我正在一个网站上工作,我正在尝试连接到SQL数据库并在<div>
内发布一个SQL字段。 <div>
位于while
循环内。
我想我已成功连接到数据库。但是,我无法在标头中显示 id 字段。我是PHP&amp;的新手SQL并不能解决它。以下是我的代码:
<?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 $row['id']?> </h1>
</div>
<div class="audio-module-preview"></div>
<div class="audio-module-download">Download</div>
<div class="audio-module-tutorial">Watch Tutorial</div>
</div>
?>
答案 0 :(得分:1)
您将代码置于<?php ?>
之外,例如
?>
while($row = mysql_fetch_assoc($queryResult)) {
然后将HTML放在<?php ?>
里面,例如
<?php
<h1> <?php echo $row['id']?> </h1>
</div>
应该是
<?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 } ?>
答案 1 :(得分:1)
- <h1> <?echo $row['id'];?> </h1> missing semi colon (;) here in this line.
- Missing } end of while loop
使用此连接。在“xxxx”部分写一个密码&amp;核实。评论你的mysql_connect&amp; mysql_select_db
$ con = mysql_connect(“localhost”,“user192”,“xxxx”)或死亡(“无法连接到sql”); $ db = mysql_select_db(“sound-library”,$ con)或die(“无法找到数据库”);
<?
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>
<?}?>