Combine JSON array into single value

Question

I am trying to concat several json key values so that I can return just a single value with all data. For example, In my code, I would like to combine address1, address2, address3 into Address. Is this possible. I have tried various methods of using the .= but nothing seems to work. Any heads up would be gratefully appreciated. Thanks

$query = "SELECT * FROM company_com";
$from = 0; 
$to = 30;
$query .= " LIMIT ".$from.",".$to;

$result = mysql_query($query) or die("SQL Error 1: " . mysql_error());
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
    $customers[] = array(
        'Address1' => $row['address1_com'],
        'Address2' => $row['address2_com'],
        'Address3' => $row['address3_com']
      );
}

echo json_encode($customers);

Shankar Narayana Damodaran · Accepted Answer · 2014-04-22 12:12:41Z

3

You can very well rewrite like this..

$customers = array(); $i=0; 
$result = mysql_query($query) or die("SQL Error 1: " . mysql_error());
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
    $customers[$i]['Address1'] = $row['address1_com'];
    $customers[$i]['Address2'] = $row['address2_com'];
    $customers[$i]['Address3'] = $row['address3_com'];
    $i++;
}
echo json_encode($customers);

Changes done :

The $customers array declaration moved outside of the while
A flag $i=0; is set outside of the while.
The $customers array is made 2-Dimensional as you can see $i is passed as the key.
Finally, $i is incremented..

EDIT :

<?php
$customers = array(); $i=0;
$result = mysql_query($query) or die("SQL Error 1: " . mysql_error());
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
    $customers[$i]['Address'] = $row['address1_com']." ".$row['address2_com']." ".$row['address3_com'];
    $i++;
}
echo json_encode($customers);

edited Apr 22, 2014 at 12:12

answered Apr 22, 2014 at 12:00

Shankar Narayana Damodaran

68.6k43 gold badges102 silver badges129 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

user1532468 Over a year ago

Thank you for that. However, I think I need to return something like: Address so I can use as datafield in jquery. var source = { datatype: "json", datafields: [ { name: 'Address', type: 'string'} ], url: 'data.php', cache: false };

Shankar Narayana Damodaran Over a year ago

You want to combine all three addresses inside a single value ?

user1532468 Over a year ago

Thanks very much Shankar. Very grateful for your time.

Shankar Narayana Damodaran Over a year ago

@user1532468, Anytime mate :)

Kayla · Accepted Answer · 2014-04-22 12:08:02Z

0

I don't quite understand your problem.

Make sure that the column names are correct. address1_com, address2_com, etc.

Test the code by using print_r($row); on every iteration. print_r shows the contents of an array.

Also, what I would recommend you doing is, combining every address into an array. For example...

'Address' => array(
    $row['address1_com'],
    $row['address2_com'],
    $row['address3_com']
)

answered Apr 22, 2014 at 12:08

Kayla

4755 silver badges14 bronze badges

1 Comment

user1532468 Over a year ago

Sorry, not quite understand your answer. All fields are correct and print_r shows correct values. I just need to combine into a single name ie; Address for inclusion in jquery as datafield. Thanks

Collectives™ on Stack Overflow

Combine JSON array into single value

2 Answers 2

Changes done :

4 Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Changes done :

4 Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest