convert my result array into string in php

Question

i want to convert my array into string i mean i want to echo only array value

i use this code

$query = "SELECT `message` FROM `appstickers`";
$result    = $mysqli->query($query) or die($mysqli->error.__LINE__);
$jsonArray  = array();
 while($row = $result->fetch_assoc()) {
          $jsonArray[]= $row;             
 }     
 //echo $string;
echo json_encode($jsonArray);

i get this output

[{"message":"welcome to team with us"},{"message":"this is for light dispatch"}]

but i want this output

[welcome to team with us,this is for light dispatch"]

so please give me proper solution with example

you required output makes no sense at all, it's just a concatinated string as the single element of an array — NDM
– NDM, Commented Mar 2, 2016 at 10:34

Halayem Anis · Accepted Answer · 2016-03-02 13:10:36Z

3

So why are you using json_encode() ? use implode() instead:

// ... previous code
while($row = $result->fetch_assoc()) {
          $jsonArray[]= $row;             
 }     
echo '[' . implode(',', $jsonArray) . ']';

edited Mar 2, 2016 at 13:10

answered Mar 2, 2016 at 10:32

Halayem Anis

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1 Comment

Trushar narodia Over a year ago

i want to send this result to android so i use json_encode

NDM · Accepted Answer · 2016-03-02 10:36:28Z

0

I'm going on a limb here, and assuming your required array is not what you actually want, but you do want this:

$jsonArray  = array();
while($row = $result->fetch_assoc()) {
    $jsonArray[]= $row['messsage'];
}
echo json_encode($jsonArray);

which will result in

["welcome to team with us", "this is for light dispatch"]

answered Mar 2, 2016 at 10:36

NDM

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Comments

Manjeet Barnala · Accepted Answer · 2016-03-02 10:37:40Z

0

<?php 
 echo implode(" ",$jsonArray);
?>

answered Mar 2, 2016 at 10:37

Manjeet Barnala

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Comments

RJParikh · Accepted Answer · 2016-03-02 10:37:43Z

0

Just you need to use implode() function.

$query = "SELECT `message` FROM `appstickers`";
$result    = $mysqli->query($query) or die($mysqli->error.__LINE__);
$jsonArray  = array();
 while($row = $result->fetch_assoc()) {
          $jsonArray[]= $row;             
 }     

//echo $string;
$ans = implode(",",$jsonArray);
echo $ans;

answered Mar 2, 2016 at 10:37

RJParikh

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