您好我正在尝试学习PHP。我正在使用phpmyadmin和xampp。我想根据以下查询收听:
"$add='SELECT * FROM movielibrary WHERE username="$libraryuser"';
但是mylibrary.php页面没有元素。当我尝试回显$ libraryuser时,它是正确的但是当我写入sql查询时,它不匹配,我的表是空白的。我的用户名是“info”,如果我写的话
"$add='SELECT * FROM movielibrary WHERE username="info"';
这是正确的,但我需要做$ libraryuser。
我该怎么办?
这是我的问题的屏幕截图:
this is when i write to username='info' it is okey
this is when i write to $libraryuser there is no listing
我也试过
'username'=$libraryuser , "username"=$libraryuser , username='$libraryuser , username="$libraryuser
但它不起作用。
<?php
require_once("config.php");
session_start();
$libraryuser=$_SESSION['user'];
echo $libraryuser;
$add='SELECT * FROM movielibrary WHERE username="$libraryuser"';
echo "<table>
<tr>
<td width=2%></td>
<td height=175 width=20%>
<a href=index.php>
<img src=images/Logo.png width=400></a>
</td>
<td width=55%></td>
<td width=12% valign=top >
<td width=12% valign=top >
<div class=row>
<div class='col-md-3 col-sm-3 col-xs-6'> <a href=logout.php class='btn btn-sm animated-button victoria-two'>Logout</a> </div></td>
</tr>
</table>";
echo "<center><font size=10><b>My Library</b></font></center> <table>";
if($result=mysqli_query($conn,$add)) {
echo "</br></br></br></br><table width=%70 border='1' align=center cellspacing='0' cellpadding='6'>
<tr align='center' bgcolor='#999966'>
<td>
<b>Movie Name</b>
</td>
<td>
<b>Director </b>
</td>
<td>
<b>Cast</b>
</td>
<td>
<b>Year</b>
</td>
<td>
<b>Runtime</b>
</td>
<td>
<b>Movie Genre</b>
</td>
</tr>";
while($read=mysqli_fetch_array($result)){
echo "
<tr align='center' bgcolor='#c2c2a3'>
<td>
$read[movie_name]
</td>
<td>
$read[director]
</td>
<td>
$read[movie_cast]
</td>
<td>
$read[year]
</td>
<td>
$read[runtime]
</td>
<td>
$read[movie_genre]
</td>";
}
}
echo "</tr>
</table>";
?>
答案 0 :(得分:-2)
您需要替换查询
'SELECT * FROM movielibrary WHERE username="$libraryuser"';
与
"SELECT * FROM movielibrary WHERE username='$libraryuser'";
然后它会起作用。