Display database value in array PHP MYSQLi

Question

I"m attempting to display some data I've sent from ajax to a php file, however for some reason its not displaying it on the page. The way it works it I enter a search term into a input field, and a ajax script post the value to a php script, which return the database value requested back.

error_reporting(E_ALL); 
ini_set('display_errors', '1');


if (isset($_POST['name']) === true && empty($_POST['name']) === false) {
    //require '../db/connect.php';

    $con = mysqli_connect("localhost","root","root","retail_management_db"); 
    $name = mysqli_real_escape_string($con,trim($_POST['name']));    
    $query = "SELECT `names`.`location` FROM `names` WHERE`names`.`name` = {$name}";
    $result = mysqli_query($con, $query);
    if (mysqli_num_rows($result) > 0) {
        while ($row = mysqli_fetch_array($result)) {
            $loc = $row['location'];
             echo $loc;   
        }//close While loop                            
    } else {
        echo $name . "Name not Found";
    }
}

html form:

<!DOCTYPE html> 
<html>
    <head>
    <meta charset="utf-8">
    <title>Retail Management Application</title>
    </head> 
    <body>             
        Name: <input type="text" id="name">
         <input type="submit" id="name-submit" value="Grab">
            <div id="name-data"></div>
        <script src="http://code.jquery.com/jquery-1.11.2.min.js"></script>
        <script src="js/global.js"></script>
    </body>
</html>

What are is the response you are getting from the AJAX call? — Chip Dean
– Chip Dean, Commented Apr 23, 2015 at 1:52
You have mysqli_error() in you query stiring. it should be - query = mysqli_query("SELECT `names`.`location` FROM `names` WHERE `names`.`name` = {$name}" ) or die(mysqli_error($con)); — Sean
– Sean, Commented Apr 23, 2015 at 1:53
and then you are doing a query on your previous query $result = mysqli_query($con, $query);, which is really $result = mysqli_query($con, mysqli_query("SELECT names.location` FROM names WHEREnames.name = {$name}" . mysqli_error($con)); );`. — Sean
– Sean, Commented Apr 23, 2015 at 1:55
$name is a string, so it needs to be quoted - ... WHERE `names`.`name` = '{$name}'" — Sean
– Sean, Commented Apr 23, 2015 at 2:28

Funk Forty Niner · Accepted Answer · 2015-04-23 02:35:26Z

2

You're appending a MySQL error result to your query, and you're trying to query a query result, try the following:

$query = "SELECT `names`.`location` FROM `names` WHERE`names`.`name` = '$name'";
$result = mysqli_query($con, $query);
if (mysqli_num_rows($result) > 0) {

Edit:

{$name} that is a string and should be quoted instead.

change it to '$name' in the where clause.

Using:

$result = mysqli_query($con, $query) or die(mysqli_error($con));

will provide you with the reason as to why your query failed.

edited Apr 23, 2015 at 2:35

Funk Forty Niner

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answered Apr 23, 2015 at 1:57

scrowler

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9 Comments

Yacub Ali Over a year ago

Hi scrowler, ive made the changes but it seems it still says Name not found even though the records exists in the database.

scrowler Over a year ago

Have you debugged $_POST in your PHP file to ensure that the value for $name is being correctly passed through?

Yacub Ali Over a year ago

Yh it shows this: br /> Warning: mysqli_real_escape_string() expects exactly 2 parameters, 1 given in /Applications/MAMP/htdocs/ajax/name.php on line 8 Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, string given in /Applications/MAMP/htdocs/ajax/name.php on line 12

Funk Forty Niner Over a year ago

@YacubAli someone in comments noted about {$name} that should read as '{$name}' either that or just do '$name' or '".$name."' in your where clause. that's the most likely reason now.

Funk Forty Niner Over a year ago

@YacubAli you're welcome, glad to have been of help, cheers

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