CODEWITHSUNDEEP
phparraysjsongetjson
Klausos Klausos
Klausos Klausos

Reputation: 16050

Joining two data fields

The $datax contains numbers 1 and 2 (see json output), however it must contain entries similar to 1 / XXX, 2 / XXX, etc. ($datax[] = $row['solution_id'] + " / " + $row['Type'];)

Where is the error?

<?php
include_once 'include/connect_db.php';

$query="SELECT A.solution_id, A.Type,A.Time2,B.Time1 
           FROM Table1 A 
           INNER JOIN Table2 B 
           ON A.Type=B.Type;";

$result=ejecutar_query($query);

$datax = array();
$datay1 = array();
$datay2 = array();

while($row=mysql_fetch_assoc($result)) {
    $datax[] = $row['solution_id'] + " / " + $row['Type'];
    $datay1[] = $row['Time1'];
    $datay2[] = $row['Time2'];
}

echo json_encode(array('x' => $datax, 'y1' => $datay1, 'y2' => $datay2));
die();

?>

JSON

{"x":[1,2,1,2,1,2,1,2,1,2,1,2],"y1":["2013-05-29 17:24:00","2013-05-29 17:24:00","2013-05-29 17:22:00","2013-05-29 17:22:00","2013-05-29 17:18:00","2013-05-29 17:18:00","2013-05-29 17:16:00","2013-05-29 17:16:00","2013-05-29 17:11:00","2013-05-29 17:11:00","2013-05-29 17:11:00","2013-05-29 17:11:00"],"y2":["2013-05-29 17:56:26","2013-05-29 18:03:38","2013-05-29 17:48:12","2013-05-29 17:42:53","2013-05-29 17:10:32","2013-05-29 17:52:08","2013-05-29 17:08:00","2013-05-29 17:10:18","2013-05-29 17:42:53","2013-05-29 17:06:12","2013-05-29 17:05:39","2013-05-29 18:09:00"]}

Upvotes: 0

Views: 46

Answers (2)

benestar
benestar

Reputation: 562

You have to use the dot operator . in php. So $row['solution_id'] . " / " . $row['Type'] will help.

Upvotes: 1

Nikhil Mohan
Nikhil Mohan

Reputation: 891

In PHP, . is the concatenation operator

while($row=mysql_fetch_assoc($result)) {
    $datax[] = $row['solution_id'] . " / " . $row['Type'];
    $datay1[] = $row['Time1'];
    $datay2[] = $row['Time2'];
}

http://php.net/manual/en/language.operators.string.php

Upvotes: 2

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