ajax JSON call returning array of strings instead of objects

Question

When I do the ajax call I am parsing the json_encoded data and when I log the data to the console it's actually an array of strings instead of objects. It's showing this.

[  
"{"   todoText":"dgdgdfgdfgdf",
"completed":false,
"editable":false
 }",

"{  
"todoText":"test 2",
"completed":false,
"editable":false
}",

"{  
"todoText":"test 3",
"completed":false,
"editable":false
}", 

"{  
"todoText":"sdfsdf",
"completed":false,
"editable":false
}"
]

This is the code I used to make the call to retrieve the data.

$(document).ready(function() {


$.get("php/listtasks.php", function(data){

var parsed = JSON.parse(data);


  $('#directions').html(parsed[0]);

  console.log(parsed);
})

});

This is the php code i used to encode the data and echo it back to the javascript.

$query  = "SELECT * FROM list";
$result = $conn->query($query);
if (!$result) die ("Database access failed: " . $conn->error);

$rows = $result->num_rows;


for ($j = 0 ; $j < $rows ; ++$j)
{
$result->data_seek($j);
$row = $result->fetch_array(MYSQLI_NUM);


$x[$j] = $row[2];


}

echo json_encode($x);

have you tried: setting mimetype in php to json? and/or console.log(data) before/without JSON.parse? — Jeff
– Jeff, Commented Jul 19, 2016 at 23:44
@Jeff that outputs almost the same thing only now it has foward slashes throughout it — nj51
– nj51, Commented Jul 19, 2016 at 23:51

Ismail RBOUH · Accepted Answer · 2016-07-19 23:53:27Z

2

Apparently, your $row[2] is a JSON object so you need to decode it like this:

$x[$j] = json_decode($row[2]);

I hope this will help you.

answered Jul 19, 2016 at 23:53

Ismail RBOUH

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2 Comments

nj51 Over a year ago

Thanks a lot, this works! I am just wondering if you can explain what happened. So first I stringifyed an object in javascript and sent an ajax request to send it to the server side.I then inserted this into a row in a mysql table. Why is that i have to decode that row just to encode it again to send it back?

Ismail RBOUH Over a year ago

$row[2] is a field, not row and is stored as a string (stringifyed object). I need to see your code in order to provide a detailed explanation !

Nisse Engström · Accepted Answer · 2016-07-20 03:27:00Z

0

Replace

$x[$j] = $row[2];

With

$x[$j] = json_decode($row[2]);

edited Jul 20, 2016 at 3:27

Nisse Engström

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answered Jul 19, 2016 at 23:54

BeetleJuice

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