mysql_num_rows error...

[luckysanj]

$sql_result=mysql_query($query) or die("Error in Checking User".mysql_error()); 
        echo $no=mysql_num_rows($sql_result);
             if($no<>1)
                {
                return(false);
                }
            else
                {
                return (true);
                }
 

In this code i have get this type of error:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in D:\xampp\htdocs\my\test\query\function.php on line 2
 

Can any one guide me. How is error occured?

Thank you

[bilibytes]

You are not doing the query well.

you should pass a mysql connection to the query so that it knows where to query.

$link = mysql_connect('hostName', 'userName', 'passWord');
 
//then select the db
 
mysql_select_db('databaseName', $link);
 
//then you can make your queries passing the $link
 
$resultSet = mysql_query('SELECT * FROM Table1', $link);
 
//now you can check if there are results
if(mysql_num_rows(0 < $resultSet)){
    echo 'yeah coolm, i have a result'
}
else{
     echo 'wtf nothing from db';
}
 

[Markus]

Your mysql_query() doesn't return a valid resource - what is the query?

Also, you if/else block can be shortened like so:

return ($no !== 1);
 
// as opposed to this:
if($no <> 1) {
    return FALSE;
}
else {
    return TRUE;
}
 

[Markus]

Billibytes beat me to it.

Anyway, couple of things: You do not have to pass a link resource to the mysql_* functions - if you do not, mysql will use the last opened connection.

Also:

if(mysql_num_rows(0 < $resultSet)){
    echo 'yeah coolm, i have a result'
}
else{
     echo 'wtf nothing from db';
}
 

will not work, but I think that's just a typing error on your part ;)

if(mysql_num_rows($resultSet) < 0){
    echo 'yeah coolm, i have a result'
}
else{
     echo 'wtf nothing from db';
}
 

[bilibytes]

@Markus
right, i have answered too quickly...

[luckysanj]

ok yet i have not solved my problem. so i send my three php page content.
1. index.php

<form id="form1" name="form1" method="post" action="check.php?id=tbl1">
  <table width="200" border="1">
    <tr>
      <td colspan="2">Pesonal Info Table 1 </td>
    </tr>
    <tr>
      <td width="72">Name:</td>
      <td width="112"><input name="name" type="text" id="name" value="Ashok" /></td>
    </tr>
    <tr>
      <td>Address</td>
      <td><input name="address" type="text" id="address" value="Kathmandu" /></td>
    </tr>
    <tr>
      <td><input name="tbl1" type="submit" id="tbl1" value="Submit" /></td>
      <td><input type="reset" name="Submit2" value="Reset" /></td>
    </tr>
  </table>
</form>
 
 

2. check.php

    <?php
    foreach($_POST as $key=>$value)
        { 
        //echo "</br>";
        $key.":".$$key=$value;
        //echo "</br>".$key;
        }
        include_once("function.php");
        $tblname= $_GET['id'];
        $db=new Functions;
        $db->dbconnection();
    if(isset($tbl1))
        {
        $field=array('name','address');    
        $value=array($name,$address);
        $y=$result=$db->InsertData($tblname,$field,$value);
        /*if($y==true)
            {
                echo "Success";
            }
            else
            {
            echo "Failed";
            }*/
        }
    if(isset($tbl2))
        {
        $field=array('phone','email');    
        $value=array($phone,$email);
        $result=$db->InsertData($tblname,$field,$value);
        }
?>
 

3. function.php

<?php 
class Database
{
function dbconnection() 
    {
    $link=mysql_connect("localhost","root","") or die("Failed Connecting to Database");
    mysql_select_db("query",$link) or die("Failed Connecting To Database");    
    }
}    
class Functions extends Database
{
function InsertData($TblName,$Fields,$Values)  
        {  
        $query="INSERT INTO `{$TblName}` (`".(is_array($Fields)?implode("`,`",$Fields):$Fields)."`) 
                VALUES (".(is_array($Values)?"'".implode("','",$Values)."'":$Values).")"; 
        $sql_result=mysql_query($query) or die("Error in Checking User".mysql_error()); 
        echo $no=mysql_num_rows($sql_result);
             if($no<>1)
                {
                return(false);
                }
            else
                {
                return (true);
                }
        } 
}    
?>
 
 

But Still Error is same

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in D:\xampp\htdocs\my\test\query\function.php on line 17
 

[Canabeez]

Try this as function.php

class Database
{
    private $link;
    function dbconnection() 
    {
        $this->link=mysql_connect("localhost","root","") or die("Failed Connecting to Database");
        mysql_select_db("query",$this->link) or die("Failed Connecting To Database");    
    }
}    
class Functions extends Database
{
    function InsertData($TblName,$Fields,$Values)  
    {  
        $query="INSERT INTO `{$TblName}` (`".(is_array($Fields)?implode("`,`",$Fields):$Fields)."`) 
        VALUES (".(is_array($Values)?"'".implode("','",$Values)."'":$Values).")"; 
        $sql_result=mysql_query($query, $this->link) or die("Error in Checking User".mysql_error());
 
        /* This line is to debug your query */ echo "<!-- QUERY: {$query} -->";
 
        $no = mysql_num_rows($sql_result);
        echo ($no != 1);
    }
}
 

[bilibytes]

I suggest that you go step by step.

first check that you get the data from the brower, and try to echo it.

then try to connect to the db and make a query with data you write in your file, not from $_POST

and try to output it.

and so on, you have to narrow down the search until you know where the problem comes from.

once you know where your code fails, come back and we will guide you.

[luckysanj]

This error is occur Canabeez sir,

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in D:\xampp\htdocs\my\test\query\function.php on line 18
Error in Checking User
 

[Markus]

echo() the $query variable, and post the output here. Your SQL is wrong, most likely.