How to encode Json Array in php?

Question

Hi I am new in php and i dont know how can i do this?

I have four rows in my data base and i want to encode a Json array of these rows using php how can i do this.

below is my code please look into this and give me a suggestion--

<html>
<head>
<title>First</title>
<body>
<?php
$con = mysql_connect("localhost","root");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("ContactDB", $con);

$result = mysql_query("SELECT * FROM mycontacts");

echo "<table border='1'>
<tr>
<th>PhoneNumber</th>
<th>Name</th>
<th>Lastname</th>
</tr>";

 while($row = mysql_fetch_array($result))
  {
  $data=array("PhoneNumber"=>$row['PhoneNumber'],"Name"=>$row['Name']); 
print_r (json_encode(array_chunk($data, 1, true)));  
}


mysql_close($con);
?>

</body>
</head>
</html>

I am getting response like this

[{"PhoneNumber":"1234567"},{"Name":"Ujjwal"}][{"PhoneNumber":"765423"},{"Name":"ABC"}][{"PhoneNumber":"098765123"},{"Name":"A"}]

but i want it in this format

[{"PhoneNumber":"1234567", "Name":"X"},{"PhoneNumber":"765423", "Name":"ABC"},{"PhoneNumber":"098765123", "Name":"A"}]

possible duplicate of Opendir function gives me multiple arrays instead of just one — mario
– mario, Commented Mar 6, 2012 at 13:33

user1252065 · Accepted Answer · 2012-03-06 13:32:47Z

1

Replace

while($row = mysql_fetch_array($result))
{
  $data=array("PhoneNumber"=>$row['PhoneNumber'],"Name"=>$row['Name']); 
  print_r (json_encode(array_chunk($data, 1, true)));  
}

with

$data = array();
while($row = mysql_fetch_array($result))
{
  $data[] = array("PhoneNumber"=>$row['PhoneNumber'],"Name"=>$row['Name']); 
}
print_r (json_encode($data));

answered Mar 6, 2012 at 13:32

user1252065

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Comments

Doug McK · Accepted Answer · 2012-03-06 13:32:34Z

0

Dont use array_chunk

json_encode($data) should do it.

You may also want to declare data as as array before the loop and then user $data[] = json_encode(.....)

answered Mar 6, 2012 at 13:32

Doug McK

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Comments

johndodo · Accepted Answer · 2012-03-06 13:33:19Z

0

Try this:

$json = json_encode($data)

answered Mar 6, 2012 at 13:33

johndodo

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Comments

mav · Accepted Answer · 2012-03-06 13:34:36Z

0

Try replacing

$data= array("PhoneNumber"=>$row['PhoneNumber'],"Name"=>$row['Name']);

with

$data= (object) array("PhoneNumber"=>$row['PhoneNumber'],"Name"=>$row['Name']);

answered Mar 6, 2012 at 13:34

mav

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Comments

Kurt · Accepted Answer · 2012-03-06 13:36:08Z

0

Example of how to do it below. Can I also suggest the use of AS in the MySQL query (and avoid the *; just get the data you need).

$q = mysql_query("SELECT ... AS ... ...");
$rows = array();
while($r = mysql_fetch_assoc($q)) {
    $rows[] = $r;
}
print json_encode($rows);

answered Mar 6, 2012 at 13:36

Kurt

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