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我无法在 mysql 中的不同查询中显示,请帮助。

<?php
mysql_connect("localhost","root","123");
mysql_select_db("sarangani");
$result = mysql_query("SELECT DISTINCT year(posted) AS year FROM news ORDER BY posted");

while($row = mysql_fetch_array($result)) { 
  echo "$row[year]";
}
?>
4

1 回答 1

2

如果您确定查询应该提供结果,则需要使用正确的方法来echo输出结果:

while ($row = mysql_fetch_array($result)) {
    echo $row['year']; // So don't put it inside double quotes
}
于 2013-09-18T14:45:40.890 回答