Question

$i_id = $_GET['iiSL'];   

require_once('../include/dbc.php');    

$sql = "SELECT invite_id FROM invite_requests WHERE invite_id = '$i_id'";
$result = mysql_query($sql);
if(mysql_num_rows($result == 1))
{
echo 'GOOD ID EXISTS';
//ECHO IS JUST TO TEST  
} 
else
{
echo 'BAD ID IS NOT IN DB';
//ECHO IS JUST TO TEST
}

为什么这不起作用？这让我疯了。

错误Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource

所有拼写，语法，语法和大小写都是正确的。 URL正在传递$i_id变量。它回声正确。

我做错了什么？

Answer 1

在这种情况下有一个错字..这个..

if(mysql_num_rows($result) == 1)

您正在将$result == 1的结果传递给mysql_num_rows，期望结果资源为mysql_query() ..：）

Answer 2

将代码更改为

$query1=mysql_query("SELECT count(invite_id) as total FROM invite_requests WHERE invite_id = '$i_id';");
$row = mysql_fetch_array($query1);
if ($row["total"]>"0")
{
echo 'GOOD ID EXISTS';
//ECHO IS JUST TO TEST  
} 
else
{
echo 'BAD ID IS NOT IN DB';
//ECHO IS JUST TO TEST
}           {

试试这个

在PHP中使用mySQL_num_rows很困难

2 个答案: