Question

我有一个mysql数据库，我需要其内容的RSS源。我在json中要求相当于这个：

mysql_select_db("mydb", $con);
$result = mysql_query("select date, title, description, url from blah where type = 'OFFERS' order by ref desc");

echo '<?xml version="1.0"?>
<rss version="2.0">
<channel>

<title></title>
<description></description>
<link></link>';

while($row = mysql_fetch_array($result))
  {
  echo "
  <item>
  <title>" . $row['title'] . "</title>
  <description>" . $row['description'] . "</description>
  <link>" . $row['url'] . "</link>
  <image>" . $row['date'] . "</image>
  </item>";
  }

到目前为止，我最接近工作的是：

<?php

$host="localhost"; 
$username="username"; 
$password="password"; 
$db_name="mydb"; 

$con=mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");
$sql = "select date, title, description, url from blah where type = 'OFFERS' order by ref desc"; 
$result = mysql_query($sql);
$json = array();
if(mysql_num_rows($result)){
while($row=mysql_fetch_row($result)){
$json['title'][]=$row;
}
}

mysql_close($db_name);

echo json_encode($json); 

?>

任何帮助将不胜感激

Answer 1

while ($row = mysql_fetch_assoc($result)) {
    $json[] = $row;
}

echo json_encode($json);

Answer 2

由于mysql_fetch_row返回两个数组键和索引，因此最好使用mysql_fetch_assoc。这只会返回键。

此外，您正在获取项目，而不是标题。因此，请使用$json['items']代替$json['title']。

除此之外，rss中还有其他信息。您也可以在$json中添加它们。

结果代码为，

$json=array();
$json['title'] = 'My JS/RSS';
$json['link']  = "http://". $_SERVER['HTTP_HOST']. $_SERVER["REQUEST_URI"];
$json['description'] = "";
$json['itmes'] = array();
while($row=mysql_fetch_assoc($result)){
    $json['items'][]=$row;
}

来自json的mysql结果

2 个答案: