2015-09-23 121 views
0

我正在一個站點上工作,我試圖連接到SQL數據庫並在<div>內發佈SQL字段。 <div>位於while循環內。 我想我已成功連接到數據庫。但是,我無法在頭中顯示ID字段。我是新手到PHP & SQL,無法弄清楚。這裏是我下面的代碼:無法使用while循環插入mysql字段到div使用while循環

<?php 

    $db_host = "localhost"; 
    $db_username = "user192"; 
    $db_pass = "xxxx"; 
    $db_name = "sound-library"; 

    @mysql_connect("$db_host","$db_username","$db_pass") or die ("couldnt connect to sql"); 
    @mysql_select_db ("$db_name") or die ("cant find database"); 

    $query = "select * from sound-library"; 
    $queryResult=mysql_query($query); 
    $numrows=mysql_num_rows($queryResult); 

?> 

    while($row = mysql_fetch_assoc($queryResult)) { 
     <div class="audio-module-parent"> 
      <div class="audio-module-header"> 
<?php 
       <h1> <?php echo $row['id']?> </h1> 
      </div> 
      <div class="audio-module-preview"></div> 
      <div class="audio-module-download">Download</div> 
      <div class="audio-module-tutorial">Watch Tutorial</div> 
     </div> 
?> 
+0

您的'while'循環在您的''標籤之外? – henrikstroem

回答

1

你把代碼之外這裏<?php ?>

?> 
while($row = mysql_fetch_assoc($queryResult)) { 

,然後把HTML裏面<?php ?>如這裏

<?php 
<h1> <?php echo $row['id']?> </h1> 
</div> 

應該

<?php 
error_reporting(E_ALL); //Enable Error Reporting 
ini_set('display_errors',1); //change value to 0 to disable the error views 
$db_host = "localhost"; 
$db_username = "user192"; 
$db_pass = "xxxx"; 
$db_name = "sound-library"; 
@mysql_connect("$db_host","$db_username","$db_pass") or die ("couldnt connect to sql"); 
@mysql_select_db ("$db_name") or die ("cant find database"); 
$query = "select * from sound-library"; 
$queryResult=mysql_query($query) or die(mysql_error()); 
$numrows=mysql_num_rows($queryResult); 
while($row = mysql_fetch_assoc($queryResult)) { 
?> 
<div class="audio-module-parent"> 
    <div class="audio-module-header"> 
     <h1> <?php echo $row['id']?> </h1> 
    </div> 
    <div class="audio-module-preview"></div> 
    <div class="audio-module-download">Download</div> 
    <div class="audio-module-tutorial">Watch Tutorial</div> 
</div> 
<?php } ?> 
+0

我已經嘗試過你的建議,但它仍然沒有迴應我想要的東西:( –

+0

進行此更改'$ queryResult = mysql_query($ query)或死(mysql_error());' – Shehary

+0

檢查更新的答案,啓用頁面 – Shehary

1
- <h1> <?echo $row['id'];?> </h1> missing semi colon (;) here in this line. 
- Missing } end of while loop 

使用此連接。在'xxxx'部分寫上你的密碼&檢查它。你的mysql_connect & mysql_select_db

$ con = mysql_connect(「localhost」,「user192」,「xxxx」)或die(「couldnt connect to sql」); $ db = mysql_select_db(「sound-library」,$ con)或die(「cant find database」);

<? 
    while($row = mysql_fetch_assoc($queryResult)) 
    {?> 
    <div class="audio-module-parent"> 
     <div class="audio-module-header"> 
      <h1> <?echo $row['id'];?> </h1> 
     </div> 
     <div class="audio-module-preview"> 

     </div> 
     <div class="audio-module-download">Download</div> 
     <div class="audio-module-tutorial">Watch Tutorial</div> 
    </div> 
    <?}?> 
+0

仍然沒有得到我想要的結果。也許問題在於我寫的while循環。我試圖達到的目的是爲了創建HTML爲「ID」的任何實例,不知道我是否使用了正確的邏輯或語法。 –

+0

你確定嗎?您的$ db_username是user192。用$ db_username ='root'&$ db_password作爲「」(意思是空白,沒有密碼)來檢查它@BenLiger –

+0

我是,但有沒有辦法來檢查查詢可能失敗? –