Receiving no text output from a PHP file when passing data to SQL database for JSON output

Question

A PHP file has been set up in order to contact a SQL database and retrieve the article with the corresponding ArticleID which I append to the end of the URL.

For example, www.example.com/retrieveArticle.php?ArticleID=1

Currently, the webpage does not show any text output whereas I expected to find one row output as a JSON format.

Note that my table is called Customer.

Code

<?php
if(issert($_REQUEST["ArticleID"]))
    {
    $con = mysql_connect("localhost","createyo_james","password");
    if(!$con)
    {
    die("Could not connect: " .mysql_error());
    }
    mysql_select_db("createyo_TestDatabase", $con);

    $articleID = $_REQUEST["ArticleID"];

    $result = mysql_query("SELECT * FROM Customer WHERE ID = "$articleID" ") or die("Errant query:");

    while($row = mysql_fetch_assoc($result))
    {
    $output[]=$row;
    }

    print(json_encode($output));

    mysql_close($con);

    }

    else
    {
    $output = "not found";
    print(json_encode($output));
    }

?>

Edit:

Since many of the solutions have not worked it may be due to incorrect database information. I know the connection details are correct as I have similar PHP files connected to the same DB. However, I might be wrong about the table structure when getting ID etc...

Database

Code Attempt 2

<?php
if(isset($_GET["ArticleID"]))
{
$con = mysql_connect("localhost","createyo_james","password");
if(!$con)
{
die("Could not connect: " .mysql_error());
}
mysql_select_db("createyo_TestDatabase", $con);
$articleID = mysql_real_escape_string($_GET["ArticleID"]);

$result = mysql_query("SELECT * FROM Customer WHERE ID = "$ArticleID"") or die("Errant query:");

while($row = mysql_fetch_assoc($result)) { array_push($output,$row); } echo json_endode($output);die;

mysql_close($con);

}

else
{
$output = "not found";
echo json_encode($output);
}

?>

It should be echo json_encode($output); or print json_encode($output); try to use it to check output print_r(json_encode($output)); also change to it if(isset($_REQUEST["ArticleID"])) — Kavin
– Kavin, Commented Jul 14, 2014 at 9:21
@Rajlaksh, parentheses are not mandatory, but can be used with print and echo. — BudwiseЯ
– BudwiseЯ, Commented Jul 14, 2014 at 9:24
@Rajlaksh I do not think this is the issue since i have used in in a similar PHP file with success. I have tried your recommendation however, but still no output is being shown. — jamesgates1
– jamesgates1, Commented Jul 14, 2014 at 9:26

joesch · Accepted Answer · 2014-07-14 09:55:48Z

1

You have a typo on the first line:

if(issert($_REQUEST["ArticleID"]))

Should be:

if(isset($_REQUEST["ArticleID"]))

Just noticed something else. Your query:

$result = mysql_query("SELECT * FROM Customer WHERE ID = "$articleID" ") or die("Errant query:");

Shouldn't include double quotes around $articleID:

$result = mysql_query("SELECT * FROM Customer WHERE ID = '$articleID' ") or die("Errant query:");

edited Jul 14, 2014 at 9:55

answered Jul 14, 2014 at 9:22

joesch

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3 Comments

ka_lin Over a year ago

It was like "Where is Waldo?"

jamesgates1 Over a year ago

Thanks, I had to copy the PHP code from a YouTube video by hand. However, this has not fixed the problem as I still receive the same output.

joesch Over a year ago

I've edited my original post as your query also contains an error which you've introduced to your second code example.

Confused · Accepted Answer · 2014-07-14 11:20:46Z

0

Edited Code

<?php

if (isset($_GET["id"])) {
    $con = mysql_connect("localhost", "USER_NAME", "PASSWORD");
    if (!$con) {
        die("Could not connect: " . mysql_error());
    }
    if (!mysql_select_db("DB_NAME", $con)) {
        die("Database error " . mysql_error());
    }
    $id = $_GET["id"];
    $query="SELECT * FROM table WHERE ID =".$id;
    $result = mysql_query($query) or die("Errant query:");
    $output = Array();
    while ($row = mysql_fetch_assoc($result)) {
        $output[] = $row;
    }
    echo json_encode($output);
    mysql_close($con);
    die;
} else {
    $output = "not found";
    echo (json_encode($output));
    die;
}
?>

edited Jul 14, 2014 at 11:20

answered Jul 14, 2014 at 9:25

Confused

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12 Comments

Confused Over a year ago

Edited code :-) Check In both cases i am returning array.

jamesgates1 Over a year ago

I have tried your code but still receiving same result.

Confused Over a year ago

use var_dump($result) to check the query result.. Are you sure you are getting query result

jamesgates1 Over a year ago

Tried var_dump($result) but no success. What do you mean by "Are you sure you are getting query result"?

Confused Over a year ago

Means, it is possible that your query is wrong and you are not getting any result.So you will not get any result in jquery

|

Kavin · Accepted Answer · 2014-07-14 12:32:03Z

0

<?php
if(isset($_GET["ArticleID"]))
    {
    $con = mysql_connect("localhost","createyo_james","password");
    if(!$con)
    {
    die("Could not connect: " .mysql_error());
    }
    mysql_select_db("createyo_TestDatabase", $con);

    $articleID = mysql_real_escape_string($_GET["ArticleID"]);

    $result = mysql_query("SELECT * FROM Customer WHERE ID = '$articleID' ") or die("Errant query:");

    while($row = mysql_fetch_assoc($result))
    {
$id = $row['id'];
    $output[]= array('id' => "$id");
    }

    echo json_encode($output);

    mysql_close($con);

    }

    else
    {
    $output = "not found";
    echo json_encode($output);
    }

?>

edited Jul 14, 2014 at 12:32

answered Jul 14, 2014 at 9:36

Kavin

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