Php 警告:mysql\u fetch\u assoc():提供的参数不是有效的mysql结果
更新时,我想从另一个表中选择route的值,当我尝试执行此操作时,它显示Php 警告:mysql\u fetch\u assoc():提供的参数不是有效的mysql结果,php,mysql,Php,Mysql,更新时,我想从另一个表中选择route的值,当我尝试执行此操作时,它显示警告:mysql\u fetch\u assoc():提供的参数不是有效的mysql结果 这是我的代码: <?php $dbHost = 'localhost'; // usually localhost $dbUsername = 'xxxxxxx'; $dbPassword = 'xxxxxxxxxx'; $dbDatabase = 'fms'; $db = mysql_connect($dbHost, $dbUs
警告:mysql\u fetch\u assoc():提供的参数不是有效的mysql结果
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'];
}
?>
请帮帮我,提前谢谢这样写:
$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))
您需要提供从函数返回到的#资源
注意:PHP5.3中不推荐使用Mysql。因此应该避免。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'];
}
$result=mysql\u query($sql\u select)$rows=mysql\u fetch\u assoc($result);请参阅url:此代码易受SQL注入的影响
$rownum
和$client\u id
必须被转义,或者您应该使用预先准备好的语句(因此首先不要使用mysql\u
。@Jack:同意您的看法。