Trying to return row from DB with matching column

Question

I have tried ready various articles to resolve the issue to no avail. I using a form to post data to my php file. Then I'm trying to find the row that has the matching value. I keep getting these errors: Undefined index: barcode & Trying to get property of non-object Can someone help me resolve this issue?

//barcode is serialized data from a form and equals 'barcode=2147483647'

$barcode = $_GET['barcode'];


$sql = "SELECT id FROM the_DB WHERE barcode = '%$barcode%'";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
        echo 'id:' . $row["id"]. '<br>Name: ' . $row["item_name"]. '<br> Barcode ' . $row["barcode"]. '<br><img src="'.$row["image"].'" height="100px">';
    }
} else {
    echo "0 results";
}

If you trying to fetching equal record then try this $sql = "SELECT id FROM the_DB WHERE barcode =".$barcode; or you can use Like query to get keyword matching result like $sql = "SELECT id FROM the_DB WHERE barcode like %".$barcode."%"; — Android Leo, Mar 06 '16 at 15:46

score 0 · Answer 1 · answered Mar 06 '16 at 15:40

0

change query as follows

 $sql = "SELECT * FROM the_DB WHERE barcode = '%$barcode%'";

answered Mar 06 '16 at 15:40

Vigikaran

725
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This returns this error: Undefined index – Rafael Mar 06 '16 at 15:42
check your table column name with parameters – Vigikaran Mar 06 '16 at 15:47
and in query instead of = use like ($sql = "SELECT * FROM the_DB WHERE barcode LIKE '%$barcode%'";) – Vigikaran Mar 06 '16 at 15:49

Vasim Shaikh · Accepted Answer · 2016-03-06T16:18:04.660

0

Following Possibility for this error:

Check your variable in scope.
If you get request from other page check your request method.
If you are use session must start before use.
echo and exit to check your variable have value or not.

You are fetching an column that are not in your result set.

    $barcode = $_GET['barcode'];
if(isset($barcode))
{

$sql = "SELECT * FROM the_DB WHERE barcode = '%$barcode%'";
$result = $conn->query($sql);

$rows = array();
while ($row = mysql_fetch_object($result)) {
    $rows[] = $row;

}

}

edited Mar 06 '16 at 16:18

answered Mar 06 '16 at 15:42

Vasim Shaikh

4,485
2
23
52

1

May I know reason? for down vote – Vasim Shaikh Mar 06 '16 at 15:45
this states "$barcode = $_GET['barcode']; " is an unidentified index – Rafael Mar 06 '16 at 15:45
@Raf,You can use $_Request['barcode'] – Vasim Shaikh Mar 06 '16 at 15:48
1

check your object name is barcode or not that passed when you submit form – Vasim Shaikh Mar 06 '16 at 15:48
I used $_Request and the object is named barcode. – Rafael Mar 06 '16 at 15:51
then what you get @raf? – Vasim Shaikh Mar 06 '16 at 15:54
I get $_Resquest is an unidentified index – Rafael Mar 06 '16 at 16:01
echo $_Request['barcode']; exit; – Vasim Shaikh Mar 06 '16 at 16:02
Undefined variable: barcode in C:\wamp\www\mypath\index.php on line xx – Vasim Shaikh Mar 06 '16 at 16:03
It's working now! Thanks for all your help! – Rafael Mar 06 '16 at 16:05
may I know whats problem? – Vasim Shaikh Mar 06 '16 at 16:07
@Raf,May I know whats problem?And If solved mark as complete so this is helpful.Any one else. – Vasim Shaikh Mar 06 '16 at 16:12
I change request to post – Rafael Mar 06 '16 at 16:14
$_REQUEST is use in both cases thats why I am always prefer – Vasim Shaikh Mar 06 '16 at 16:15

Trying to return row from DB with matching column

2 Answers2