我只是想在MySQL (PHP)中显示一个咨询结果。守则是:
$example = mysql_query("SELECT count(*) as text FROM table WHERE name = '$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中有效的MySQL结果资源
发布于 2013-03-10 19:13:17
改变这个
while($data = mysql_fetch_array($qtd)){
$count = $data["text"];
}至
while($data = mysql_fetch_array($example)){
$count = $data["text"];
}编辑:,您不需要在这里执行EDIT循环。
你就应该这样做
$example = mysql_query("SELECT count(*) as text FROM table WHERE name = '$name'");
$data = mysql_fetch_array($example) ;
$count = $data["text"];
echo "<h3>".$count."</h3>";发布于 2013-03-10 19:26:32
这应该是可行的,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;https://stackoverflow.com/questions/15326335
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