如何检查 while($row = mysql_fetch_array in PHP
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How to check if results on while($row = mysql_fetch_array in PHP
提问by jrutter
Im trying to figure out how to handle this is no results are returned, how would I code that?
我试图弄清楚如何处理这个没有返回结果的问题，我将如何编码？
while($row = mysql_fetch_array($Result))
while($row = mysql_fetch_array($Result))
So like if there a results: print them out
所以就像如果有结果：打印出来
else: show a link
否则：显示链接
回答by mpen
http://ca3.php.net/manual/en/function.mysql-num-rows.php
http://ca3.php.net/manual/en/function.mysql-num-rows.php
if(mysql_num_rows($result) > 0) {
while($row = mysql_fetch_array($result)) { ... }
} else {
// show link
}
回答by Sampson
You can use mysql_num_rows()to tell you how many results are found. Using that with a simple if-statement, and you can determine what action to take.
您可以使用mysql_num_rows()来告诉您找到了多少结果。使用简单的if-statement，您可以确定要采取的操作。
if (mysql_num_rows($result) > 0) {
// do while loop
} else {
// show link
}
回答by Bill Karwin
Others suggest using mysql_num_rows()but you should be aware that that function works only if you use a buffered query. If you query using mysql_unbuffered_query(), the number of rows in the result is not available.
其他人建议使用，mysql_num_rows()但您应该知道该功能仅在您使用缓冲查询时才有效。如果使用 查询mysql_unbuffered_query()，则结果中的行数不可用。
I would use a simple flag variable:
我会使用一个简单的标志变量：
$found_row = false;
while ($row = mysql_fetch_array($result)) {
$found_row = true;
. . .
}
if ($found_row == false) {
// show link
}
It may seem redundant to set $found_rowto true repeatedly, but assigning a literal value to a variable ought to be an insignificant expense in any language. Certainly it is small compared to fetching and processing an SQL query result.
重复设置$found_row为 true似乎是多余的，但为变量分配文字值在任何语言中都应该是微不足道的开销。当然，与获取和处理 SQL 查询结果相比，它很小。
回答by Binyamin
Use even shorter syntax without insignificant mysql_num_rowsto save processor time:
使用更短的语法mysql_num_rows来节省处理器时间：
if($result) {
// return db results
} else {
// no result
}
回答by VolkerK
This can be done without mysql_num_rows() or an additional (flag) variable
这可以在没有 mysql_num_rows() 或附加（标志）变量的情况下完成
if ( false===($row=mysql_fetch_array($result, MYSQL_ASSOC)) ) {
echo 'no rows in result set';
}
else {
do {
echo $row['X'];
} while ( false===($row=mysql_fetch_array($result, MYSQL_ASSOC)) );
}
but it duplicates the actual fetch command (one in the if-statement and one in the while-clause).
但它复制了实际的 fetch 命令（一个在 if 语句中，一个在 while 子句中）。
回答by jrutter
I might have figured it out:
我可能已经想通了：
if (!($row = mysql_fetch_array($descResult)))
{
echo "<tr><td>Add Link</td></tr>";
}
