從javascript生成URL的問題

我試圖通過從數據庫中獲取值來生成多個鏈接。我可以連接到沒有任何問題的數據庫，並且我可以毫無問題地提取變量，但是當我嘗試生成我的html代碼時，一切都變成了亂碼。從javascript生成URL的問題

<script type="text/javascript"> 
//Generate carousel 
var mymovies = "<div id='tumbnailtest' class = 'carousel'><ul><li><a href='watchmovie.php?path="; 
for (var i = 1; i < 10; i++){ 
    <?php 
    //Get movie List 
    $result = mysql_query("SELECT Title, Filepath, Posterpath FROM movies WHERE ID = $count"); 
    while ($row = mysql_fetch_array($result)) { 
     //for generating url 
     $mpath = $row["Filepath"]; 
     $mmname = $row["Title"]; 
     $ppath = $row["Posterpath"]; 
     $count = $count + 1; 
    } 
    ?> 
    mymovies += "<li><a href=watchmovie.php?path=<?php echo $mpath;?>" 
    mymovies += "&mname=<?php echo $mmname;?><img src=<?php echo $ppath;?> width='200' height='300' border='0' style='padding: 0 0px 0 0px;' title='<?php echo $mmname; ?>'></a></li>"; 
    } 
    mymovies += "</ul></div>"; 
document.write(mymovies); 

</script>

路段端部這樣看

<a href="watchmovie.php?path=&lt;li&gt;&lt;a href=watchmovie.php?path=Movies/iRobot.mp4&amp;mname=iRobot&lt;img src=images/iRobot.jpeg width=" 200'="" height="300" border="0" style="padding: 0 0px 0 0px;" title="iRobot"></a>

當應該結束這樣看

<li><a href='watchmovie.php?path=<?php mname ?>&mname=<?php echo $mname;?>' target="_blank"><img id='' src='<?php echo $ppath; ?>' width="200" height="300" border="0" style="padding: 0 0px 0 0px;" title="<?php echo $mname; ?>"></a></li>

哪裏MNAME是iRobot公司當$計數= 1。這哪裏是多餘的<li> < a href = watchmovie.php？path = from？好像我正在將隨機垃圾寫入我的鏈接。

更新文件在這裏。

<?php 
//connect to database server 
$row; 
    $dbcnx1 = mysql_connect("localhost", "root", "wouldntyouliketoknow"); 
    if (!$dbcnx1){ 
     echo("<P>Unable to connect to the database server at this time.</P>"); 
     exit(); 
    } 
    else{ 
     echo("<P>Successfully connected to the database!</P>"); 
    } 
    //select database 
    mysql_select_db("movies", $dbcnx1); 
    if (! @mysql_select_db("movies")) { 
     echo("<P>Unable to locate the movies database at this time.</P>"); 
     exit(); 
    } else { 
     echo("<P>Successfully connected to the movies database!</P>"); 
    } 
    //Get movie List 
    $count = 1; 
?> 
<html> 
<link rel="stylesheet" type="text/css" href="css/JMyCarousel.css" /> 
<script type='text/javascript' src="js/jquery.js"></script> 
<script type='text/javascript' src='js/scollingcarouselunminified1.js'></script> 
<script type="text/javascript"> 
$(document).ready(function() { 
    $('#tumbnailtest').scrollingCarousel({ 
    autoScroll: false 
    }); 
}); 
</script> 
<?php 
$mymovies = "<div id='tumbnailtest' class = 'carousel'><ul>"; 
for ($i = 1; $i <= 10; i++) { 
//Get movie List 
$result = mysql_query("SELECT Title, Filepath, Posterpath FROM movies WHERE ID = ".$i); 
while ($row = mysql_fetch_array($result)) { 
    //for generating url 
    $mpath = $row["Filepath"]; 
    $mmname = $row["Title"]; 
    $ppath = $row["Posterpath"]; 
} 
$mymovies .= "<li><a href='watchmovie.php?path=".$mpath; 
$mymovies .= "&mname=".$mmname."'><img src='".$ppath."' width='200' height='300' border='0' style='padding: 0 0px 0 0px;' title='".$mmname."'></a></li>"; 
} 
$mymovies .= "</ul></div>"; 
echo $mymovies; 
?> 

</html>

來源

2013-03-24 Mr404servererror

更改 JS 代碼：

<html> 
<head> 
    <link rel="stylesheet" type="text/css" href="css/JMyCarousel.css" /> 
</head> 
<body> 
<?php 
//connect to database server 
$row; 
    $dbcnx1 = mysql_connect("localhost", "root", "wouldntyouliketoknow"); 
    if (!$dbcnx1){ 
     echo("<P>Unable to connect to the database server at this time.</P>"); 
     exit(); 
    } 
    else{ 
     echo("<P>Successfully connected to the database!</P>"); 
    } 
    //select database 
    mysql_select_db("movies", $dbcnx1); 
    if (! @mysql_select_db("movies")) { 
     echo("<P>Unable to locate the movies database at this time.</P>"); 
     exit(); 
    } else { 
     echo("<P>Successfully connected to the movies database!</P>"); 
    } 
?> 
<script type='text/javascript' src="js/jquery.js"></script> 
<script type='text/javascript' src='js/scollingcarouselunminified1.js'></script> 
<script type="text/javascript"> 
$(document).ready(function() { 
    $('#tumbnailtest').scrollingCarousel({ 
    autoScroll: false 
    }); 
}); 
</script> 
<?php 
$mpath; 
$mmname; 
$ppath; 
$mymovies = "<div id='tumbnailtest' class = 'carousel'><ul>"; 
for ($i = 1; $i <= 10; $i++) { 
//Get movie List 
$result = mysql_query("SELECT Title, Filepath, Posterpath FROM movies WHERE ID = '".$i."'"); 
while ($row = mysql_fetch_array($result)) { 
    //for generating url 
    $mpath = $row["Filepath"]; 
    $mmname = $row["Title"]; 
    $ppath = $row["Posterpath"]; 
} 
$mymovies .= "<li><a href='watchmovie.php?path=".$mpath; 
$mymovies .= "&mname=".$mmname."'><img src='".$ppath."' width='200' height='300' border='0' style='padding: 0 0px 0 0px;' title='".$mmname."'></a></li>"; 
} 
$mymovies .= "</ul></div>"; 
echo $mymovies; 
?> 
</body> 
</html>

~~仍然存在與JS和PHP的一個問題，雖然，你可以不喜歡這樣的搭配它來實現什麼，我認爲你正在嘗試做...~~ 但試試這個，看看它是否工作:)

來源

2013-03-24 09:12:52 aNewStart847

好得多！謝謝！但是， Mr404servererror 2013-03-24 09:20:44

Try again, just edited – aNewStart847 2013-03-24 09:25:27

Awwww yeah! You are the best. – Mr404servererror 2013-03-24 09:29:15

從javascript生成URL的問題

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