php 如果返回 0 行，则 MySQL 返回结果
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MySQL return result if 0 rows returned
提问by Tom
I have this PHP code:
我有这个 PHP 代码：
$query = "SELECT name, COUNT(message) FROM guestbook_message WHERE name='".$req_user_info['username']."' GROUP BY name";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
echo "Messages posted: ". $row['COUNT(message)'] ."";
echo "<br />";
}
Which will show the amount of comments a user has posted.
这将显示用户发布的评论数量。
How do I make it return a value if there is no messages posted from that user? Currently is displays nothing at all. But I want it to show "Messages posted: 0"
如果没有来自该用户的消息，我如何让它返回一个值？目前是什么都不显示。但我希望它显示“发布的消息：0”
Any ideas?
有任何想法吗？
回答by shamittomar
Check the number of results returned in result using mysql_num_rows.
使用 来检查 result 中返回的结果数mysql_num_rows。
$query = "SELECT name, COUNT(message) FROM guestbook_message WHERE name='".$req_user_info['username']."' GROUP BY name";
$result = mysql_query($query) or die(mysql_error());
if(mysql_num_rows($result) > 0)
while($row = mysql_fetch_array($result))
{
echo "Messages posted: ". $row['COUNT(message)'] ."";
echo "<br />";
}
else
echo "NO Messages posted. <br />";
回答by mellamokb
if ($row = mysql_fetch_array($result)) {
echo "Messages posted: ". $row['COUNT(message)'] . "";
echo "<br />";
}
else {
echo "Messages posted: 0";
echo "<br />";
}
回答by eykanal
Replace $row['COUNT(message)']with the tertiary operator:
替换$row['COUNT(message)']为三级运算符：
( $row['COUNT(message)'] > 0 ? $row['COUNT(message)'] : 0 )
This is basically a compacted if... else...statement.
这基本上是一个压缩if... else...语句。
