0

我只是想在 MySQL (PHP) 中显示一个咨询的结果。代码是:

$example = mysql_query("SELECT count(*) as text FROM table WHERE name = '$name'");
$qtd = mysql_num_rows($example);
while($data = mysql_fetch_array($qtd)){
$count = $data["text"];
}

echo "<h3>($count)</h3>";

错误:警告:mysql_num_rows():提供的参数不是第 9 行 C:\AppServ\www\Site\index.php 中的有效 MySQL 结果资源

4

2 回答 2

1

改变这个

 while($data = mysql_fetch_array($qtd)){
 $count = $data["text"];
 }

while($data = mysql_fetch_array($example)){
 $count = $data["text"];
  }

编辑: 你不需要在这里做一个while循环。

你应该这样做

     $example = mysql_query("SELECT count(*) as text FROM table WHERE name = '$name'");
     $data = mysql_fetch_array($example) ;
     $count = $data["text"];
     echo "<h3>".$count."</h3>";
于 2013-03-10T19:13:17.763 回答
0

这应该有效,您不需要使用 while 语句。

$query = mysql_query("SELECT * FROM table WHERE name = '$name'");
$count = mysql_num_rows($query);
echo $count;
于 2013-03-10T19:26:32.510 回答