AAlright, so here goes nothing.
the URL: http://tandemenvoy.michaeldvinci.com/forum/repliesJSON.php?rID=2returns me absolutely nothing. if I hard code the SQL query to
"SELECT * FROM replies WHERE replyID = 2"
I get the correct JSON formatted request, but when I try passing it as a parameter in the URL. is just displays nothing -
now the reason I'm doing this is because I'm working on an iOS app that is transversing multiple tables and I need it to only display the matching 'replyTopic's to the 'categoryID's and I'm guessing if I can just append the categoryIDs to the end of the url with the correct query, it'll be a heck of a lot easier
repliesJSON.php:
<?php
$con=mysqli_connect("localhost","","","");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$rID = $GET_['rID'];
$sql = "SELECT * FROM replies WHERE replyID = '".$rID."' ";
if ($result = mysqli_query($con, $sql))
{
$resultArray = array();
$tempArray = array();
while($row = $result->fetch_object())
{
$tempArray = $row;
array_push($resultArray, $tempArray);
}
echo json_encode($resultArray);
}
mysqli_close($con);
?>
Because your Mysql statement is wrapped in double quotes, PHP will parse it for variables, meaning that
'".$rID."'
is outputted as '2' (with the single quotes still there) because PHP ignores the double quotes and concatenates the single quotation marks and the number 2.
Simply changing:
WHERE replyID = '".$rID."'
to: WHERE replyID = '$rID'
will fix this problem.
Edit:
you can also leave out the single quotes altogether as @steven has mentioned:
WHERE replyID = $rID
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