CODEWITHSUNDEEP
javascriptphpjqueryjsonajax
nj51
nj51

Reputation: 15

ajax JSON call returning array of strings instead of objects

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);

Upvotes: 0

Views: 241

Answers (2)

BeetleJuice
BeetleJuice

Reputation: 40936

Replace

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

With

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

Upvotes: 0

Ismail RBOUH
Ismail RBOUH

Reputation: 10470

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.

Upvotes: 2

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