0

我有这个代码我正在尝试输出这个：

{
    title:"<?php echo $sender_fullname; ?>",
    mp3:"link",
},

在 php 中使用它在 javascript 中显示它

//database include
require_once "db.php";

//get email from session 
$email = $_SESSION['username'];

//fetch user fullname and  id based on session
$name_query = mysql_query("SELECT fullname,id FROM users WHERE email = '$email'");
$name = mysql_fetch_object($name_query);

//fecth sender id, receiver id, audioclip, fullname and email
$query = "SELECT m.sender,m.receiver, m.audioclip, u.fullname, u.email 
                     FROM `users` AS u 
                     JOIN `messages` AS m ON m.receiver = u.id 
                     WHERE u.email = '".$email."'";

$result = mysql_query($query);

这是循环我应该怎么做才能输出相同的内容 

while(
        $row = mysql_fetch_assoc($result)) { 
        $sender = $row['sender'];
        $sender_name_query = mysql_query("SELECT fullname FROM users WHERE id = '$sender'");
        $sender_name = mysql_fetch_object($sender_name_query);
        $sender_fullname = $sender_name->fullname;
        echo '{<br/>title:".'$sender_fullname'.",<br/>mp3:"link",<br/>},';      
    }
4

3 回答 3

0

错误是由最后一行引起的：

echo '{<br/>title:".'$sender_fullname'.",<br/>mp3:"link",<br/>},';

应该是这样的：

echo '{<br/>title:"' . $sender_fullname . '",<br/>mp3:"link",<br/>},';

注1

你可以这样写：

$sender_fullname = $sender_name->fullname;
echo '{<br/>title:"' . $sender_fullname . '",<br/>mp3:"link",<br/>},';

容易为：

echo '{<br/>title:"' . $sender_name->fullname. '",<br/>mp3:"link",<br/>},';

笔记2

请不要mysql_*在新代码中使用函数。它们不再被维护并被正式弃用。看到红框了吗？改为了解准备好的语句，并使用PDO或MySQLi -本文将帮助您决定使用哪个。如果您选择 PDO，这里有一个很好的教程。

来自http://brightmeup.info/comment.html

于 2013-02-04T17:49:25.657 回答
0

将 php 转换为 javascript 的最佳方法是通过 php 的 json_encode使用 JSON 。

$rows = array();
while($row = mysql_fetch_assoc($result)) { 
    $sender = $row['sender'];
    $sender_name_query = mysql_query("SELECT fullname FROM users WHERE id = '$sender'");
    $sender_name = mysql_fetch_object($sender_name_query);
    $sender_fullname = $sender_name->fullname;
    $row['sender'] = $sender_fullname;
    $rows[] = $row;
}   

echo "<script type='text/javascript'>";
echo "var rows = " . json_encode($rows) . ";";
echo "</script>";
于 2013-02-04T17:46:18.177 回答
0

你可以做一些像：

while(
        $row = mysql_fetch_assoc($result)) { 
        $sender = $row['sender'];
        $sender_name_query = mysql_query("SELECT fullname FROM users WHERE id = '$sender'");
        $sender_name = mysql_fetch_object($sender_name_query);
        $sender_fullname = $sender_name->fullname;
        echo '{<br/>title:".'$sender_fullname'.",<br/>mp3:"link",<br/>},';
        print "<script>alert(\"$sender_fillname\")</script>";
    }

然后脚本标记中的代码像 javascript 代码一样运行，如果你想将 php 变量的值放入 javascript 变量中，你可以这样做：

<?
    $mivar = "hola mundo";
    print "<script>";
    print "var mivar = \"$mivar\"";
    print "</script>";
?>
于 2013-02-04T17:45:06.067 回答