-2

INPUT:

echo json_encode($json_array);

Output:

{
   "subject_id": "10",
   "subjectName": "Mathematics",
   "subject_icon": "http:\/\/www.unfoldu.com\/admin\/images\/maths-icon.png"
}

I need a url in proper format like:

www.domain.com/admin/images/maths-icon.png
Sangeeta Dhiman
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    [json_decode](http://php.net/json_decode). – Jonnix Mar 27 '17 at 10:59
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    Possible duplicate of [How do I extract data from JSON with PHP?](http://stackoverflow.com/questions/29308898/how-do-i-extract-data-from-json-with-php) – Sahil Gulati Mar 27 '17 at 11:00
  • $classid = $_GET['classid']; $sub_query ="SELECT * FROM tablename WHERE subjectClass = '".$classid. "'"; $sub_sql = mysql_query($sub_query); //create an array $subarray = array(); while($row =mysql_fetch_assoc($sub_sql)) { //$subarray[] = $row; $json_array['subject_id'] = $row['subject_id']; $json_array['subjectName'] = $row['subjectName']; $json_array['subject_icon'] = "domain.com/admin/images/".$row['subject_icon']; //array_push($subarray,$json_array); } echo json_encode($json_array); – Sangeeta Dhiman Mar 27 '17 at 11:01
  • do you need only icon name from entire URL in json? – Rana Ghosh Mar 27 '17 at 11:04
  • when I decode the data than my url output like this:http:\/\/www.unfoldu.com\/admin\/images\/maths-icon.png – Sangeeta Dhiman Mar 27 '17 at 11:09

1 Answers1

0

Yes, you only need to use the JSON_UNESCAPED_SLASHES flag:

<?php 

$arr = array('subject_id'=>'10','subjectName'=>'Mathematics','subject_icon'=>'http://www.example.com/admin/images/maths-icon.png');

echo json_encode($arr, JSON_UNESCAPED_SLASHES);

?>
Mittul At TechnoBrave
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