从php中的数据库中获取结果并将其作为循环输出以在javascript中显示

i have this code im working on im trying to output this :

我有这个代码我正在尝试输出这个:

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

using this in php to display it in javascript

在php中使用它来在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);

here's the loop what should i do to output the same

这是循环我应该怎么做输出相同

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/>},';      
    }

3 个解决方案

#1

you can do some like :

你可以这样做:

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>";
    }

and then the code into the script tag run like javascript code if you wants puts the value of a php variable into a javascript variable you can do :

如果你想将php变量的值放入一个javascript变量中,那么代码进入脚本标记就像javascript代码一样运行:

<?
    $mivar = "hola mundo";
    print "<script>";
    print "var mivar = \"$mivar\"";
    print "</script>";
?>

#2

The best way to get php into javascript is to use JSON via php's json_encode.

将php导入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>";

#3

The error is caused by the last line:

错误是由最后一行引起的:

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

It should be this:

它应该是这样的:

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

Note 1

You can write this:

你可以这样写:

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

easily as:

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

Note 2

Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial.

请不要在新代码中使用mysql_ *函数。它们不再维护,并且已被正式弃用。看到红色的盒子?转而学习预备语句,并使用PDO或MySQLi - 本文将帮助您确定哪些。如果您选择PDO,这是一个很好的教程。

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#1