mysqli_fetch_assoc() expects parameter 1 to be mysqli_result,

ausgezeichnete · #1 (**permalink**) 10-27-07, 07:04 AM

Hello There
i got this error with my script and i dont know why???????

PHP Code:


			
<html>

<head>

<title>

Welcome to the search Page

</title>

</head>

<body>

<form action="2.php" method="post">

Search:

<input type="text" name="search" >

<input type="submit" name="submit" >

</form>

</body>

</html>

PHP Code:


			


<?php

    //connect to database server and select a database called cars

    $conn = mysql_connect("localhost", "root", "");

    mysql_select_db("cars",$conn);

    $query= "select * from cartable where name like '%".$search."%'";

    $result=mysql_query($query);

    if($result === false) {

die(mysqli_error());

}

    while($record=mysqli_fetch_assoc($result)){

    

    while(list($name,$value)=each($record)){

        echo $name.$value;

    }

echo "</br>";

}

Nico · #2 (**permalink**) 10-27-07, 07:14 AM

You're mixing mysql with mysqli.

Replace all mysqli_* occurrences with the appropriate mysql_* equivalent.

ausgezeichnete · #3 (**permalink**) 10-27-07, 07:33 AM

and it gave me that:
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in D:\wamp\www\Maha\2.php on line 8

ausgezeichnete · #4 (**permalink**) 10-27-07, 08:09 AM

no it worked now thnxxxxx

PHP Code:


			
<?php
    //connect to database server and select a database called cars
    $conn = mysqli_connect("localhost", "root", "","cars");
    
    $query= "select * from cartable where make like '%".$search."%'";
    $result=mysqli_query($conn,$query);
    if (!$result) {
    die('Could not query:' . mysql_error());
}

    while($record=mysqli_fetch_assoc($result)){
    
    while(list($name,$value)=each($record)){
        echo $name.":".$value."</br>";
    }
echo "</br>";
}    
    ?>
    
    <a href="1.php">Back</a>