php - Checking SQL Database for value -
i trying use php check database see if value exists. main goal use value
$_get['udid']
and if equal value in database return
echo 'found';
i using code:
<?php $servername = "*****"; $username = "*****"; $password = "*****"; $dbname = "*****"; $connect = new mysqli($servername, $username, $password, $dbname); if ($connect->connect_error) { die("connection failed: " . $connect->connect_error); } $udid = $_get['udid']; $id = mysqli_real_escape_string($connect, $udid); $result = mysqli_query($connect, "select udid data udid = '$id'"); if($result === false) { die("error: " . mysqli_error($result)); } else { while ($row = mysqli_fetch_array($result)) { if($row['udid'] == $udid) { $results = 'your device registered on our servers.'; $results2 = 'please click install button below.'; $button = 'install'; $buttonlink = 'https://**link here**'; } else { $results = 'your device not registered on our servers'; $results2 = 'please click request access button below.'; $button = 'request access'; $buttonlink = 'https://**link here**'; } } } ?>
but reason not working, sure on looking something. appreciated.
try this:
$sql = mysqli_query($connect, "select udid data udid = '" .$udid. "'");
and also, make sure set value 'get' $udid. should this:
$udid = $_get['udid'];
we can use mysqli_fetch_array() instead result row. include error handling. code must :
$udid = $_get['udid']; $id = mysqli_real_escape_string($connect, $udid); $result = mysqli_query($connect, "select `udid` `wmaystec_wmt-ss`.`data` = '$id'"); if($result === false) { die(mysqli_error("error message user")); //error handling } else { while ($row = mysqli_fetch_array($result)) { echo "found :" .$row['thefieldnameofudidfromyourdb']; } }
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