,"; 
      $query.= "usrEmail As EmailAddress,usrGender AS Gender,DATE(usrDOB) As DOB,usrBelt AS BeltId,ggName As Groupname "; 
     $query.= "FROM user LEFT JOIN gyg ON user.usrIndex = gyg.usrIndex;"; 
     $result = mysql_query($query); 
echo mysql_error(); 
if($result) 
{ 
$row= mysql_fetch_array($result); 
if($row) 
{  
    $fullname = $row['FullName']; 
    $nickname = $row['Nickname']; 
    $emialid = $row['EmailAddress']; 
    $gender = $row['Gender']; 
    $Dateofbirth = $row['DOB']; 
    $belt = $row['BeltId']; 
    $group = $row['Groupname'];  
} 

    } 

來爲表格式和HTML代碼來是這樣的

<table height= "600" width="800"> 
    <tr style="vertical-align: top; text-align:top display:inline-block"> 
    <thead> 
      <td>FUll name</td><td> Nickname<?php echo $nickname ?></td><td>Email Address<?php echo $emialid ?></td><td>Gender<?php echo $gender ?></td><td>DOB <?php echo $Dateofbirth ?><td>BELT ID <?php echo $belt ?></td><td>GROUP <?php echo $group ?></td> 
     </thead> 
    </tr> 
</table> 

我想說明這樣

fullname nickname emailid gender dob beltid group 
     xxxxx  xxxxx  xxxxx  xxx xxx xxx xxxx 
     xxxxx  xxxxx  xxxxx  xxx xxx xxx xxxx 

但它顯示這樣

  fullname xxxxx nickname xxxxx emailid xxxxx gender xxxxx dobxxxxx beltid xxxxx groupxxxxx 

和我有四行未來形式的數據庫,但它是隻顯示一行

如何我解決這個問題,可以在這個任何幫助......

修改後的代碼:它顯示像這樣

fullname nickname emailid gender dob beltid group 
     xxxxx  xxxxx  xxxxx  xxx xxx xxx xxxx 
     xxxxx  xxxxx  xxxxx  xxx xxx xxx xxxx 

我必須做什麼請幫助

回答

2
<?php 
$rows = array(); 
if($result) 
{ 
     while($row=mysql_fetch_assoc($result)){ 
      $rows[] = $row; 
     } 
} 

<table height= "600" width="800"> 
<tr style="vertical-align: top; text-align:top display:inline-block"> 
<thead> 
     <tr> 
     <td>FUll name</td> 
     <td> Nickname</td> 
     <td>Email Address</td> 
     <td>Gender</td> 
     <td>DOB</td> 
     <td>BELT ID</td> 
     <td>GROUP</td> 
     </tr> 
    </thead> 
    <?php foreach ($rows as $row){?> 
    <tr> 
     <td><?php echo $row['fullname']?></td> 
     <td><?php echo $row['nickname']?></td> 
     <!--other fields here--> 
     <td><?php echo $row['GROUP']?></td> 
    </tr> 
    <?php }?> 
</tr> 

+0

HI JOHN了這樣警告的錯誤:mysql_fetch_assoc()預計至少有一個參數,0 – user682417

+0

給出哎呀,對不起,愚蠢的錯誤... :)編輯... – J0HN

+0

喜約翰·請看到我的修改代碼......任何想法都會非常有幫助 – user682417