Question

更新时，我想在尝试执行此操作时从另一个表格中选择路线的值，显示Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result

这是我的代码：

<?php
$dbHost = 'localhost'; // usually localhost
$dbUsername = 'xxxxxxx';
$dbPassword = 'xxxxxxxxxx';
$dbDatabase = 'fms';
$db = mysql_connect($dbHost, $dbUsername, $dbPassword) or die ("Unable to connect to Database Server.");
mysql_select_db ($dbDatabase, $db) or die ("Could not select database.");

$client_id=$_POST['clientid'];
$feild=$_POST['field'];
        $data= $_POST['value'];
        $rownum=$_POST['rowid'];  
        $sql="UPDATE $client_id SET ".$feild." = '".$data."' WHERE net_id = ".$rownum."";

         print $sql;

        mysql_query($sql);  


//Select route from client Table

$sql_select="select route from $client_id WHERE net_id = ".$rownum."";
mysql_query($sql_select);
 print $sql_select;
 print mysql_error();
$i=1;
while($rows=mysql_fetch_assoc($sql_select))
{

    $route=$rows['route'];

}

?>

请帮助我，提前谢谢

Answer 1

这样写：

$sql_select="select route from $client_id WHERE net_id = ".$rownum."";
$queryRes = mysql_query($sql_select);
 print $sql_select;
 print mysql_error();
$i=1;
while($rows=mysql_fetch_assoc($queryRes))

您需要将mysql_query()函数返回的#Resource提供给mysql_fetch_assoc()。

注意：PHP5.3不推荐使用Mysql_ *。因此应该避免。

Answer 2

试试这个

$sql_select="select route from $client_id WHERE net_id = ".$rownum."";
$result = mysql_query($sql_select);
 print $result;
 print mysql_error();
$i=1;
while($rows=mysql_fetch_assoc($result))
{

    $route=$rows['route'];

}

警告：mysql_fetch_assoc（）：提供的参数不是有效的MySQL结果

2 个答案: