CODEWITHSUNDEEP
javascriptphpmysqlajax
hammad rashid
hammad rashid

Reputation: 27

querying values from the table

I am trying to get values from table the problem I am facing is that regardless of what the input is it gives me the result of both rows. I want to match the input ID with the ID in the table. This is the code I am trying to useImage of the table is attached the file name is linked with the javascript file where I have a click function and this javascript(global.js) file is linked with another file having text box. My main file code is below

  <!DOCTYPE html>
<html>
    <head>
        <title>products</title>
    </head>
    <body>
            Item: 
            <input type="text" required id="item">
            <input type="submit" id="item-submit" value="Grab"> 
            <div id="item-data"></div>

            <script src="http://code.jquery.com/jquery-2.2.0.min.js"></script> 
            <script src="js/global.js"></script> 

    </body>
</html>

global.js file code

 $(document).ready(function(){
    $('input#item-submit').click(function(){     

        var item = $('input#item').val();    
        if ($.trim(item) != ''){             
            $.post('ajax/name.php',{item:name}, function(data){ 
                $('div#item-data').text(data);  

            });

        }

    })
});

name.php file code is below

 require '../db/connect.php';

    $sql = "SELECT BarcodeID, shirts, price FROM clothes WHERE BarcodeID = ($_POST['name']) ";

    $result = mysqli_query($con,$sql);
        while($row = mysqli_fetch_array($result)) {
            echo " " . $row["BarcodeID"]. " shirt color:  " . $row["shirts"]. " price: " . $row["price"];
        }
    mysqli_close($con);

Upvotes: 0

Views: 77

Answers (2)

safin chacko
safin chacko

Reputation: 1390

    $id=$_POST['item']; // your post variable
    $sql = "SELECT BarcodeID, shirts, price FROM clothes WHERE BarcodeID=".$id;

     $result = mysqli_query($con,$sql);
if(mysqli_num_rows($result)>0)
{
        while($row = mysqli_fetch_array($result)) {
            echo " " . $row["BarcodeID"]. " shirt color:  " . $row["shirts"]. " price: " . $row["price"];
        }
}
else
{
echo "No results found";
}
    mysqli_close($con);

Upvotes: 0

Mr. Engineer
Mr. Engineer

Reputation: 3515

Try this query : 

"SELECT BarcodeID, shirts, price FROM clothes WHERE BarcodeID = '{$_POST['name']}'"

And why are you comparing BarcodeID with name?

Side note : Your query is unsafe. Read this How can I prevent SQL injection in PHP?.

Upvotes: 1

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